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Electricity PYQs – Physics, SSC & Railway

187 Electricity (Physics) questions asked in recent SSC and Railway exams, newest first, with exam, date & shift, answers and solutions – page 2 of 2.

📚 187 questionsSSC 17 · Railway 170
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RRB NTPC UG CBT-1 2024-25 – 22 August 2025 Shift 3 · Electric field, flux and Gauss's law

Q101. Which law is most frequently applied when calculating the electric field, particularly when high symmetry is present?

  1. (a) Coulomb's law
  2. (b) Gauss's Law
  3. (c) Faraday's Law
  4. (d) Lenz's Law
Answer & solution

Answer: (b) Gauss's Law

SSC Selection Post 2025 (Phase XIII) – 1 August 2025 Shift 2 · Resistance, resistivity and their factors

Q102. If the length of a wire is doubled and its thickness (cross-sectional area) is reduced to half, what will be the effect on its electrical resistance?

  1. (a) It becomes one-fourth
  2. (b) It remains unchanged
  3. (c) It becomes four times
  4. (d) It becomes half
Answer & solution

Answer: (c) It becomes four times

Correct Answer: Option 3 – It becomes four times
The formula for electrical resistance is: R = rho x L / A
Where rho = resistivity, L = length, A = cross-sectional area.
New R = rho x (2L) / (A/2) = rho x 2L x 2 / A = 4 x (rho x L / A) = 4R
So, when length is doubled and cross-sectional area is halved, resistance becomes 4 times the original.
Option 1 (One-fourth): Would occur if L is halved and A is doubled.
Option 2 (Unchanged): Would occur only if both L and A change proportionally.
Option 4 (Half): Incorrect; doesn't match the formula calculation.
SSC Selection Post 2025 (Phase XIII) – 31 July 2025 Shift 2 · Resistance, resistivity and their factors

Q103. What happens to the resistance of a conductor as temperature increases?

  1. (a) It decreases
  2. (b) It increases
  3. (c) It remains the same
  4. (d) It fluctuates
Answer & solution

Answer: (b) It increases

Correct Answer: Option 2 – It increases
For metallic conductors, electrical resistance increases with increasing temperature. This is because: as temperature rises, the metal atoms/ions in the crystal lattice vibrate with greater amplitude; these increased vibrations impede the flow of free electrons (charge carriers) more frequently; this results in greater opposition to current flow = higher resistance. The relationship is approximately linear: R = R₀(1 + αΔT), where α is the temperature coefficient of resistance (positive for metals). This property is used in resistance thermometers (RTDs like platinum thermometers) to measure temperature.
Option 1 (It decreases): This is true for semiconductors (like silicon, germanium) — for semiconductors, resistance DECREASES with temperature because more electron-hole pairs are generated. But for metallic conductors, resistance increases.
Option 3 (It remains the same): Only true for special alloys like Nichrome or Manganin at specific temperature ranges (used in standard resistors).
Option 4 (It fluctuates): Resistance of a pure metallic conductor changes predictably with temperature — it does not fluctuate randomly.
RRB NTPC GL CBT-1 2024-25 – 23 June 2025 Shift 1 · Heating effect of current and Joule's law

Q104. Which of the following materials is most suitable for making filaments in electric bulbs?

  1. (a) Copper
  2. (b) Iron
  3. (c) Aluminium
  4. (d) Tungsten
Answer & solution

Answer: (d) Tungsten

RRB NTPC GL CBT-1 2024-25 – 17 June 2025 Shift 3 · Conductors, insulators and semiconductors

Q105. Which metals are generally used for electrical transmission lines?

  1. (a) Iron and nickel
  2. (b) Silver and gold
  3. (c) Zinc and lead
  4. (d) Copper and aluminium
Answer & solution

Answer: (d) Copper and aluminium

RRB NTPC GL CBT-1 2024-25 – 16 June 2025 Shift 3 · Electric circuits and circuit symbols

Q106. Why is a voltmeter connected in parallel in a circuit?

  1. (a) To measure the difference in potential between two points
  2. (b) To allow maximum current through it
  3. (c) To measure the total resistance
  4. (d) To reduce the voltage drop
Answer & solution

Answer: (a) To measure the difference in potential between two points

RRB NTPC GL CBT-1 2024-25 – 12 June 2025 Shift 3 · Electric circuits and circuit symbols

Q107. Which component is commonly used in a circuit to vary the resistance?

  1. (a) Transformer
  2. (b) Diode
  3. (c) Rheostat
  4. (d) Voltmeter
Answer & solution

Answer: (c) Rheostat

RRB Technician Grade III 2024-25 – 30 December 2024 Shift 1 · Electric power and electrical energy

Q108. The electric power consumed by a resistor of resistance R, carrying current I, is _________.

  1. (a) I²R
  2. (b) IR
  3. (c) I/R
  4. (d) IR²
Answer & solution

Answer: (a) I²R

Correct Answer: Option 1 — I²R
Electric power is the rate at which electrical energy is converted into heat in a resistor. Starting from
P = V × I
and substituting Ohm's law V = IR, we get
P = (IR) × I = I²R
Similarly, putting I = V/R gives the third standard form P = V²/R. So the three equivalent expressions are P = VI = I²R = V²/R, all measured in watts.
The form P = I²R is the one behind Joule's law of heating, H = I²Rt. Notice that the heat depends on the square of the current — doubling the current makes the heating four times greater. This is why transmission lines carry electricity at high voltage and low current, to keep I²R losses small, and why fuse wire melts when the current becomes excessive.
❌ Why the other options are incorrect
✗ IR — by Ohm's law this is simply the potential difference V across the resistor, measured in volts, not power.
✗ I/R — current divided by resistance has no physical meaning here; power must increase with resistance at a fixed current, not decrease.
✗ IR² — squaring the resistance instead of the current is a common slip; it is dimensionally wrong for power.
📌 SSC & Railway Exam Facts
• P = VI = I²R = V²/R; SI unit watt (W).
• Joule's law of heating: H = I²Rt joules.
• For a series combination the same current flows, so the largest resistance produces the most heat.
• Electrical energy is billed in kWh; 1 kWh = 3.6 × 10⁶ J = 1 "unit".
• A fuse is a short wire of low melting point and high resistance, always connected in series with the live wire.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Numericals on electricity

Q109. Two resistors of 2Ω each are connected in series to a battery of 4V. The current in Ampere flowing through the resistors will be:

  1. (a) 4
  2. (b) 1/2
  3. (c) 1
  4. (d) 2
Answer & solution

Answer: (c) 1

Correct Answer: Option 3 — 1
📖 Given
• First resistor, R₁ = 2Ω
• Second resistor, R₂ = 2Ω
• They are joined in series
• Battery voltage, V = 4V
✍️ Formula
Series combination: R(eq) = R₁ + R₂
Ohm's law: I = V/R(eq)
🔢 Solution
• R(eq) = 2 + 2
• R(eq) = 4Ω
• I = 4/4
• I = 1 A
In a series circuit the same current of 1 A flows through both resistors.
❌ Why the other options are incorrect
✗ 4 — this comes from dividing 4V by a single 2Ω resistor and then doubling, or from using R = 1Ω; the total resistance here is 4Ω, not 1Ω.
✗ 1/2 — this is the answer for a parallel combination reversed: it uses R(eq) = 8Ω, which never occurs here.
✗ 2 — this is what you get if you use only one 2Ω resistor (4/2 = 2 A) and forget the second one in series.
📌 SSC & Railway Exam Facts
• Ohm's law: V = IR, valid at constant temperature.
• Series: R(eq) = R₁ + R₂ + R₃… — current is the same, voltage divides.
• Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃… — voltage is the same, current divides.
• Two equal resistances R in parallel give R/2; in series they give 2R.
• SI unit of resistance is the ohm (Ω); of current the ampere (A); 1 A = 1 C/s.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Electric current and its unit

Q110. The correct relation between charge(Q) following through a conductor, time(t) for which the charge follows and the current (I) following through the conductor is:

  1. (a) I = Q t²
  2. (b) I = Q/t
  3. (c) I = Qt
  4. (d) I = Q² t
Answer & solution

Answer: (b) I = Q/t

Correct Answer: Option 2 — I = Q/t
Electric current is defined as the rate of flow of electric charge — that is, the amount of charge that crosses any cross-section of the conductor in one second.
I = Q/t
So if 6 coulombs of charge pass a point in 2 seconds, the current is I = 6/2 = 3 A. Rearranged, the same relation gives Q = I × t, which is how you find the total charge that has flowed. The SI unit of current is the ampere (A), and 1 ampere = 1 coulomb per second. Since 1 electron carries 1.6 × 10⁻¹⁹ C, one ampere corresponds to about 6.25 × 10¹⁸ electrons crossing a section every second.
❌ Why the other options are incorrect
✗ I = Q t² — dimensionally wrong; this would make the current grow as time passes even for a fixed charge, which is absurd.
✗ I = Qt — this is the product, not the rate; the longer the same charge takes, the *smaller* the current should be, so t must be in the denominator.
✗ I = Q² t — squaring the charge has no physical basis; current is directly (first power) proportional to charge.
📌 SSC & Railway Exam Facts
• I = Q/t, hence Q = It; SI unit of charge is the coulomb (C).
• 1 A = 1 C/s; current is a scalar quantity though it has direction of flow.
• Conventional current flows from + to − outside the cell, opposite to the flow of electrons.
• Current is measured by an ammeter, connected in series and having very low resistance.
• Charge on one electron = −1.6 × 10⁻¹⁹ C; charge is quantised: Q = ne.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Electric circuits and circuit symbols

Q111. In an electric circuit, the device that acts as a variable resistor is called ____________.

  1. (a) ammeter
  2. (b) rheostat
  3. (c) galvanometer
  4. (d) voltmeter
Answer & solution

Answer: (b) rheostat

Correct Answer: Option 2 — rheostat
A rheostat is a variable resistor — a component whose resistance can be changed smoothly while the circuit is running. It is made of a long nichrome or manganin wire wound on an insulating cylinder, with a sliding contact (jockey). Moving the slider changes the length of wire included in the circuit, and since R = ρl/A, a longer length means more resistance.
Because resistance controls current through I = V/R, a rheostat is used to regulate the current in a circuit without changing the source — exactly how a fan regulator changes fan speed or a dimmer changes lamp brightness. In circuit diagrams it is shown as a rectangle with an arrow across it.
❌ Why the other options are incorrect
✗ ammeter — a measuring instrument for current; connected in series and made with very low resistance so it does not disturb the circuit.
✗ galvanometer — detects the presence and direction of a very small current; its needle deflects either way from the centre zero.
✗ voltmeter — measures potential difference; connected in parallel across a component and made with very high resistance.
📌 SSC & Railway Exam Facts
• Rheostat = variable resistance, used to control current; based on R = ρl/A.
• Ammeter — series, low resistance; Voltmeter — parallel, high resistance.
• An ideal ammeter has zero resistance and an ideal voltmeter has infinite resistance.
• Nichrome is used for heating elements and rheostats: high resistivity, high melting point, does not oxidise readily.
• Ohm's law: V = IR; resistance R = ρl/A, where ρ is resistivity.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Electric circuits and circuit symbols

Q112. The device used for measuring potential difference across the ends of a conductor is:

  1. (a) Battery
  2. (b) Voltmeter
  3. (c) Galvanometer
  4. (d) Ammeter
Answer & solution

Answer: (b) Voltmeter

Correct Answer: Option 2 — Voltmeter
A voltmeter is the instrument that measures the potential difference (voltage) between two points of a circuit, in volts (V). Potential difference is defined as the work done in moving a unit charge from one point to the other:
V = W/Q, so 1 volt = 1 joule per coulomb
A voltmeter is always connected in parallel across the component whose voltage is to be measured, and it is built with a very high resistance so that it draws almost no current and leaves the circuit undisturbed. An ideal voltmeter has infinite resistance.
❌ Why the other options are incorrect
✗ Battery — a battery is a source of potential difference, not a measuring device; it supplies the emf that drives the current.
✗ Galvanometer — it only detects a small current and shows its direction; by adding a high resistance in series it can be *converted into* a voltmeter, but by itself it is not one.
✗ Ammeter — it measures current in amperes, is connected in series, and has a very low resistance.
📌 SSC & Railway Exam Facts
• Voltmeter — parallel, high resistance; Ammeter — series, low resistance.
• 1 volt = 1 joule/coulomb; the volt is named after Alessandro Volta.
• A galvanometer + high series resistance = voltmeter; a galvanometer + low shunt resistance = ammeter.
• Ohm's law: V = IR, where V is potential difference, I current and R resistance.
• Current always flows from higher to lower potential in the external circuit.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Electric power and electrical energy

Q113. The power input in a circuit P can be expressed in terms of voltage V and current I in the circuit as follows:

  1. (a) P = V/I
  2. (b) P = V²I
  3. (c) P = V²/I
  4. (d) P = VI
Answer & solution

Answer: (d) P = VI

Correct Answer: Option 4 — P = VI
When a charge Q is driven through a potential difference V, the work done by the source is W = VQ. Power is the rate of doing that work, so
P = W/t = V × (Q/t) = V × I
because I = Q/t. Hence the electric power supplied to (or consumed by) any circuit element is
P = V I
measured in watts when V is in volts and I in amperes. Using Ohm's law V = IR, the same expression can be rewritten as P = I²R and P = V²/R. That is why a 220 V, 100 W bulb draws a current of I = P/V = 100/220 ≈ 0.45 A.
❌ Why the other options are incorrect
✗ P = V/I — by Ohm's law V/I is simply the resistance R, measured in ohms, not power.
✗ P = V²I — squaring the voltage *and* multiplying by current gives far too large a quantity; the correct squared forms are I²R and V²/R, each involving resistance.
✗ P = V²/I — dividing by current instead of multiplying is dimensionally wrong; at a fixed voltage, more current must mean more power drawn, not less.
📌 SSC & Railway Exam Facts
• P = VI = I²R = V²/R; SI unit watt (W); 1 W = 1 J/s.
• Ohm's law: V = IR, valid at constant temperature.
• Joule's law of heating: H = I²Rt.
• A bulb marked 220 V, 60 W has resistance R = V²/P = 220²/60 ≈ 807 Ω.
• Electrical energy consumed = P × t; billed in kWh, where 1 kWh = 3.6 × 10⁶ J.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Resistors in series and parallel

Q114. Consider a parallel combination of 3 resistors R₁ , R₂ and R₃ with currents I₁ , I₂ and I₃ , respectively. Which of the following is true? (I = total current, R(eq) = equivalent resistance of the circuit)

  1. (a) R(eq) = R₁ + R₂R₃/(R₂ + R₃)
  2. (b) I = I₁ + I₂ + I₃
  3. (c) 1/I = 1/I₁ + 1/I₂ + 1/I₃
  4. (d) R(eq) = R₃ + R₂R₁/(R₂ + R₁)
Answer & solution

Answer: (b) I = I₁ + I₂ + I₃

Correct Answer: Option 2 — I = I₁ + I₂ + I₃
In a parallel combination all three resistors are connected between the same two points, so each of them has the same potential difference V across it. The main current arriving at the junction divides among the three branches and joins up again afterwards.
By the conservation of charge (Kirchhoff's junction rule), the current entering a junction equals the total current leaving it:
I = I₁ + I₂ + I₃
Since I = V/R(eq) and I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃, dividing throughout by V gives the standard result
1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃
The equivalent resistance of a parallel group is therefore always smaller than the smallest individual resistance.
❌ Why the other options are incorrect
✗ R(eq) = R₁ + R₂R₃/(R₂ + R₃) — this is for a mixed circuit, where R₂ and R₃ are in parallel and that pair is in series with R₁; it does not describe three resistors all in parallel.
✗ 1/I = 1/I₁ + 1/I₂ + 1/I₃ — the reciprocal rule applies to the resistances in parallel, never to the currents; currents simply add.
✗ R(eq) = R₃ + R₂R₁/(R₂ + R₁) — again a series–parallel mixed combination (R₁ ∥ R₂, in series with R₃), not a pure parallel group.
📌 SSC & Railway Exam Facts
• Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃; voltage same, currents add.
• Series: R(eq) = R₁ + R₂ + R₃; current same, voltages add.
• Two resistors in parallel: R(eq) = R₁R₂/(R₁ + R₂); n equal resistances R in parallel give R/n.
• Household appliances are wired in parallel so each gets the full 220 V and can be switched independently.
• Ohm's law: V = IR; I = Q/t; SI unit of resistance is the ohm (Ω).
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Resistance, resistivity and their factors

Q115. If the cross-sectional area of a conductor is doubled, other factors remaining the same, the resistivity of the conductor will be:

  1. (a) doubled
  2. (b) one fourth
  3. (c) halved
  4. (d) same
Answer & solution

Answer: (d) same

Correct Answer: Option 4 — same
You must separate two different quantities here — resistance and resistivity.
Resistance does depend on the shape of the wire:
R = ρ l / A
So doubling the area A would halve the resistance.
But resistivity ρ is a characteristic property of the material itself. It is defined as the resistance of a conductor of that material of unit length and unit cross-sectional area, and it depends only on the nature of the material and its temperature — not on the length, the area, or the shape of the specimen. A thick copper bus-bar and a thin copper wire have very different resistances but exactly the same resistivity.
Therefore doubling the cross-sectional area leaves the resistivity unchanged.
❌ Why the other options are incorrect
✗ doubled — nothing about ρ increases with area; this confuses resistivity with a geometry-dependent quantity.
✗ one fourth — there is no inverse-square dependence on area anywhere in R = ρl/A, let alone in ρ.
✗ halved — this is what happens to the resistance, not to the resistivity; it is the classic trap in this question.
📌 SSC & Railway Exam Facts
• R = ρl/A; resistivity ρ = RA/l; SI unit of ρ is the ohm metre (Ω m).
• ρ depends only on the material and temperature; R depends additionally on length and area.
• Resistivity of metals increases with temperature; that of semiconductors decreases.
• Silver has the lowest resistivity, followed by copper; nichrome has a high resistivity and is used in heating elements.
• Conductivity σ = 1/ρ, unit Ω⁻¹ m⁻¹ (S m⁻¹).
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Numericals on electricity

Q116. In a circuit, suppose the three resistors having the values 5 Ω, 10 Ω, 20 Ω are connected in parallel; then equivalent resistance of the circuit will be_____________.

  1. (a) 35 Ω
  2. (b) 20/7 Ω
  3. (c) 40/7 Ω
  4. (d) 20 Ω
Answer & solution

Answer: (b) 20/7 Ω

Correct Answer: Option 2 — 20/7 Ω
📖 Given
• R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 20 Ω
• All three are connected in parallel
✍️ Formula
1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃
🔢 Solution
• 1/R(eq) = 1/5 + 1/10 + 1/20
• Taking LCM = 20: 1/R(eq) = 4/20 + 2/20 + 1/20
• 1/R(eq) = 7/20
• R(eq) = 20/7 Ω
• As a decimal this is about 2.86 Ω.
Notice the useful check: in a parallel combination the equivalent resistance is always smaller than the smallest resistor, and here 20/7 ≈ 2.86 Ω is indeed less than 5 Ω.
❌ Why the other options are incorrect
✗ 35 Ω — this is 5 + 10 + 20, the formula for a series connection, not parallel.
✗ 40/7 Ω — the LCM step has been done wrongly; the correct sum of reciprocals is 7/20, giving 20/7, not 40/7.
✗ 20 Ω — this is simply the largest resistor copied down; a parallel combination can never equal or exceed it.
📌 SSC & Railway Exam Facts
• Series: R(eq) = R₁ + R₂ + R₃, the same current flows through each resistor.
• Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃, the same voltage is across each resistor.
• For two resistors in parallel there is a shortcut: R(eq) = R₁R₂/(R₁ + R₂).
• n equal resistors each of R in parallel give R/n; in series they give nR.
• Ohm's law: V = IR; domestic appliances are connected in parallel so that each gets the full 220 V and can be switched independently.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Resistance, resistivity and their factors

Q117. A resistor has a resistivity ρ. This resistor is now cut into three equal parts. The resistivity of each part will be:

  1. (a) ρ
  2. (b) 3ρ
  3. (c) 6ρ
  4. (d) ρ/3
Answer & solution

Answer: (a) ρ

Correct Answer: Option 1 — ρ
📖 Given
• Resistivity of the original resistor = ρ
• The resistor is cut into 3 equal parts
✍️ Formula
R = ρL/A, so ρ = RA/L
🔢 Solution
• Resistivity ρ is a property of the material, not of the size or shape of the conductor. It depends only on the nature of the material and its temperature.
• Cutting the wire changes only L and hence R — for each piece the new length is L/3, so the new resistance is R/3.
• Check with the formula: new ρ = (R/3 × A)/(L/3) = RA/L = ρ.
• Therefore the resistivity of each part remains ρ.
In plain words: breaking a copper wire into pieces does not turn it into a different material — each piece is still copper, so each piece has the same resistivity.
❌ Why the other options are incorrect
✗ 3ρ — this is the mistake of thinking resistivity behaves like resistance when three pieces are joined in series.
✗ 6ρ — has no physical basis; it comes from randomly multiplying the given value.
✗ ρ/3 — this is the value of the resistance of each part (R/3), wrongly written for resistivity.
📌 SSC & Railway Exam Facts
• SI unit of resistivity is ohm metre (Ω m); SI unit of resistance is ohm (Ω).
• R ∝ L and R ∝ 1/A — a thin, long wire has more resistance.
• Silver is the best conductor; copper and aluminium are used for wires because they are cheaper.
• Resistivity of metals increases with temperature but of semiconductors decreases with temperature.
• Nichrome has a very high resistivity, so it is used in heaters, toasters and irons.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Electric power and electrical energy

Q118. Which of the following is NOT the correct relation between electric power (P), electric current(I), electric potential difference (V) and resistance(R):

  1. (a) P = VI
  2. (b) P = V²/R
  3. (c) P = I²R
  4. (d) P = IR
Answer & solution

Answer: (d) P = IR

Correct Answer: Option 4 — P = IR
P = IR is not a power relation at all. By Ohm's law, the product IR equals the potential difference V, not the power. So the expression P = IR is dimensionally and physically wrong — it actually gives volts, not watts.
The correct starting point is P = VI (power = potential difference × current). Substituting Ohm's law V = IR into it gives the other two standard forms:
• Put V = IR in P = VI → P = (IR) × I = I²R
• Put I = V/R in P = VI → P = V × (V/R) = V²/R
So P = VI, P = I²R and P = V²/R are all correct, and only P = IR is wrong.
❌ Why the other options are incorrect
✗ P = VI — the basic definition of electric power; correct.
✗ P = V²/R — obtained by substituting I = V/R; correct, and used when V and R are known.
✗ P = I²R — obtained by substituting V = IR; correct, and it is the basis of Joule's law of heating.
📌 SSC & Railway Exam Facts
• SI unit of power is the watt (W); 1 W = 1 J s⁻¹ = 1 V × 1 A.
• Ohm's law: V = IR, valid at constant temperature; given by Georg Simon Ohm.
• Joule's law of heating: H = I²Rt — heat produced is proportional to I², R and t.
• Commercial unit of electrical energy = kilowatt-hour (kWh); 1 kWh = 3.6 × 10⁶ J.
• In a series circuit the current is the same everywhere; in a parallel circuit the potential difference is the same across each branch.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Heating effect of current and Joule's law

Q119. An electric water heater converts electrical energy into ________.

  1. (a) potential energy
  2. (b) wind energy
  3. (c) light energy
  4. (d) heat energy
Answer & solution

Answer: (d) heat energy

Correct Answer: Option 4 — heat energy
An electric water heater (geyser / immersion rod) works on the heating effect of electric current, also called Joule heating. When current passes through the heating element, the free electrons collide repeatedly with the atoms of the wire; the electrical energy they lose appears as heat energy, which raises the temperature of the water.
The heat produced is given by Joule's law of heating, H = I²Rt, where I is the current, R the resistance of the element and t the time. Because H is proportional to R, the element is made of an alloy of high resistivity and high melting point — usually nichrome (nickel + chromium), which also does not oxidise readily when red-hot.
So the energy conversion in a water heater is simply electrical energy → heat energy.
❌ Why the other options are incorrect
✗ potential energy — this is stored energy due to position or configuration, as in water in an overhead tank or a stretched spring; a heater does not raise the water's position.
✗ wind energy — the kinetic energy of moving air, used in windmills; nothing in a heater produces wind.
✗ light energy — an electric bulb converts electrical energy mainly into light (and heat), but a water heater produces no useful light.
📌 SSC & Railway Exam Facts
• Joule's law of heating: H = I²Rt = VIt = (V²/R)t.
• Heating element alloy — nichrome; the filament of a bulb is tungsten (melting point ≈ 3380 °C).
• A fuse wire has low melting point and high resistance; it is always connected in series in the live wire.
• Devices based on the heating effect: electric iron, geyser, toaster, room heater, electric kettle.
• In domestic wiring: live — red/brown, neutral — black/blue, earth — green/yellow.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Numericals on electricity

Q120. A current of 1 A is drawn by a filament of an electric bulb for 5 minutes. Then the amount of electric charge that flows through the circuit will be _____________.

  1. (a) 5 C
  2. (b) 10 C
  3. (c) 300 C
  4. (d) 150 C
Answer & solution

Answer: (c) 300 C

Correct Answer: Option 3 — 300 C
📖 Given
• Current, I = 1 A
• Time, t = 5 minutes
• To find: charge Q flowing through the circuit
✍️ Formula
I = Q/t, therefore Q = I × t
🔢 Solution
• First convert the time into seconds, because the ampere is defined per second: t = 5 × 60 = 300 s.
• Q = I × t
• Q = 1 × 300
• Q = 300 C
• So 300 coulomb of charge passes through the filament in those 5 minutes.
In plain language, a current of 1 ampere means 1 coulomb of charge flows every second; in 300 seconds that is simply 300 coulomb.
❌ Why the other options are incorrect
✗ 5 C — this is the mistake of using t = 5 minutes directly as seconds, forgetting the conversion 1 min = 60 s.
✗ 10 C — no combination of 1 A with the given time gives 10 C; it appears to come from an arbitrary 10 s.
✗ 150 C — this comes from halving the time (using 150 s instead of 300 s), i.e. taking 1 minute as 30 seconds.
📌 SSC & Railway Exam Facts
• Q = It; SI unit of charge is the coulomb (C), of current the ampere (A); 1 A = 1 C/s.
• Charge on one electron = −1.6 × 10⁻¹⁹ C; so 1 coulomb ≈ 6.25 × 10¹⁸ electrons.
• Current is measured by an ammeter, connected in series; potential difference by a voltmeter, connected in parallel.
• Conventional current flows from + to − outside the cell, opposite to the direction of electron flow.
• Ohm's law: V = IR; electric power P = VI = I²R = V²/R.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Numericals on electricity

Q121. An electric geyser of resistance 4 Ω produces 100 J of heat per sec. The current flowing through it will be:

  1. (a) 5 A
  2. (b) 25 A
  3. (c) 2.5 A
  4. (d) 50 A
Answer & solution

Answer: (a) 5 A

Correct Answer: Option 1 — 5 A
📖 Given
• Resistance of the geyser, R = 4 Ω
• Heat produced per second, H/t = 100 J/s, i.e. power P = 100 W
• To find: current I
✍️ Formula
H = I²Rt (Joule's law of heating), so P = H/t = I²R
🔢 Solution
• P = I²R
• 100 = I² × 4
• I² = 100/4
• I² = 25
• I = √25 = 5 A
In plain language, "100 joule of heat per second" is just another way of saying the geyser consumes 100 watt of power; putting that into P = I²R with R = 4 Ω gives a current of 5 ampere.
❌ Why the other options are incorrect
✗ 25 A — this is the value of I², not of I; the final square root step was forgotten.
✗ 2.5 A — this comes from wrongly using P = 2IR or from halving the correct answer; check: 2.5² × 4 = 25 W, not 100 W.
✗ 50 A — far too large; 50² × 4 = 10,000 W, a hundred times the given power.
📌 SSC & Railway Exam Facts
• Joule's law of heating: H = I²Rt — heat ∝ I², ∝ R and ∝ t.
• Electric power: P = VI = I²R = V²/R; SI unit watt (W).
• Heating appliances (geyser, heater, iron, toaster) use nichrome, because of its high resistivity and high melting point, and because it does not oxidise easily.
• A fuse wire works on the same heating effect; it has a low melting point and is connected in series with the live wire.
• 1 unit of electricity = 1 kWh = 3.6 × 10⁶ J.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 1 · Electric power and electrical energy

Q122. The relation between power ( P ), voltage ( V ) and resistance ( R ) is given by ____.

  1. (a) P = R²/V
  2. (b) P = V²R
  3. (c) P = R/2V²
  4. (d) P = V²/R
Answer & solution

Answer: (d) P = V²/R

Correct Answer: Option 4 — P = V²/R
📖 Given
• Electric power = P
• Potential difference (voltage) = V
• Resistance = R
✍️ Formula
P = VI (power = voltage × current)
V = IR (Ohm's law), so I = V/R
🔢 Solution
• Start with P = V × I
• Substitute I = V/R from Ohm's law
• P = V × (V/R)
• P = V²/R
So for a fixed voltage, the power consumed is inversely proportional to the resistance — a lower resistance appliance draws more power. Using V = IR the other way round also gives the companion form P = I²R.
❌ Why the other options are incorrect
✗ P = R²/V — this has resistance and voltage interchanged; its units work out to Ω²/V, which is not the watt.
✗ P = V²R — this comes from multiplying by R instead of dividing by it; it would wrongly mean a higher resistance gives more power at fixed voltage.
✗ P = R/2V² — an invented expression; there is no factor of 2 anywhere in the derivation of electric power.
📌 SSC & Railway Exam Facts
• The three forms of electric power: P = VI = I²R = V²/R.
• SI unit of power is the watt (W); 1 W = 1 J/s and 1 horsepower = 746 W.
• Ohm's law: V = IR; SI unit of resistance is the ohm (Ω).
• Commercial unit of electrical energy is the kilowatt-hour (kWh); 1 kWh = 3.6 × 10⁶ J.
• Heat produced H = I²Rt — this is Joule's law of heating, used in heaters, irons and fuses.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 1 · Resistors in series and parallel

Q123. What is the equivalent resistance of a series combination of resistors?

  1. (a) The product of the individual resistances.
  2. (b) The sum of the individual resistances.
  3. (c) The difference between the individual resistances.
  4. (d) The reciprocal of the sum of the individual resistances.
Answer & solution

Answer: (b) The sum of the individual resistances.

Correct Answer: Option 2 — The sum of the individual resistances.
In a series combination the resistors are joined end to end, so there is only one path for the current. Therefore the same current I flows through every resistor, while the total voltage is shared among them:
V = V₁ + V₂ + V₃
Using Ohm's law (V = IR) for each part:
IR(eq) = IR₁ + IR₂ + IR₃
Cancelling the common I gives
R(eq) = R₁ + R₂ + R₃
So the equivalent resistance of a series combination is simply the sum of the individual resistances, and it is always larger than the largest single resistance in the set. Connecting resistors in series is therefore the way to increase the total resistance of a circuit.
❌ Why the other options are incorrect
✗ The product of the individual resistances. — multiplying resistances gives units of Ω², which is not a resistance at all. The product appears only in the two-resistor parallel shortcut R = R₁R₂/(R₁+R₂).
✗ The difference between the individual resistances. — resistances always add; a difference could even give a negative value, which is impossible.
✗ The reciprocal of the sum of the individual resistances. — this is a confusion with the parallel rule, where 1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃ (the reciprocal of the sum of the reciprocals).
📌 SSC & Railway Exam Facts
• Series: R(eq) = R₁ + R₂ + R₃, current is the same, voltage divides.
• Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃, voltage is the same, current divides.
• In parallel, R(eq) is less than the smallest resistance in the combination.
• Domestic appliances are connected in parallel so that each gets the full 220 V and works independently.
• Ohm's law V = IR; resistance R = ρl/A, where ρ is the resistivity in Ω m.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 1 · Resistors in series and parallel

Q124. In a parallel combination of two different resistors, which of the following is/are correct?
(A) The potential difference across each resistor has the same value.
(B) The current flowing through each resistor has the same value.

  1. (a) 'A' is incorrect and 'B' is correct.
  2. (b) Both 'A' and 'B' are incorrect.
  3. (c) Both 'A' and 'B' are correct.
  4. (d) 'A' is correct and 'B' is incorrect.
Answer & solution

Answer: (d) 'A' is correct and 'B' is incorrect.

Correct Answer: Option 4 — 'A' is correct and 'B' is incorrect.
In a parallel combination, both ends of every resistor are joined to the same two points of the circuit. Since the two points have fixed potentials, every resistor gets exactly the same potential difference V. So statement (A) is correct.
The current, however, has more than one path to choose from. The main current divides at the junction and, by Ohm's law, each branch carries
I = V/R
Because the two resistors are different, their R values are different, so the currents through them must also be different — the smaller resistance carries the larger current. So statement (B) is incorrect.
The total current is I = I₁ + I₂, and the equivalent resistance follows 1/R(eq) = 1/R₁ + 1/R₂.
❌ Why the other options are incorrect
✗ 'A' is incorrect and 'B' is correct. — exactly the reverse of the truth; the equal quantity in parallel is the voltage, not the current.
✗ Both 'A' and 'B' are incorrect. — statement (A) is definitely correct, since all branches lie between the same two nodes.
✗ Both 'A' and 'B' are correct. — (B) would be true only if the two resistors were equal; the question clearly says they are different.
📌 SSC & Railway Exam Facts
• Parallel: V same, current divides, 1/R(eq) = 1/R₁ + 1/R₂; for two resistors R(eq) = R₁R₂/(R₁+R₂).
• Series: I same, voltage divides, R(eq) = R₁ + R₂.
• In parallel the equivalent resistance is smaller than the smallest branch resistance.
• House wiring uses parallel connection so that each appliance gets the full 220 V and can be switched separately.
• In parallel, current is inversely proportional to resistance: I₁/I₂ = R₂/R₁.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 3 · Numericals on electricity

Q125. If a current of 1 A is flowing through a conductor for 10 secs, then the charge flowing through it (in coulombs) will be:

  1. (a) 0.1
  2. (b) 100
  3. (c) 10
  4. (d) 1
Answer & solution

Answer: (c) 10

Correct Answer: Option 3 — 10
📖 Given
• Current, I = 1 A
• Time, t = 10 s
• To find: charge Q in coulombs
✍️ Formula
I = Q / t, therefore Q = I × t
🔢 Solution
• Q = I × t
• Q = 1 × 10
• Q = 10 C
So 10 coulomb of charge passes through the conductor. This follows straight from the definition of current: 1 ampere is 1 coulomb of charge flowing per second, so in 10 seconds a 1 A current carries 10 coulomb.
❌ Why the other options are incorrect
✗ 0.1 — this is what you get if you divide the current by the time (1 ÷ 10) instead of multiplying; that operation has no physical meaning here.
✗ 100 — this comes from using t = 100 s or from multiplying 10 by 10; the time given is only 10 s.
✗ 1 — this is the value of the current, not the charge; the charge grows with time and after 10 s it is ten times larger.
📌 SSC & Railway Exam Facts
• I = Q/t; SI unit of charge is the coulomb (C), of current the ampere (A).
• 1 A = 1 C/s, and 1 C is the charge of about 6.25 × 10¹⁸ electrons.
• Charge on one electron = −1.6 × 10⁻¹⁹ C; on one proton = +1.6 × 10⁻¹⁹ C.
• Current is measured with an ammeter, connected in series; voltage with a voltmeter, connected in parallel.
• Conventional current flows from positive to negative terminal, opposite to the actual flow of electrons.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 3 · Numericals on electricity

Q126. An electric lamp, whose resistance is 10 Ω, and a conductor of 2 Ω resistance are connected to a 24 V battery in series. Then the current through the circuit will be _______________.

  1. (a) 1 A
  2. (b) 4 A
  3. (c) 2 A
  4. (d) 3 A
Answer & solution

Answer: (c) 2 A

Correct Answer: Option 3 — 2 A
📖 Given
• Resistance of the lamp, R₁ = 10 Ω
• Resistance of the conductor, R₂ = 2 Ω
• Supply voltage, V = 24 V
• The two are connected in series
✍️ Formula
Series: R(eq) = R₁ + R₂, and Ohm's law: I = V / R(eq)
🔢 Solution
• R(eq) = 10 + 2
• R(eq) = 12 Ω
• I = V / R(eq) = 24 / 12
• I = 2 A
So a current of 2 ampere flows through the whole circuit. In a series circuit the same current passes through every component, so both the lamp and the conductor carry 2 A. The voltage divides as 20 V across the lamp and 4 V across the conductor.
❌ Why the other options are incorrect
✗ 1 A — this is obtained if the total resistance is wrongly taken as 24 Ω (for example by adding the voltage to the resistances).
✗ 4 A — this comes from taking R(eq) = 6 Ω, for example by wrongly combining 10 Ω and 2 Ω in parallel-like fashion instead of adding them; in series the total is 10 + 2 = 12 Ω.
✗ 3 A — this comes from taking R(eq) = 8 Ω, i.e. wrongly subtracting 2 from 10 instead of adding, which is the parallel-style mistake.
📌 SSC & Railway Exam Facts
• Ohm's law: V = IR; SI unit of resistance is the ohm (Ω).
• Series: R(eq) = R₁ + R₂ + …, current same; Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + …, voltage same.
• R = ρl/A — resistance is directly proportional to length and inversely proportional to area of cross-section.
• Electric power P = VI = I²R = V²/R; here P = 24 × 2 = 48 W.
• Ohm's law is valid only at constant temperature; Georg Simon Ohm gave it in 1827.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 3 · Resistors in series and parallel

Q127. The effective resistance of the parallel combination of two resistors is always _________.

  1. (a) equal to the sum of the resistance of the resistors
  2. (b) equal to the product of resistance of the individual resistors
  3. (c) greater than the resistance of the individual resistors
  4. (d) less than the resistance of the individual resistors
Answer & solution

Answer: (d) less than the resistance of the individual resistors

Correct Answer: Option 4 — less than the resistance of the individual resistors
For two resistors joined in parallel,
1/R(eq) = 1/R₁ + 1/R₂, which gives R(eq) = R₁R₂ / (R₁ + R₂)
Since we are adding the reciprocals, the reciprocal of R(eq) is larger than either 1/R₁ or 1/R₂, so R(eq) must be smaller than both R₁ and R₂ — smaller even than the smallest resistance in the combination.
A quick check with numbers: for R₁ = 6 Ω and R₂ = 3 Ω,
R(eq) = (6 × 3)/(6 + 3) = 18/9 = 2 Ω, which is indeed less than 3 Ω.
The physical reason is that connecting resistors in parallel gives the current more paths to flow through, which is like widening a road — the effective area of cross-section increases, so the opposition to current falls.
❌ Why the other options are incorrect
✗ equal to the sum of the resistance of the resistors — this is the rule for a series combination, R(eq) = R₁ + R₂, the opposite arrangement.
✗ equal to the product of resistance of the individual resistors — the product appears only in the numerator of R₁R₂/(R₁+R₂); by itself it has the wrong unit, Ω².
✗ greater than the resistance of the individual resistors — again the reverse of the truth; only a series joining can raise the total resistance above each individual value.
📌 SSC & Railway Exam Facts
• Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + …; for two resistors R(eq) = R₁R₂/(R₁+R₂).
• n equal resistors of R in parallel give R/n; in series they give nR.
• In parallel the voltage is the same across each branch and the current divides.
• Household appliances are wired in parallel so each receives the full 220 V and works independently.
• Ohm's law V = IR; R = ρl/A, where ρ is resistivity with unit Ω m.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 1 · Numericals on electricity

Q128. A current of 2 A is flowing through a conductor of resistance 12 Ω. Then the potential difference across the conductor is:

  1. (a) 35 V
  2. (b) 24 V
  3. (c) 12 V
  4. (d) 6 V
Answer & solution

Answer: (b) 24 V

Correct Answer: Option 2 — 24 V
📖 Given
• Current, I = 2 A
• Resistance, R = 12 Ω
• Temperature is constant, so the conductor is ohmic
✍️ Formula
Ohm's law: V = IR
🔢 Solution
• V = I × R
• V = 2 × 12
• V = 24 V
So a potential difference of 24 volts must be maintained across this conductor to push a steady current of 2 amperes through it.
❌ Why the other options are incorrect
✗ 35 V — does not come from any correct step; 35 is neither 2 × 12 nor 12/2.
✗ 12 V — this is just the resistance value written as a voltage, forgetting to multiply by the current.
✗ 6 V — this is R/I = 12/2, the division taken the wrong way round; Ohm's law needs I × R, not R ÷ I.
📌 SSC & Railway Exam Facts
• Ohm's law: V = IR, given by Georg Simon Ohm in 1827.
• 1 Ω = 1 V/A; 1 A = 1 C/s; 1 V = 1 J/C.
• A V–I graph of an ohmic conductor is a straight line through the origin.
• Voltmeter is connected in parallel; ammeter in series.
• Electric power P = VI = I²R = V²/R; SI unit watt (W).
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 2 · Potential difference and EMF

Q129. Which of the following represents voltage?

  1. (a) Work done × Charge
  2. (b) Work done × Charge × Time
  3. (c) Work done/(Current × Time)
  4. (d) Work done/(Current × Charge)
Answer & solution

Answer: (c) Work done/(Current × Time)

Correct Answer: Option 3 — Work done/(Current × Time)
📖 Given
• The quantity required is voltage (potential difference), V
• Charge, Q, current, I, and time, t
✍️ Formula
V = W/Q and Q = I × t
🔢 Solution
• Potential difference is the work done per unit charge: V = W/Q
• Charge that flows in time t is Q = I t
• Substituting, V = W/(I × t)
• In words, Voltage = Work done/(Current × Time)
So a 12 V battery does 12 joules of work on every coulomb of charge it drives through the circuit.
❌ Why the other options are incorrect
✗ Work done × Charge — multiplying instead of dividing; this gives units of J C, which is not volt.
✗ Work done × Charge × Time — even further from the definition; voltage is work per unit charge, never work times charge.
✗ Work done/(Current × Charge) — the denominator must be charge (I × t), not current × charge, which would give J/(A C).
📌 SSC & Railway Exam Facts
• 1 volt = 1 joule/coulomb; 1 A = 1 C/s; 1 Ω = 1 V/A.
• Ohm's law: V = IR, given by Georg Simon Ohm.
• Voltage is measured by a voltmeter in parallel; current by an ammeter in series.
• Electrical energy W = VIt and electric power P = VI = I²R = V²/R.
• Charge on one electron = 1.6 × 10⁻¹⁹ C; 1 coulomb ≈ 6.25 × 10¹⁸ electrons.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 2 · Resistance, resistivity and their factors

Q130. Which of the following statements is/are NOT correct?
(i) Resistance of a conductor is independent of temperature.
(ii) The more the voltage across a resistor, the more is the current flowing through it.
(iii) The S.I unit of resistance is ohm.

  1. (a) Only (ii)
  2. (b) Only (i)
  3. (c) Both (i) and (ii)
  4. (d) Both (ii) and (iii)
Answer & solution

Answer: (b) Only (i)

Correct Answer: Option 2 — Only (i)
Statement (i) is NOT correct. Resistance definitely depends on temperature. From R = ρl/A, the resistivity ρ of a material changes with temperature. For metals, ρ and hence R increase as the temperature rises, because the vibrating ions scatter the electrons more. For semiconductors, resistance decreases with a rise in temperature. So calling resistance independent of temperature is wrong.
Statement (ii) is correct. By Ohm's law, V = IR, so at constant temperature I ∝ V. Increasing the voltage across a resistor increases the current through it in the same ratio; the V–I graph of an ohmic conductor is a straight line through the origin.
Statement (iii) is correct. The SI unit of resistance is the ohm (Ω), defined as 1 Ω = 1 volt/1 ampere.
Since only statement (i) is wrong, the answer is Only (i).
❌ Why the other options are incorrect
✗ Only (ii) — statement (ii) is a direct restatement of Ohm's law and is perfectly correct.
✗ Both (i) and (ii) — (i) is indeed wrong, but (ii) is correct, so this pair is not acceptable.
✗ Both (ii) and (iii) — both of these statements are correct, so neither belongs in a list of incorrect statements.
📌 SSC & Railway Exam Facts
• Ohm's law: V = IR — valid at constant temperature for ohmic conductors.
• R = ρl/A; resistance ∝ length and ∝ 1/area of cross-section.
• SI unit of resistance is ohm (Ω); of resistivity, ohm metre (Ω m).
• In metals resistance rises with temperature; in semiconductors it falls.
• Superconductors show zero resistance below a critical temperature.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 1 · Ohm's law

Q131. A resistor has current flowing through it. If its resistance is halved while the voltage across it is kept constant, what will be the change in current?

  1. (a) It will be halved.
  2. (b) It will remain the same.
  3. (c) It will be doubled.
  4. (d) It will become zero.
Answer & solution

Answer: (c) It will be doubled.

Correct Answer: Option 3 — It will be doubled.
📖 Given
• Original resistance R, original current I = V/R
• New resistance, R′ = R/2
• Voltage V across the resistor is kept constant
✍️ Formula
Ohm's law: V = IR, so I = V/R — at constant voltage the current is inversely proportional to resistance
🔢 Solution
• I′ = V/R′
• I′ = V/(R/2)
• I′ = 2 × (V/R)
• I′ = 2I, i.e. the current is doubled
In plain words, halving the resistance removes half of the opposition to the flow of charge, so with the same push (voltage) twice as much current flows.
❌ Why the other options are incorrect
✗ It will be halved. — this would be right only if current were directly proportional to resistance; Ohm's law says the opposite, I ∝ 1/R.
✗ It will remain the same. — the current could stay the same only if both V and R were unchanged, or if both were halved together.
✗ It will become zero. — current becomes zero only for an open circuit (R → ∞) or when V = 0. Reducing the resistance actually increases the current.
📌 SSC & Railway Exam Facts
• Ohm's law: V = IR, valid at constant temperature; given by Georg Simon Ohm.
• I ∝ 1/R at constant V; I ∝ V at constant R.
• R = ρl/A; SI unit of resistance ohm (Ω), of current ampere (A), of voltage volt (V).
• 1 A = 1 C/s; current is measured by an ammeter in series, voltage by a voltmeter in parallel.
• Electric power P = VI = I²R = V²/R; SI unit watt (W).
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 1 · Heating effect of current and Joule's law

Q132. Which of the following is/are related to the practical applications of electricity in daily life?
(A) Electric oven.
(B) Electric laundry iron.
(C) Solar water heater

  1. (a) (A), (B) and (C)
  2. (b) Only (A) and (B)
  3. (c) Only (B) and (C)
  4. (d) Only (A) and (C)
Answer & solution

Answer: (b) Only (A) and (B)

Correct Answer: Option 2 — Only (A) and (B)
An electric oven and an electric laundry iron both work on the heating effect of electric current (Joule heating). A current I passing through a coil of high resistance R for time t produces heat H = I²Rt. The coils are made of nichrome, an alloy chosen because it has a high resistivity, a high melting point and does not oxidise easily when red hot. Both are therefore genuine practical applications of electricity.
A solar water heater, however, does not run on electricity at all. It absorbs solar (radiant) energy directly in a black collector plate and converts it into heat in the water. The energy conversion is solar energy → heat energy, with no electric current involved, so (C) does not belong with the other two.
❌ Why the other options are incorrect
✗ (A), (B) and (C) — wrong because it wrongly includes the solar water heater, which uses sunlight, not electric current.
✗ Only (B) and (C) — it drops the electric oven, a clear example of the heating effect of current, and keeps the solar heater instead.
✗ Only (A) and (C) — it drops the electric laundry iron, which is one of the most familiar electrical heating appliances of all.
📌 SSC & Railway Exam Facts
• Joule's law of heating: H = I²Rt; used in electric iron, oven, heater, toaster, geyser and the electric bulb filament.
• Nichrome is used in heating coils; tungsten (melting point 3380 °C) is used in bulb filaments.
• An electric fuse works on the heating effect and is connected in the live wire in series.
• A solar water heater and a solar cooker use solar thermal energy; a solar cell converts sunlight directly into electricity.
• Electric power P = VI = I²R = V²/R; commercial unit of energy 1 kWh = 3.6 × 10⁶ J.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 1 · Electric charge and Coulomb's law

Q133. Charge on how many electrons is equivalent to 1 coulomb of electric charge?

  1. (a) 6 × 10¹⁸
  2. (b) 9.1 × 10³¹
  3. (c) 4.2 × 10¹⁶
  4. (d) 1.6 × 10¹⁹
Answer & solution

Answer: (a) 6 × 10¹⁸

Correct Answer: Option 1 — 6 × 10¹⁸
📖 Given
• Total charge, Q = 1 C
• Charge on one electron, e = 1.6 × 10⁻¹⁹ C
• Number of electrons required, n = ?
✍️ Formula
Q = n × e, so n = Q/e
🔢 Solution
• n = 1 / (1.6 × 10⁻¹⁹)
• n = (1/1.6) × 10¹⁹
• n = 0.625 × 10¹⁹
• n = 6.25 × 10¹⁸ ≈ 6 × 10¹⁸ electrons
So one coulomb — the charge that flows in one second when the current is 1 ampere — is made up of about six billion billion electrons. That shows how extremely small the charge on a single electron is.
❌ Why the other options are incorrect
✗ 9.1 × 10³¹ — this figure comes from the mass of an electron, 9.1 × 10⁻³¹ kg, with the sign of the power changed; it is a mass, not a count.
✗ 4.2 × 10¹⁶ — no step of the calculation gives this; 4.2 is the mechanical equivalent of heat (1 cal = 4.2 J), placed here as a distractor.
✗ 1.6 × 10¹⁹ — this is the value of 1/e written with the wrong digits; it repeats the 1.6 of the electronic charge instead of dividing by it.
📌 SSC & Railway Exam Facts
• Q = ne; e = 1.6 × 10⁻¹⁹ C, measured by R. A. Millikan's oil-drop experiment.
• 1 C ≈ 6.25 × 10¹⁸ electrons; charge is quantised — it comes only in whole multiples of e.
• I = Q/t; 1 ampere = 1 coulomb/second.
• Mass of an electron = 9.1 × 10⁻³¹ kg; the electron was discovered by J. J. Thomson (1897).
• SI unit of charge is the coulomb (C), named after Charles-Augustin de Coulomb.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 1 · Potential difference and EMF

Q134. What is the unit of electro motive force (EMF)?

  1. (a) Ampere
  2. (b) Volt
  3. (c) Newton
  4. (d) Sec
Answer & solution

Answer: (b) Volt

Correct Answer: Option 2 — Volt
Despite its name, the electromotive force (EMF) is not a force at all — it is the energy supplied by a cell or battery to each unit charge that it drives round the circuit. Since it is energy per unit charge, its unit is joule per coulomb, which is given the special name volt (V):
EMF = Work done / Charge, so 1 V = 1 J/C
A cell of EMF 1.5 V therefore gives 1.5 joule of energy to every 1 coulomb of charge passing through it. EMF is measured when no current is being drawn; once current flows, the terminal voltage is a little less because of the cell's internal resistance r: V = EMF − Ir.
❌ Why the other options are incorrect
✗ Ampere — the SI unit of electric current, the rate of flow of charge; 1 A = 1 C/s, a different quantity altogether.
✗ Newton — the SI unit of mechanical force. The word "force" in EMF is only historical; EMF is energy per charge, not a push.
✗ Sec — the second is the SI unit of time; it has no connection with the energy given to a charge.
📌 SSC & Railway Exam Facts
• EMF and potential difference are both measured in volt (V); 1 V = 1 J/C.
• V = EMF − Ir, where r is the internal resistance of the cell.
• EMF is measured accurately by a potentiometer, potential difference by a voltmeter (connected in parallel).
• A dry cell gives about 1.5 V; a lead-acid car battery about 12 V.
• Ohm's law V = IR; P = VI; SI unit of charge coulomb (C), of current ampere (A).
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 1 · Resistance, resistivity and their factors

Q135. How does resistance depend on the cross-sectional area A of a conductor?

  1. (a) Inversely proportional to A
  2. (b) Proportional to A
  3. (c) Does not depend on A
  4. (d) Proportional to A²
Answer & solution

Answer: (a) Inversely proportional to A

Correct Answer: Option 1 — Inversely proportional to A
The resistance of a uniform conductor is given by
R = ρl/A
where ρ is the resistivity of the material, l the length and A the area of cross-section. Since A sits in the denominator, the resistance is inversely proportional to the area of cross-section: R ∝ 1/A.
Physically, a thicker wire offers a wider path for the electrons, so more of them can move through side by side and the opposition to the current falls. Doubling the area halves the resistance; making the wire four times thicker in area reduces the resistance to one-fourth. This is exactly why thick wires are used for heavy-current appliances such as an air conditioner or a geyser, while thin wires are used in heating coils where high resistance is wanted.
❌ Why the other options are incorrect
✗ Proportional to A — this would mean a thicker wire resists current more, which contradicts both R = ρl/A and everyday wiring practice.
✗ Does not depend on A — resistance clearly changes with thickness; only resistivity ρ is independent of l and A.
✗ Proportional to A² — no such relation exists. A square appears only when the radius changes, because A = πr², so R ∝ 1/r².
📌 SSC & Railway Exam Facts
• R = ρl/A; R ∝ l, R ∝ 1/A; SI unit of resistance ohm (Ω).
• Resistivity ρ depends only on material and temperature; SI unit ohm metre (Ω m).
• If a wire is stretched to n times its length, its resistance becomes n²R (volume stays the same).
• Silver has the lowest resistivity; nichrome has a high resistivity and is used in heating elements.
• Ohm's law V = IR; series R(eq) = R₁ + R₂, parallel 1/R(eq) = 1/R₁ + 1/R₂.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Potential difference and EMF

Q136. Which of the following is equal to the value of 1 Volt? (assume J-joule, C-coulomb, s-second)

  1. (a) 1 C/s
  2. (b) 1 Cs
  3. (c) 1 Js
  4. (d) 1 J/1C
Answer & solution

Answer: (d) 1 J/1C

Correct Answer: Option 4 — 1 J/1C
📖 Given
• The quantity to be expressed is potential difference, V
• J = joule (unit of work), C = coulomb (unit of charge), s = second
✍️ Formula
V = W/Q — potential difference = work done per unit charge
🔢 Solution
• Putting the units: V = joule / coulomb
• So 1 volt = 1 J / 1 C
• In words: 1 volt is the potential difference between two points when 1 joule of work is done to move a charge of 1 coulomb between them.
This is why a 12 V battery gives 12 joules of energy to every coulomb of charge it pushes round the circuit.
❌ Why the other options are incorrect
✗ 1 C/s — coulomb per second is the ampere (A), the unit of electric current, not voltage.
✗ 1 Cs — coulomb × second is not a standard electrical unit at all.
✗ 1 Js — joule × second is the unit of Planck's constant, not of potential difference.
📌 SSC & Railway Exam Facts
• 1 A = 1 C/s; 1 Ω = 1 V/A; 1 W = 1 J/s.
• Potential difference is measured with a voltmeter, connected in parallel.
• Current is measured with an ammeter, connected in series.
• Ohm's law: V = IR, given by Georg Simon Ohm.
• Charge on one electron = 1.6 × 10⁻¹⁹ C.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Numericals on electricity

Q137. If 400 J of heat is produced each second in a 4 Ω resistance, then the current flowing across the resistor is:

  1. (a) 20 A
  2. (b) 30 A
  3. (c) 40 A
  4. (d) 10 A
Answer & solution

Answer: (d) 10 A

Correct Answer: Option 4 — 10 A
📖 Given
• Heat produced, H = 400 J in time t = 1 s
• So power, P = H/t = 400 J/s = 400 W
• Resistance, R = 4 Ω
✍️ Formula
H = I²Rt (Joule's law of heating), so P = I²R
🔢 Solution
• 400 = I² × 4 × 1
• I² = 400/4
• I² = 100
• I = √100
• I = 10 A
So a current of 10 amperes flowing through the 4 Ω resistor turns 400 joules of electrical energy into heat every second.
❌ Why the other options are incorrect
✗ 20 A — you get this if you wrongly take I² = 400 and forget to divide by R.
✗ 30 A — does not satisfy the equation; 30² × 4 = 3600 W, nine times too large.
✗ 40 A — this is simply 400/10, a random division that has no basis in Joule's law.
📌 SSC & Railway Exam Facts
• Joule's law of heating: H = I²Rt — heat depends on the square of the current.
• Electric power: P = VI = I²R = V²/R.
• Heating effect is used in the electric iron, heater, geyser, toaster and electric bulb.
• The filament of a bulb is made of tungsten, melting point about 3380 °C.
• A fuse wire has high resistance and low melting point, and is always connected in series with the live wire.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Resistance, resistivity and their factors

Q138. Which of the following would increase the resistance of a metallic conductor?

  1. (a) Using a material with higher conductivity.
  2. (b) Decreasing the current flowing through the conductor.
  3. (c) Decreasing its area of cross-section.
  4. (d) Decreasing its length.
Answer & solution

Answer: (c) Decreasing its area of cross-section.

Correct Answer: Option 3 — Decreasing its area of cross-section.
📖 Given
• A metallic conductor of length l, area of cross-section A and resistivity ρ
• Temperature is unchanged
✍️ Formula
R = ρl/A
🔢 Solution
• R is directly proportional to l — a longer wire has more resistance.
• R is inversely proportional to A — a thinner wire has more resistance.
• So if A is decreased, R increases.
That is exactly why a thin wire glows in a heater while the thick connecting wires stay cool.
❌ Why the other options are incorrect
✗ Using a material with higher conductivity — conductivity σ = 1/ρ, so higher conductivity means lower resistivity and hence lower resistance.
✗ Decreasing the current flowing through the conductor — at a fixed temperature, resistance is a property of the conductor and does not change with the current.
✗ Decreasing its length — since R ∝ l, a shorter wire has less resistance, not more.
📌 SSC & Railway Exam Facts
• Resistance depends on length, area of cross-section, nature of the material and temperature.
• SI unit of resistance is the ohm (Ω); of resistivity, the ohm metre (Ω m).
• Silver is the best conductor; copper and aluminium are used for wires.
• Nichrome has high resistivity and is used in heating elements.
• For metals, resistance increases with temperature; for semiconductors it decreases.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Resistance, resistivity and their factors

Q139. Select the correct pairs from the following options.

  1. (a) (a) Resistance - volt/ampere (b) Resistivity - ohm
  2. (b) (a) Resistance - ohm (b) Resistivity - volt/metre
  3. (c) (a) Resistance - ohm (b) Resistivity - ohm metre
  4. (d) (a) Resistance - ampere metre (b) Resistivity - volt/metre
Answer & solution

Answer: (c) (a) Resistance - ohm (b) Resistivity - ohm metre

Correct Answer: Option 3 — (a) Resistance - ohm, (b) Resistivity - ohm metre
Resistance (R) is the opposition a conductor offers to the flow of current. From Ohm's law, R = V/I, so its unit is volt/ampere, which is given the special name ohm (Ω). Resistance depends on the shape and size of the conductor as well as its material.
Resistivity (ρ) is the resistance of a conductor of unit length and unit area of cross-section. From R = ρl/A we get ρ = RA/l, so its unit is Ω × m²/m = ohm metre (Ω m). Resistivity is a property of the material alone and does not change with the length or thickness of the wire.
❌ Why the other options are incorrect
✗ Option 1 — resistance in volt/ampere is correct, but resistivity is ohm metre, not ohm.
✗ Option 2 — resistance in ohm is correct, but volt/metre is the unit of electric field, not resistivity.
✗ Option 4 — ampere metre is not a unit of resistance at all, and volt/metre is again wrong for resistivity.
📌 SSC & Railway Exam Facts
• R = V/I (ohm) and ρ = RA/l (ohm metre).
• Resistivity of metals ≈ 10⁻⁸ to 10⁻⁶ Ω m; of insulators ≈ 10¹² to 10¹⁷ Ω m.
• Conductance = 1/R, unit siemens (S) or mho.
• Series: R(eq) = R₁ + R₂ + R₃; Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃.
• An alloy like nichrome or manganin has higher resistivity than its constituent metals and does not oxidise easily.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 3 · Resistance, resistivity and their factors

Q140. Which of the following factors do NOT affect the resistance of a conductor (at constant temperature):
(i) Area of cross-section of the conductor
(ii) Current flowing through the conductor
(iii)Length of the conductor

  1. (a) Both (i) and (ii)
  2. (b) Both (i) and (iii)
  3. (c) Only (iii)
  4. (d) Only (ii)
Answer & solution

Answer: (d) Only (ii)

Correct Answer: Option 4 — Only (ii)
The resistance of a metallic conductor is given by R = ρl/A, where ρ is the resistivity of the material, l the length and A the area of cross-section.
• Length (iii) — R ∝ l. A longer wire offers more resistance, so length does affect R.
• Area of cross-section (i) — R ∝ 1/A. A thicker wire offers less resistance, so area does affect R.
• Current (ii) — resistance is a property of the conductor itself. At a constant temperature, changing the current does not change R; by Ohm's law V = IR, a larger current simply means a larger potential difference across the same resistance.
Hence the only factor that does not affect resistance is (ii), the current.
❌ Why the other options are incorrect
✗ Both (i) and (ii) — area of cross-section certainly affects resistance, so (i) cannot be included.
✗ Both (i) and (iii) — both length and area are the two geometrical factors that decide resistance.
✗ Only (iii) — length is one of the strongest factors: double the length and the resistance doubles.
📌 SSC & Railway Exam Facts
• Resistance depends on length, area of cross-section, nature of material and temperature.
• SI unit of resistance is the ohm (Ω); of resistivity, the ohm metre (Ω m).
• Ohm's law: V = IR, valid at constant temperature.
• Silver has the lowest resistivity; nichrome has a high resistivity and is used in heaters.
• For a metal, resistance increases with rise in temperature.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 3 · Numericals on electricity

Q141. A current of 2A flows through a wire when the two ends of the wire are maintained at a potential difference of 4 V. The resistance of the wire will be:

  1. (a) 8 Ω
  2. (b) 2 Ω
  3. (c) 1/2 Ω
  4. (d) 4 Ω
Answer & solution

Answer: (b) 2 Ω

Correct Answer: Option 2 — 2 Ω
📖 Given
• Current through the wire, I = 2 A
• Potential difference across its ends, V = 4 V
• Temperature is constant
✍️ Formula
Ohm's law: V = IR, so R = V/I
🔢 Solution
• R = V/I
• R = 4/2
• R = 2 Ω
So this wire opposes the flow of charge with a resistance of 2 ohms: every 2 volts applied across it drives 1 ampere of current through it.
❌ Why the other options are incorrect
✗ 8 Ω — this comes from multiplying V by I (4 × 2) instead of dividing; V × I gives power in watts, not resistance.
✗ 1/2 Ω — this is I/V = 2/4, the ratio taken upside down; that quantity is the conductance in siemens.
✗ 4 Ω — this is simply the voltage value copied as the answer, ignoring the current altogether.
📌 SSC & Railway Exam Facts
• Ohm's law: V = IR, given by Georg Simon Ohm; valid at constant temperature.
• 1 Ω = 1 V/A; conductance G = 1/R, unit siemens (S).
• A V–I graph for an ohmic conductor is a straight line through the origin, its slope giving 1/R.
• Ammeter is connected in series, voltmeter in parallel.
• Series resistance: R(eq) = R₁ + R₂; parallel: 1/R(eq) = 1/R₁ + 1/R₂.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 1 · Resistors in series and parallel

Q142. Which of the following holds for the parallel combination of resistors?

  1. (a) The total resistance is equal to the smallest resistance.
  2. (b) The total resistance is always less than the smallest resistance.
  3. (c) The total resistance is always greater than the smallest resistance.
  4. (d) The total resistance is equal to the highest resistance.
Answer & solution

Answer: (b) The total resistance is always less than the smallest resistance.

Correct Answer: Option 2 — The total resistance is always less than the smallest resistance.
In a parallel combination the reciprocals of the resistances add up:
1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃ + …
Adding more paths gives the current more ways to flow, exactly like opening extra doors in a hall, so the equivalent resistance keeps falling. Because 1/R(eq) is larger than each individual reciprocal, R(eq) must be smaller than every one of the resistances, including the smallest.
For example, 3 Ω and 6 Ω in parallel give R(eq) = (3 × 6)/(3 + 6) = 18/9 = 2 Ω, which is less than 3 Ω, the smaller of the two. In parallel the voltage across each resistor is the same while the current divides between them.
❌ Why the other options are incorrect
✗ The total resistance is equal to the smallest resistance — it is strictly less than the smallest; equality would need only one resistor in the circuit.
✗ The total resistance is always greater than the smallest resistance — that is true of a series combination, where R(eq) = R₁ + R₂ + R₃ and is greater than every resistance.
✗ The total resistance is equal to the highest resistance — in parallel the highest resistance carries the least current and cannot decide the total; the answer is far below it.
📌 SSC & Railway Exam Facts
• Series: R(eq) = R₁ + R₂ + R₃, same current through each, greater than the largest resistance.
• Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃, same voltage across each, less than the smallest resistance.
• For two resistors in parallel, R(eq) = R₁R₂/(R₁ + R₂); n equal resistors R in parallel give R/n.
• Ohm's law V = IR; SI unit of resistance is the ohm (Ω).
• Household appliances are connected in parallel so that each gets the full 220 V and works independently.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 3 · Resistance, resistivity and their factors

Q143. If the length of a conductor is doubled, then the resistance of the conductor will be: (other parameters are kept same)

  1. (a) the same
  2. (b) halved
  3. (c) one fourth
  4. (d) doubled
Answer & solution

Answer: (d) doubled

Correct Answer: Option 4 — doubled
📖 Given
• Original length l, original resistance R = ρl/A
• New length, l′ = 2l
• Resistivity ρ and area of cross-section A are unchanged
✍️ Formula
R = ρl/A — resistance is directly proportional to length
🔢 Solution
• R′ = ρl′/A = ρ(2l)/A
• R′ = 2 × (ρl/A)
• R′ = 2R, i.e. the resistance is doubled
Making the wire twice as long simply gives the electrons twice as much material to push through, so the opposition to the current becomes twice as large.
❌ Why the other options are incorrect
✗ the same — this would be true only if resistance did not depend on length at all, which contradicts R = ρl/A.
✗ halved — halving happens when the area of cross-section is doubled (a thicker wire), or when the length is halved, not when the length is doubled.
✗ one fourth — this would need R ∝ 1/l²; no such relation exists. A quarter appears if the radius is doubled, since A = πr² then becomes 4 times.
📌 SSC & Railway Exam Facts
• R = ρl/A; R ∝ l and R ∝ 1/A; SI unit of resistance ohm (Ω).
• Resistivity ρ depends only on the material and temperature, not on l or A; SI unit Ω m.
• Ohm's law: V = IR; given by Georg Simon Ohm.
• Series: R(eq) = R₁ + R₂ + …; Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + ….
• Silver is the best conductor; nichrome has high resistivity and is used in heating elements.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 1 · Numericals on electricity

Q144. A current of 1 A is drawn by a filament of an electric bulb. The number of electrons passing through a cross-section of the filament in 8 mins would be equal to ______.
(Given, charge of electron (e) is equal to 1.6 × 10⁻¹⁹ C)

  1. (a) 3 × 10²¹
  2. (b) 3 × 10¹⁹
  3. (c) 3 × 10²²
  4. (d) 3 × 10²⁰
Answer & solution

Answer: (a) 3 × 10²¹

Correct Answer: Option 1 — 3 × 10²¹
📖 Given
• Current through the filament, I = 1 A
• Time, t = 8 mins = 8 × 60 = 480 s
• Charge on one electron, e = 1.6 × 10⁻¹⁹ C
• To find: number of electrons, n
✍️ Formula
Q = I × t and Q = n × e, so n = Q/e = It/e
🔢 Solution
• t = 8 × 60 = 480 s
• Q = I × t = 1 × 480 = 480 C
• n = Q/e = 480/(1.6 × 10⁻¹⁹)
• n = 300 × 10¹⁹
• n = 3 × 10²¹ electrons
In plain language: a current of 1 A means 1 coulomb of charge passes every second, so in 480 seconds 480 coulomb flows, and since each electron carries only 1.6 × 10⁻¹⁹ C, about 3 × 10²¹ electrons must cross the filament.
❌ Why the other options are incorrect
✗ 3 × 10¹⁹ — a hundred times too small. This is the slip of forgetting to convert minutes into seconds properly, or of dividing by 1.6 × 10⁻¹⁷ instead of 1.6 × 10⁻¹⁹.
✗ 3 × 10²² — ten times too large, from a power-of-ten slip while dividing by 1.6 × 10⁻¹⁹.
✗ 3 × 10²⁰ — ten times too small, again a decimal-place error in the same division.
📌 SSC & Railway Exam Facts
• I = Q/t, so Q = It; SI unit of charge is the coulomb (C) and of current the ampere (A).
• Charge is quantised: Q = ne, where e = 1.6 × 10⁻¹⁹ C — the electron was discovered by J. J. Thomson.
• 1 coulomb ≈ 6.25 × 10¹⁸ electrons.
• Conventional current flows opposite to the direction of electron flow.
• Current is measured with an ammeter, connected in series in the circuit.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 1 · Ohm's law

Q145. Which of the following is equal to 1 Ohm (If V and A are volt and ampere, respectively)?

  1. (a) 1V * 1A
  2. (b) 1V/1A
  3. (c) 1V/2A
  4. (d) 1A/1V
Answer & solution

Answer: (b) 1V/1A

Correct Answer: Option 2 — 1V/1A
According to Ohm's law, the potential difference V across a conductor is directly proportional to the current I flowing through it, provided its physical conditions such as temperature remain constant:
V ∝ I, so V = IR, where R is the resistance of the conductor.
Rearranging gives R = V/I. Putting the SI units into this relation gives the definition of the ohm:
1 ohm = 1 volt / 1 ampere
So a conductor has a resistance of 1 Ω if a potential difference of 1 V across its ends drives a current of exactly 1 A through it. The unit is named after the German physicist Georg Simon Ohm.
❌ Why the other options are incorrect
✗ 1V * 1A — volt multiplied by ampere gives power in watts (P = VI), not resistance.
✗ 1V/2A — this works out to 0.5 Ω, not 1 Ω; the definition needs the current to be exactly 1 A.
✗ 1A/1V — this is the reciprocal, ampere per volt, which is the unit of conductance, the siemens (S) or mho.
📌 SSC & Railway Exam Facts
• Ohm's law: V = IR; R = V/I; SI unit ohm (Ω) = volt/ampere.
• Resistance R = ρl/A — it increases with length and decreases with area of cross-section; ρ is the resistivity.
• Resistivity of metals increases with temperature; that of semiconductors decreases.
• Conductance = 1/R, SI unit siemens (S); conductivity = 1/ρ.
• Power P = VI = I²R = V²/R; SI unit watt (W).
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 2 · Resistors in series and parallel

Q146. How are three resistances 3 Ω, 4 Ω, and 6 Ω connected in combination, so that their total resistance will be 13 Ω?

  1. (a) Does not depend on any combination
  2. (b) All in series
  3. (c) 3 Ω in series with 6 Ω and 4 Ω in parallel
  4. (d) All in parallel
Answer & solution

Answer: (b) All in series

Correct Answer: Option 2 — All in series
📖 Given
• Three resistances, R₁ = 3 Ω, R₂ = 4 Ω, R₃ = 6 Ω
• Required total resistance, R(eq) = 13 Ω
✍️ Formula
Series: R(eq) = R₁ + R₂ + R₃
Parallel: 1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃
🔢 Solution
• Series check: R(eq) = 3 + 4 + 6
• R(eq) = 13 Ω ✓ — this matches the required value
• Parallel check: 1/R(eq) = 1/3 + 1/4 + 1/6 = (4 + 3 + 2)/12 = 9/12
• R(eq) = 12/9 = 1.33 Ω, which is far from 13 Ω
So the only way to obtain 13 Ω is to connect all three resistors in series.
❌ Why the other options are incorrect
✗ Does not depend on any combination — the equivalent resistance depends very strongly on the combination; series and parallel give completely different values.
✗ 3 Ω in series with 6 Ω and 4 Ω in parallel — here 6 Ω ∥ 4 Ω = (6 × 4)/(6 + 4) = 2.4 Ω, so total = 3 + 2.4 = 5.4 Ω, not 13 Ω.
✗ All in parallel — as calculated above this gives only 1.33 Ω, the smallest possible value.
📌 SSC & Railway Exam Facts
• In series the current is the same through each resistor and resistances simply add up.
• In parallel the voltage is the same across each resistor and the equivalent resistance is less than the smallest one.
• For two resistors in parallel, R(eq) = R₁R₂/(R₁ + R₂).
• Ohm's law: V = IR; SI unit of resistance is the ohm (Ω).
• Household appliances are connected in parallel so that each gets the full 220 V and can be switched independently.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 3 · Ohm's law

Q147. Which law states that the current flowing through a conductor is directly proportional to the voltage applied across it, provided the temperature and other physical conditions remain constant?

  1. (a) Faraday's law
  2. (b) Newton's law
  3. (c) Joule's law
  4. (d) Ohm's law
Answer & solution

Answer: (d) Ohm's law

Correct Answer: Option 4 — Ohm's law
Ohm's law, given by the German physicist Georg Simon Ohm in 1827, states that the current (I) flowing through a conductor is directly proportional to the potential difference (V) applied across its ends, provided the temperature and other physical conditions remain constant.
So V ∝ I, which gives V = IR, where the constant of proportionality R is the resistance of the conductor, measured in ohm (Ω). For such a conductor a graph of V against I is a straight line passing through the origin, and materials that obey this relation are called ohmic conductors.
❌ Why the other options are incorrect
✗ Faraday's law — Michael Faraday's laws deal with electromagnetic induction (induced emf ∝ rate of change of magnetic flux) and with electrolysis, not with the V–I relation.
✗ Newton's law — Isaac Newton gave the three laws of motion and the law of universal gravitation, which have nothing to do with current and voltage.
✗ Joule's law — Joule's law of heating gives the heat produced by a current, H = I²Rt; it describes the heating effect, not the proportionality of V and I.
📌 SSC & Railway Exam Facts
• Ohm's law: V = IR; SI units — resistance ohm (Ω), current ampere (A), potential difference volt (V).
• 1 ohm = 1 volt/ampere; resistance R = ρl/A, where ρ is the resistivity of the material.
• Resistance increases with length and decreases with area of cross-section; for metals it increases with temperature.
• Semiconductor diodes, gases and filaments at high temperature are non-ohmic — they do not obey Ohm's law.
• Electric power P = VI = I²R = V²/R, and energy = Pt.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 3 · Numericals on electricity

Q148. A resistor whose value R = 10 Ω is connected to a 10-volt battery. What will be the power P and the amount of heat H generated in one minute?

  1. (a) P = 100 W, H = 6000 J
  2. (b) P = 20 W, H = 1200 J
  3. (c) P = 1 W, H = 60 J
  4. (d) P = 10 W, H = 600 J
Answer & solution

Answer: (d) P = 10 W, H = 600 J

Correct Answer: Option 4 — P = 10 W, H = 600 J
📖 Given
• Resistance, R = 10 Ω
• Potential difference of the battery, V = 10 V
• Time, t = 1 minute = 60 s
✍️ Formula
P = V²/R and H = Pt = I²Rt
🔢 Solution
• Current, I = V/R = 10/10 = 1 A
• P = V²/R = (10)²/10 = 100/10
• P = 10 W (check: P = VI = 10 × 1 = 10 W)
• H = P × t = 10 × 60
• H = 600 J
So the resistor converts electrical energy into heat at a steady rate of 10 joule per second, giving 600 J in one minute.
❌ Why the other options are incorrect
✗ P = 100 W, H = 6000 J — here V² = 100 has been taken as the power without dividing by R; every value then comes out ten times too large.
✗ P = 20 W, H = 1200 J — double the correct answer; it appears if the resistance is wrongly taken as 5 Ω instead of 10 Ω.
✗ P = 1 W, H = 60 J — this is the current (V/R = 1 A) mistaken for the power; 1 A is not 1 W.
📌 SSC & Railway Exam Facts
• Electric power: P = VI = I²R = V²/R; SI unit watt (W).
• Joule's law of heating: H = I²Rt = VIt = Pt, measured in joule (J).
• Always convert the time first — 1 minute = 60 s, 1 hour = 3600 s.
• 1 kWh = 3.6 × 10⁶ J is the commercial 'unit' of electrical energy.
• The heating effect of current is used in the electric iron, heater, geyser, toaster and fuse.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 3 · Numericals on electricity

Q149. An electric bulb is connected to a 200 V generator. The current drawn by the bulb is 0.1 A. What is the power of the bulb?

  1. (a) 10 W
  2. (b) 200 W
  3. (c) 2 W
  4. (d) 20 W
Answer & solution

Answer: (d) 20 W

Correct Answer: Option 4 — 20 W
📖 Given
• Potential difference of the generator, V = 200 V
• Current drawn by the bulb, I = 0.1 A
• To find: electric power of the bulb, P
✍️ Formula
P = VI
🔢 Solution
• P = 200 × 0.1
• P = 20 W
So this bulb uses 20 joule of electrical energy every second while it is switched on. Its resistance, by Ohm's law, is R = V/I = 200/0.1 = 2000 Ω.
❌ Why the other options are incorrect
✗ 10 W — half the correct value; it comes from taking the current as 0.05 A instead of 0.1 A.
✗ 200 W — this is just the voltage value copied down as the power; the current has not been multiplied at all.
✗ 2 W — a decimal-point slip: 200 × 0.01 = 2 W, using 0.01 A in place of 0.1 A.
📌 SSC & Railway Exam Facts
• Electric power: P = VI = I²R = V²/R; SI unit watt (W), named after James Watt.
• 1 watt = 1 volt × 1 ampere = 1 joule/second; 1 horsepower = 746 W.
• Electrical energy consumed = P × t; the commercial unit is 1 kWh = 1 unit = 3.6 × 10⁶ J.
• Ohm's law V = IR gives the bulb's resistance as 2000 Ω here.
• Household appliances are joined in parallel so that each gets the full supply voltage.
RRB Technician Grade III 2024-25 – 20 December 2024 Shift 3 · Numericals on electricity

Q150. In a circuit, a 10 volt battery and three resistors R1 = 2 Ω, R2 = 3 Ω, and R3 = 6 Ω are connected in parallel to each other. Which of the following is the correct value of effective resistance R(e) and current I flowing through the circuit?

  1. (a) R(e) = 1 Ω, I = 1 A
  2. (b) R(e) = 2 Ω, I = 10 A
  3. (c) R(e) = 1 Ω, I = 10 A
  4. (d) R(e) = 2 Ω, I = 1 A
Answer & solution

Answer: (c) R(e) = 1 Ω, I = 10 A

Correct Answer: Option 3 — R(e) = 1 Ω, I = 10 A
📖 Given
• Battery voltage, V = 10 V
• R1 = 2 Ω, R2 = 3 Ω, R3 = 6 Ω
• The three resistors are joined in parallel
• To find: effective resistance R(e) and total current I
✍️ Formula
For a parallel combination, 1/R(e) = 1/R1 + 1/R2 + 1/R3, and by Ohm's law, I = V/R(e)
🔢 Solution
• 1/R(e) = 1/2 + 1/3 + 1/6
• Taking LCM 6: 1/R(e) = 3/6 + 2/6 + 1/6
• 1/R(e) = 6/6 = 1
• R(e) = 1 Ω
• I = V/R(e) = 10/1
• I = 10 A
Note that the effective resistance 1 Ω is smaller than the smallest individual resistance (2 Ω) — that is always true for a parallel combination, because the current gets extra paths to flow through.
❌ Why the other options are incorrect
✗ R(e) = 1 Ω, I = 1 A — the resistance is right but the current is wrong; 10 V across 1 Ω must give 10 A, not 1 A.
✗ R(e) = 2 Ω, I = 10 A — 2 Ω is just R1; the parallel formula was not applied, and 10 V across 2 Ω would give only 5 A.
✗ R(e) = 2 Ω, I = 1 A — both values are wrong; this ignores the parallel combination as well as Ohm's law.
📌 SSC & Railway Exam Facts
• Series: R(s) = R1 + R2 + R3 (resistance increases); Parallel: 1/R(p) = 1/R1 + 1/R2 + 1/R3 (resistance decreases).
• In parallel the voltage is the same across each resistor; in series the current is the same through each.
• Branch currents here: 10/2 = 5 A, 10/3 = 3.33 A, 10/6 = 1.67 A, total = 10 A — which checks the answer.
• Ohm's law: V = IR; SI unit of resistance ohm (Ω), of current ampere (A), of potential difference volt (V).
• Domestic appliances are connected in parallel so that each gets the full 220 V and can be switched independently.
RRB Technician Grade III 2024-25 – 20 December 2024 Shift 3 · Numericals on electricity

Q151. The total amount of electric charge that flows through a circuit is 1200 C in 2 min. Then the amount of electric current that is drawn by the same electric circuit is:

  1. (a) 60 A
  2. (b) 100 A
  3. (c) 10 A
  4. (d) 600 A
Answer & solution

Answer: (c) 10 A

Correct Answer: Option 3 — 10 A
📖 Given
• Total charge, Q = 1200 C
• Time, t = 2 min
• 1 min = 60 s, so t = 2 × 60 = 120 s
• To find: electric current, I
✍️ Formula
Electric current = Charge / Time, i.e. I = Q/t
🔢 Solution
• t = 2 × 60 = 120 s
• I = 1200/120
• I = 10 A
In plain language: 10 coulomb of charge crosses any section of the circuit every second, and that rate of flow of charge is what an ammeter reads as 10 amperes.
❌ Why the other options are incorrect
✗ 60 A — this is the number of seconds in a minute, mistakenly written as the answer; it also results from dividing 1200 by 20.
✗ 100 A — obtained by wrongly using t = 12 s instead of 120 s, i.e. a decimal-place slip.
✗ 600 A — obtained by dividing 1200 by 2 and forgetting to convert minutes into seconds — the most common mistake in this question.
📌 SSC & Railway Exam Facts
• I = Q/t; SI unit of current is the ampere (A), and 1 A = 1 C/s.
• Charge on one electron = 1.6 × 10⁻¹⁹ C; so 1 C ≈ 6.25 × 10¹⁸ electrons.
• Current is a scalar quantity even though it has a direction of flow.
• Conventional current flows from the positive to the negative terminal, opposite to the direction of electron flow.
• An ammeter is connected in series (low resistance) and a voltmeter in parallel (high resistance).
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 1 · Resistance, resistivity and their factors

Q152. What happens to the resistance of most conductors as their temperature increases?

  1. (a) It decreases.
  2. (b) It remains the same.
  3. (c) It increases.
  4. (d) It becomes zero.
Answer & solution

Answer: (c) It increases.

Answer: Option 3 — It increases.
In a metal, heat makes the ions vibrate more, so the free electrons collide more often — the resistance rises with temperature.
Revision note — Factors affecting resistance
• R = ρL ÷ A — resistance is directly proportional to length and inversely proportional to area of cross-section.
• ρ (resistivity) depends on the material and temperature, not on the size of the wire; SI unit Ω m.
• For metals, resistance increases with temperature; for semiconductors (silicon, germanium) it decreases.
• Alloys (nichrome, manganin, constantan) have higher resistivity and change little with temperature — used in heating elements and standard resistors.
• Superconductors have zero resistance below a critical temperature.
• Good conductors: silver (lowest resistivity), then copper, aluminium.
Do not mix these up
• Metal vs semiconductor — heating raises a metal's resistance but lowers a semiconductor's.
Also asked from this topic: why alloys are used in toaster heating elements (higher resistivity, do not oxidise readily); the copper alloy used for standard resistors (manganin).
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 1 · Resistance, resistivity and their factors

Q153. A copper wire has a resistance of 10 Ω. If it is replaced with a wire of the same material and length but twice the cross-sectional area, its resistance will be ______.

  1. (a) 10 Ω
  2. (b) 2.5 Ω
  3. (c) 20 Ω
  4. (d) 5 Ω
Answer & solution

Answer: (d) 5 Ω

Rule — R = ρL ÷ A.
Same material ⇒ same ρ; same length.
So resistance is inversely proportional to the area of cross-section.
New resistance.
Area becomes 2A ⇒ R becomes R ÷ 2
R₂ = 10 ÷ 2 = 5 Ω
Answer: 5 Ω
Revision note — Resistance of a wire
• R ∝ L — doubling the length doubles the resistance.
• R ∝ 1 ÷ A — a thicker wire has less resistance.
• Stretching a wire to n times its length (volume fixed) makes the resistance n² times.
• ρ depends only on the material and temperature; unit Ω m.
Thicker wire, easier path — doubling the area halves the resistance, it does not double it.
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 1 · Resistors in series and parallel

Q154. If three resistors of 6 Ω, 3 Ω and 2 Ω are connected in parallel, the total equivalent resistance is ________.

  1. (a) 11 Ω
  2. (b) 2 Ω
  3. (c) 1 Ω
  4. (d) 0.5 Ω
Answer & solution

Answer: (c) 1 Ω

Formula — resistors in parallel.
The reciprocal of the equivalent resistance equals the sum of the reciprocals.
1/R = 1/R₁ + 1/R₂ + 1/R₃
Substitute.
1/R = 1/6 + 1/3 + 1/2
= 1/6 + 2/6 + 3/6
= 6/6 = 1
R = 1 Ω
Answer: 1 Ω
Revision note — Series and parallel combinations
• Series: R = R₁ + R₂ + R₃; the same current flows through each resistor.
• Parallel: 1/R = 1/R₁ + 1/R₂ + 1/R₃; the same potential difference is across each resistor.
• The parallel equivalent is always less than the smallest resistor (here 1 Ω < 2 Ω).
• Two resistors in parallel: R = (R₁ × R₂) ÷ (R₁ + R₂).
• Household wiring is in parallel — each appliance gets the full voltage and works independently.
Also asked from this topic: 4 Ω, 6 Ω and 12 Ω in parallel (2 Ω); a wire of resistance R cut in half and the halves joined in parallel (R/4); what is the same across resistors in parallel (voltage).
11 Ω is the series total — in parallel the answer must be below the smallest resistor, 2 Ω.
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 1 · Resistance, resistivity and their factors

Q155. If the length of a wire-shaped resistor is doubled, the resistivity:

  1. (a) remains the same
  2. (b) is halved
  3. (c) is quadrupled
  4. (d) is doubled
Answer & solution

Answer: (a) remains the same

Answer: Option 1 — remains the same
Resistivity (ρ) is a property of the material (and its temperature), not of the wire's size.
Doubling the length doubles the resistance, but the resistivity stays unchanged.
Revision note — Resistance and resistivity
• R = ρ l ÷ A — resistance is directly proportional to length and inversely proportional to area.
• SI unit of resistivity: ohm metre (Ω m); of resistance: ohm (Ω).
• Resistivity changes only with the material and temperature.
• Metals and alloys: 10^(−8) to 10^(−6) Ω m; insulators (glass, rubber): 10^(12) to 10^(17) Ω m.
• Alloys (nichrome) have higher resistivity than their metals and do not oxidise easily — used in heaters, irons, toasters.
• Copper and aluminium have low resistivity — used for transmission wires.
Do not mix these up
• Resistance vs resistivity — cutting, stretching or thickening a wire changes its resistance; its resistivity stays the same.
• Stretching a wire to double length: area halves, so resistance becomes 4 times; resistivity is still unchanged.
Also asked from this topic: new resistance when a 4 Ω wire is stretched to double its length (16 Ω); why nichrome is used as a heating element (high resistivity and high melting point).
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 1 · Ohm's law

Q156. The potential difference versus current plot for a conducting wire is a straight line. The slope of this line gives us the:

  1. (a) number of electrons in the wire
  2. (b) voltage across the wire
  3. (c) resistance of the wire
  4. (d) current flowing through the wire
Answer & solution

Answer: (c) resistance of the wire

Answer: Option 3 — resistance of the wire
By Ohm's law, V = IR, so V/I is constant for a given wire.
On a V (y-axis) versus I (x-axis) graph, the slope = V/I = R, the resistance.
Revision note — Ohm's law
• Ohm's law: at constant temperature, V ∝ I, i.e. V = IR.
• The V–I graph of an ohmic conductor is a straight line through the origin.
• Steeper V–I line ⇒ higher resistance.
• If I is on the y-axis and V on the x-axis, the slope is 1/R (conductance).
• SI unit of resistance: ohm (Ω); 1 Ω = 1 V/1 A.
• R depends on length (R ∝ l), area (R ∝ 1/A), material and temperature: R = ρl/A.
Do not mix these up
• V–I vs I–V graph — slope of V against I is R; slope of I against V is 1/R.
Also asked from this topic: the shape of the graph that represents Ohm's law (a straight line through the origin); the unit volt per ampere (ohm).
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 1 · Electric power and electrical energy

Q157. Which of the following will cause the greatest increase in power dissipation for a given resistor?

  1. (a) Halving the resistance
  2. (b) Doubling the voltage across the resistor
  3. (c) Doubling the resistance
  4. (d) Halving the current through the resistor
Answer & solution

Answer: (b) Doubling the voltage across the resistor

Answer: Option 2 — Doubling the voltage across the resistor
Power in a resistor: P = V²/R = I²R.
Doubling V ⇒ P = (2V)²/R = 4 times, the largest rise of the four.
Revision note — Electric power
• P = VI = I²R = V²/R; SI unit watt (W); 1 W = 1 V × 1 A.
• Commercial unit of energy: kilowatt-hour (kWh) = 3.6 × 10^(6) J = 1 "unit".
• Energy = Power × Time.
• P ∝ square of V or I, so doubling either gives 4 times the power; halving either gives one-fourth.
• At fixed V, P ∝ 1/R — halving R only doubles the power.
• Heat produced (Joule's law): H = I²Rt.
Also asked from this topic: the formula for electrical energy (Energy = Power × Time); how to reduce power without changing current (decrease the resistance); resistance of a 60 W bulb at 120 V (240 Ω).
Power goes as the square of voltage — doubling V gives four times, not two times, the power.
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 2 · Resistors in series and parallel

Q158. In a series circuit, if one of the resistors is removed or broken, the current in the circuit will ________.

  1. (a) increase
  2. (b) decrease
  3. (c) become zero
  4. (d) stay the same
Answer & solution

Answer: (c) become zero

Answer: Option 3 — become zero
A series circuit has only one path for current. Removing or breaking any resistor opens the circuit, so no current flows.
Revision note — Series and parallel combinations
• Series: R = R₁ + R₂ + R₃; same current through each resistor; voltage divides.
• Equivalent resistance in series is greater than the largest resistor.
• Parallel: 1/R = 1/R₁ + 1/R₂ + 1/R₃; same voltage across each; current divides.
• Equivalent resistance in parallel is less than the smallest resistor.
• Household wiring is parallel — each appliance gets 220 V and works independently.
• Decorative serial bulb strings: one bulb fuses → all go off (series).
Do not mix these up
• Series vs parallel break — in series the whole circuit stops; in parallel only that branch stops, the rest keep working.
Also asked from this topic: the arrangement where removing one device does not stop the others (parallel); 4 Ω, 6 Ω and 12 Ω in parallel (2 Ω).
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 2 · Heating effect of current and Joule's law

Q159. Which of the following devices operates based on the heating effect of electric current?

  1. (a) Electric motor
  2. (b) Generator
  3. (c) Transformer
  4. (d) Electric iron
Answer & solution

Answer: (d) Electric iron

Answer: Option 4 — Electric iron
An electric iron has a heating element of high resistance; current through it produces heat by Joule's law (H = I^(2)Rt).
The motor, generator and transformer work on magnetic effects of current, not on heating.
Revision note — Heating effect of electric current
• Joule's law: H = I^(2)Rt — heat ∝ square of current, ∝ resistance, ∝ time.
• Devices using heating: electric iron, heater, toaster, oven, geyser, kettle, bulb filament.
• Heating elements are made of nichrome — high resistivity and high melting point, does not oxidise easily.
• Bulb filaments are tungsten (melting point about 3380 °C); bulbs are filled with nitrogen and argon.
• A fuse works on the heating effect; its wire has a low melting point and is connected in series.
• Electric motor — magnetic force on a current-carrying conductor; generator — electromagnetic induction.
Also asked from this topic: why nichrome is used in heaters (high resistivity and high melting point); the change in heat when current is doubled (four times); the gas filled in electric bulbs (nitrogen and argon).
Transformer and generator also "get hot", but heat is their loss — the iron is the one built to use it.
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 3 · Numericals on electricity

Q160. An electric refrigerator rated 400 W operates 8 hours/day. What is the cost of the energy to operate it for 30 days at ₹3.00 per kW h?

  1. (a) ₹288
  2. (b) ₹36
  3. (c) ₹9.6
  4. (d) ₹400
Answer & solution

Answer: (a) ₹288

Energy used.
Energy (kWh) = power (kW) × time (hours)
Power = 400 W = 0.4 kW
Time = 8 h/day × 30 days = 240 h
Energy = 0.4 × 240 = 96 kWh
Cost.
Cost = energy × rate
Cost = 96 × ₹3 = ₹288
Answer: ₹288
Revision note — Electrical energy and its commercial unit
• 1 kWh = 1 kW used for 1 hour = 1 unit of electricity.
• 1 kWh = 3.6 × 10^(6) J.
• Electric power P = VI = I²R = V²/R; unit watt.
• Always change W to kW (÷ 1000) before multiplying by hours.
Also asked from this topic: the units used by a 60 W lamp for 5 hours a day for 30 days (9 units); the cost of running a 2.2 kW kettle for 30 minutes at ₹4 per unit (₹4.40).
₹36 comes from forgetting the 8 hours a day — count total hours before multiplying.
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 3 · Heating effect of current and Joule's law

Q161. An electric fan becomes warm if used continuously for longer time. This happens due to:

  1. (a) the high temperature during the summer
  2. (b) heating effect of the current flowing through the fan
  3. (c) the fast rotation of the fan
  4. (d) the slow rotation of the fan
Answer & solution

Answer: (b) heating effect of the current flowing through the fan

Answer: Option 2 — heating effect of the current flowing through the fan
The current in the fan's copper windings meets resistance.
Part of the electrical energy turns into heat (Joule heating), so the fan warms up over time.
Revision note — Heating effect of electric current
• Joule's law: H = I^(2)Rt — heat ∝ square of current, resistance and time.
• Doubling the current makes the heat four times.
• Useful applications: electric iron, heater, toaster, geyser, bulb filament, fuse.
• Heater coils use nichrome — high resistivity and high melting point.
• Bulb filament: tungsten (melting point about 3380 °C); bulbs are filled with nitrogen / argon.
• In motors, fans and transformers this heat is wasted energy.
Also asked from this topic: the change in heat when current is doubled (four times); why nichrome is used in heaters (high resistivity and high melting point).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 1 · Electric power and electrical energy

Q162. Equipment such as geysers and air conditioners, compared to light bulbs and fans, require:

  1. (a) higher power
  2. (b) less voltage
  3. (c) less power
  4. (d) higher voltage
Answer & solution

Answer: (a) higher power

Answer: Option 1 — higher power
Every appliance in a house gets the same 220 V because household wiring is in parallel.
Geysers and ACs have a far higher power rating (about 1,500–2,000 W) than a bulb (about 60 W) or a fan (about 75 W).
Since P = VI, at the same voltage they draw a larger current.
Revision note — Power of household appliances
• P = V × I; unit watt (W); 1 kW = 1000 W.
• Indian household supply: 220 V AC, 50 Hz.
• High-power appliances (geyser, AC, heater, iron) run on the 15 A circuit; bulbs and fans on the 5 A circuit.
• Heating appliances use the heating effect of current (H = I²Rt) — hence the large current.
• Energy used = power × time; 1 kWh = 1 unit = 3.6 × 10⁶ J.
Do not mix these up
• Higher power vs higher voltage — all appliances get the same voltage; the difference is in power and current.
Also asked from this topic: the current rating of the circuit used for geysers and air coolers (15 A).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 1 · Resistors in series and parallel

Q163. When several resistors are joined in series, the equivalent resistance of the combination is:

  1. (a) less than the sum of the individual resistances
  2. (b) greater than the sum of the individual resistances
  3. (c) equal to the sum of the individual resistances
  4. (d) equal to the product of the individual resistances
Answer & solution

Answer: (c) equal to the sum of the individual resistances

Answer: Option 3 — equal to the sum of the individual resistances
In series the same current flows through every resistor, and the potential differences add up.
So R_(s) = R_(1) + R_(2) + R_(3) + …
Revision note — Resistors in series and parallel
• Series: same current through each; V = V_(1) + V_(2) + V_(3); R_(s) is greater than the largest resistor.
• Parallel: same potential difference across each; 1/R_(p) = 1/R_(1) + 1/R_(2) + …
• R_(p) is less than the smallest resistor in the combination.
• n equal resistors R: series = nR, parallel = R/n; ratio R_(s) : R_(p) = n² : 1.
• Household wiring is in parallel — each appliance gets the full 220 V and can be switched separately.
• In series, if one element fails the whole circuit breaks.
Do not mix these up
• Sum vs greater than the sum — the series total is exactly the sum; it is greater than each resistor, not greater than the sum.
Also asked from this topic: current when 3 Ω, 2 Ω and 1 Ω in series are joined to a 9 V battery (1.5 A); a practical reason for a series combination (to increase total resistance and limit current).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 1 · Numericals on electricity

Q164. An electric fan is connected to a 220 V generator. The current is 0.50 A. What is the power used by the fan?

  1. (a) 440 Watt
  2. (b) 11 Watt
  3. (c) 110 Watt
  4. (d) 220 Watt
Answer & solution

Answer: (c) 110 Watt

Formula — electric power.
Power = potential difference × current
P = V × I
Substitute.
V = 220 V, I = 0.50 A
P = 220 × 0.50 = 110 W
Answer: 110 Watt
Revision note — Electric power
• P = VI = I²R = V²/R; SI unit watt (W): 1 W = 1 V × 1 A.
• Electrical energy = power × time; commercial unit kilowatt hour (kWh) = 1 unit.
• 1 kWh = 3.6 × 10^(6) J.
• An appliance rated 220 V, 100 W draws I = P ÷ V ≈ 0.45 A.
Also asked from this topic: the formula for electrical energy (Energy = Power × Time); 1 kilowatt hour in joules (3.6 × 10^(6) J); units used by a 60 W lamp run 5 hours a day for 30 days (9 units).
Multiply V by I for power; dividing gives resistance (V ÷ I = 440 Ω), not power.
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 1 · Resistance, resistivity and their factors

Q165. Which combination of factors would result in the lowest resistance for a conductor?

  1. (a) Long length and small cross-sectional area
  2. (b) Short length and large cross-sectional area
  3. (c) Long length and large cross-sectional area
  4. (d) Short length and small cross-sectional area
Answer & solution

Answer: (b) Short length and large cross-sectional area

Answer: Option 2 — Short length and large cross-sectional area
Resistance R = ρL/A: it increases with length and decreases with cross-sectional area.
So the shortest, thickest conductor has the lowest resistance.
Revision note — Factors affecting resistance
• R ∝ L (length) and R ∝ 1/A (area of cross-section); SI unit ohm (Ω).
• ρ (resistivity) depends on the material and temperature only; SI unit Ω m.
• Stretching a wire to double its length halves its area, so resistance becomes 4 times.
• Copper and aluminium (low ρ) are used for transmission wires.
• Alloys such as nichrome (high ρ, do not oxidise easily) are used in heating elements.
Also asked from this topic: the resistance when length is doubled and area halved (four times); which of two wires of the same material and length has greater resistance (the one with the smaller cross-sectional area).
Long and thin means high resistance; short and thick means low.
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 2 · Ohm's law

Q166. If the resistance in a circuit is increased while the voltage remains constant, the current will __________.

  1. (a) remain constant
  2. (b) increase
  3. (c) double
  4. (d) decrease
Answer & solution

Answer: (d) decrease

Answer: Option 4 — decrease
By Ohm's law, I = V ÷ R.
With V fixed, current is inversely proportional to resistance — more resistance means less current.
Revision note — Ohm's law
• Statement: at constant temperature, the current through a conductor is directly proportional to the potential difference across it: V = IR.
• R = V ÷ I; SI unit ohm (Ω) = 1 volt per ampere.
• At constant V: doubling R halves I; halving R doubles I.
• The V–I graph of an ohmic conductor is a straight line through the origin; its slope gives R.
• Resistance itself depends on length, area of cross-section, material and temperature, not on V or I.
Do not mix these up
• Current vs resistance — changing V changes I, but the resistance of the conductor stays the same.
Also asked from this topic: what happens to the current when the resistance is doubled at constant voltage (it halves); the quantity measured in volt per ampere (resistance, unit ohm).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 2 · Numericals on electricity

Q167. The number of 176-Ω resistors (in parallel) required to carry 5 A on a 220-V line is:

  1. (a) 4
  2. (b) 8
  3. (c) 2
  4. (d) 16
Answer & solution

Answer: (a) 4

Formula — current through each resistor.
In parallel every resistor gets the full 220 V.
Current in one resistor = V ÷ R (Ohm's law)
= 220 ÷ 176 = 1.25 A
Number of resistors.
Total current = number × current in each
n = 5 ÷ 1.25 = 4
Short trick.
Required total resistance = 220 ÷ 5 = 44 Ω
n equal resistors R in parallel give R ÷ n.
176 ÷ n = 44 ⇒ n = 4
Answer: 4 resistors
Revision note — Resistors in parallel
• 1/R_(p) = 1/R_(1) + 1/R_(2) + …; n equal resistors R give R ÷ n.
• The voltage is the same across each branch; the currents add up.
• R_(p) is less than the smallest single resistance.
• Household appliances are joined in parallel so each gets the full voltage and can be switched separately.
Multiplying 176 by 4 treats them as a series combination — in parallel the resistance falls, it does not rise.
Also asked from this topic: the total current when 4 Ω and 12 Ω in parallel are joined to a 6 V supply (2 A); how the equivalent resistance of a parallel combination compares with the resistors (less than the smallest).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 3 · Heating effect of current and Joule's law

Q168. If the resistance of a conductor is doubled while the current remains the same, how will the heat produced per unit time change?

  1. (a) It will quadruple.
  2. (b) It will remain the same.
  3. (c) It will be halved.
  4. (d) It will double.
Answer & solution

Answer: (d) It will double.

Answer: Option 4 — It will double.
By Joule's law of heating, H = I²Rt.
With the current and time unchanged, H is directly proportional to R.
So doubling R doubles the heat produced per unit time.
Revision note — Heating effect of electric current
• Joule's law: H = I²Rt — heat is proportional to I², to R and to t.
• Doubling the current makes the heat four times (because of I²).
• At a fixed voltage use H = V²t/R — there, doubling R halves the heat.
• Heating elements use nichrome (nickel–chromium alloy): high resistivity and high melting point.
• Bulb filaments are tungsten (melting point about 3380 °C); bulbs are filled with inert argon/nitrogen.
• An electric fuse works on the heating effect; its wire has a low melting point and is connected in series.
Do not mix these up
• Same current vs same voltage — with same current, heat ∝ R; with same voltage, heat ∝ 1/R.
Also asked from this topic: heat produced when the current is doubled (four times); the SI unit of heat produced in a resistor (joule); why nichrome is used in heaters (high resistivity and high melting point).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 3 · Resistance, resistivity and their factors

Q169. The measure of how much a material resists the flow of charges is known as:

  1. (a) power generated
  2. (b) potential difference
  3. (c) current
  4. (d) resistivity
Answer & solution

Answer: (d) resistivity

Answer: Option 4 — resistivity
Resistivity (ρ) is the property of a material that measures how strongly it opposes the flow of charges.
It depends only on the material and its temperature, not on the wire's length or thickness.
Revision note — Resistance and resistivity (Class 10)
• R = ρL/A — resistance rises with length (L) and falls with area of cross-section (A).
• SI unit of resistivity: ohm-metre (Ω m); of resistance: ohm (Ω).
• Resistivity = resistance of a conductor of unit length and unit cross-sectional area.
• Metals have low resistivity (10^(−8) Ω m range); insulators like glass and rubber very high (10^(12)–10^(17) Ω m).
• Alloys (nichrome, manganin) have higher resistivity than their metals and do not oxidise easily — used in heating elements.
• For metals, resistivity increases with temperature.
Do not mix these up
• Resistance vs resistivity — resistance depends on the wire's size and shape; resistivity is fixed for a material at a given temperature.
Also asked from this topic: the resistance of a conductor of unit length and unit area (resistivity); resistance when a wire is stretched to double length (four times).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 3 · Resistance, resistivity and their factors

Q170. Which of the following materials will have the least resistance at room temperature?

  1. (a) Nichrome
  2. (b) Copper
  3. (c) Iron
  4. (d) Aluminum
Answer & solution

Answer: (b) Copper

Answer: Option 2 — Copper
Among the four, copper has the lowest resistivity (about 1.62 × 10^(−8) Ω m), so for the same size it offers the least resistance.
Revision note — Resistivity of materials at 20 °C (NCERT table)
• Silver 1.60 × 10^(−8) Ω m — the best conductor.
• Copper 1.62 × 10^(−8) Ω m — used for household wiring.
• Aluminium 2.63 × 10^(−8) Ω m — used in overhead transmission lines (light and cheap).
• Iron 10.0 × 10^(−8) Ω m.
• Nichrome (Ni–Cr–Mn–Fe alloy) about 100 × 10^(−8) Ω m — heating elements.
• Manganin and constantan — alloys with high resistivity used in standard resistors.
Do not mix these up
• Copper vs aluminium — aluminium is used for long transmission lines because it is lighter and cheaper, not because it conducts better; copper conducts better.
Also asked from this topic: the wire with lower resistance between copper and iron of the same size (copper); why copper is used in household wiring (low resistance).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 3 · Resistors in series and parallel

Q171. Which of the following is correct with regard to resistances in a series circuit?

  1. (a) Total resistance is equal to the sum of all individual resistances.
  2. (b) Total resistance is equal to the product of all resistances.
  3. (c) Total resistance is less than any individual resistance.
  4. (d) Total resistance is inversely proportional to each resistance.
Answer & solution

Answer: (a) Total resistance is equal to the sum of all individual resistances.

Answer: Option 1 — Total resistance is equal to the sum of all individual resistances.
In series the same current passes through each resistor one after another.
So the resistances add up: R = R_(1) + R_(2) + R_(3).
Revision note — Series and parallel combinations
• Series — same current through every resistor; the potential difference divides among them.
• Series total resistance is greater than the largest single resistance.
• Parallel — same potential difference across each resistor; the current divides.
• Parallel: 1/R = 1/R_(1) + 1/R_(2) + 1/R_(3) — total is less than the smallest resistance.
• Domestic wiring uses parallel — each appliance gets full voltage and can be switched independently.
• In series, if one bulb fuses the whole circuit breaks.
Do not mix these up
• Series vs parallel — series keeps current the same; parallel keeps voltage the same.
Also asked from this topic: why resistors are joined in series (to increase total resistance and limit current); current when 3 Ω, 2 Ω and 1 Ω in series are joined to 9 V (1.5 A).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 1 · Resistors in series and parallel

Q172. In a series combination of resistors, how is the voltage distributed across each resistor?

  1. (a) Directly proportional to resistance
  2. (b) Unaffected by resistance values
  3. (c) The same across all resistors
  4. (d) Inversely proportional to resistance
Answer & solution

Answer: (a) Directly proportional to resistance

Answer: Option 1 — Directly proportional to resistance
In series the same current I flows through every resistor.
By Ohm's law, V = I × R, so with I fixed the larger resistor takes the larger share of voltage.
Revision note — Series and parallel combinations
• Series: same current; voltage divides as V_(1) : V_(2) = R_(1) : R_(2); R_(s) = R_(1) + R_(2) + R_(3).
• Parallel: same voltage; current divides inversely as resistance; 1/R_(p) = 1/R_(1) + 1/R_(2) + 1/R_(3).
• R_(s) is greater than the largest resistor; R_(p) is less than the smallest.
• One bulb failing in series breaks the whole circuit; in parallel the others keep working.
• Household wiring is in parallel, so every appliance gets the full 220 V.
Do not mix these up
• Series vs parallel — series shares voltage, parallel shares current.
Also asked from this topic: what each resistor receives in a parallel combination (the same voltage); why resistors are joined in series (to increase total resistance and limit current).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 1 · Numericals on electricity

Q173. What is the power rating of a device that draws 1.75 A of current when connected to a 180-V source?

  1. (a) 440 W
  2. (b) 225 W
  3. (c) 315 W
  4. (d) 105 W
Answer & solution

Answer: (c) 315 W

Formula — electric power.
Power = voltage × current
P = V × I
Substitute.
V = 180 V, I = 1.75 A
P = 180 × 1.75
P = 180 × 7/4 = 45 × 7 = 315 W
Answer: 315 W
Revision note — Electric power
• P = VI = I²R = V²/R; SI unit watt (W) = 1 J/s.
• 1 kW = 1000 W; 1 horsepower = 746 W.
• Energy = power × time; commercial unit 1 kWh = 1 unit = 3.6 × 10⁶ J.
• Power rating on a device = energy it uses per second at its rated voltage.
Also asked from this topic: the formula for electrical energy (Energy = Power × Time); 1 kilowatt hour in joules (3.6 × 10⁶ J).
Trap: 180 ÷ 1.75 gives the resistance (about 103 Ω), not the power.
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 1 · Electric power and electrical energy

Q174. The power in a circuit having voltage V and current I during time interval t is:

  1. (a) I/V
  2. (b) V/I
  3. (c) (I × t)/V
  4. (d) I × V
Answer & solution

Answer: (d) I × V

Answer: Option 4 — I × V
Work done in moving charge Q through potential difference V is W = VQ, and Q = I × t.
Power = work ÷ time = VIt ÷ t = V × I — the time interval cancels out.
Revision note — Electric power
• Electric power = rate at which electrical energy is used: P = VI.
• Using Ohm's law (V = IR): P = I²R = V²/R.
• SI unit watt (W); 1 W = 1 V × 1 A = 1 J/s.
• Electrical energy = P × t = VIt; commercial unit 1 kWh = 3.6 × 10⁶ J.
• A 100 W, 220 V bulb draws I = P/V = 0.45 A.
Do not mix these up
• V/I vs V × I — V/I is resistance (ohm); V × I is power (watt).
Also asked from this topic: the formula for electrical energy (Energy = Power × Time); 1 kilowatt hour in joules (3.6 × 10⁶ J).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 1 · Resistors in series and parallel

Q175. In a parallel circuit, how does the voltage across each resistor compare to the total voltage supplied by the battery?

  1. (a) It is the same for each resistor.
  2. (b) It is divided according to the resistance values.
  3. (c) It is inversely proportional to the resistance.
  4. (d) It is zero.
Answer & solution

Answer: (a) It is the same for each resistor.

Answer: Option 1 — It is the same for each resistor.
In parallel, every resistor is connected across the same two points of the battery.
So each one gets the full battery voltage; it is the current that divides among them.
Revision note — Series and parallel combinations
• Parallel: same voltage across each; current divides inversely as resistance; 1/R_(p) = 1/R_(1) + 1/R_(2) + 1/R_(3).
• Series: same current through each; voltage divides in proportion to resistance; R_(s) = R_(1) + R_(2) + R_(3).
• R_(p) is less than the smallest resistor; R_(s) is greater than the largest.
• In parallel, one appliance failing does not stop the others.
• Household appliances are wired in parallel, so each gets the full 220 V.
Do not mix these up
• Series vs parallel — series shares voltage, parallel shares current.
Also asked from this topic: the property unique to a parallel combination (each resistor receives the same voltage); why domestic appliances are connected in parallel (each gets equal potential difference).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 1 · Electric power and electrical energy

Q176. What is the relationship between current (I), power (P), and voltage (V)?

  1. (a) I = PV
  2. (b) I = P ÷ V
  3. (c) I = V ÷ P
  4. (d) I = P2V
Answer & solution

Answer: (b) I = P ÷ V

Answer: Option 2 — I = P ÷ V
Electric power = voltage × current (P = VI).
Dividing both sides by V gives I = P ÷ V.
Revision note — Electric power
• Power is the rate at which electrical energy is used: P = VI.
• With Ohm's law (V = IR): P = I²R = V²/R.
• SI unit watt (W); 1 W = 1 V × 1 A; 1 kW = 1,000 W.
• Electrical energy E = P × t; the commercial unit is the kilowatt-hour (kWh).
• 1 kWh = 3.6 × 10⁶ J = 1 unit on the electricity bill.
Also asked from this topic: resistance of a 60 W bulb on 120 V (R = V²/P = 240 Ω); the formula for electrical energy (Energy = Power × Time).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 2 · Ohm's law

Q177. If a conductor follows the Ohm’s law, then the graph between current and voltage will be a/an _______.

  1. (a) straight line
  2. (b) ellipse
  3. (c) parabola
  4. (d) hyperbola
Answer & solution

Answer: (a) straight line

Answer: Option 1 — straight line
Ohm's law: V = I × R; with R constant, current ∝ voltage, so the V–I graph is a straight line through the origin.
Revision note — Ohm's law
• Statement: at constant temperature, the current through a conductor is directly proportional to the potential difference across it.
• V = IR; unit of resistance ohm (Ω) = 1 V/A.
• Slope of the V–I graph (V on the y-axis) gives the resistance R; a steeper line means higher resistance.
• Conductors that obey the law are ohmic (metal wires, nichrome); diodes and filament bulbs at high temperature are non-ohmic (curved graph).
• Law given by Georg Simon Ohm (1827).
Also asked from this topic: the graph for Ohm's law for a resistor at constant temperature (a straight line through the origin); what happens to current if resistance is doubled at the same voltage (it halves).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 2 · Heating effect of current and Joule's law

Q178. If a resistor of 15 Ω carries a current of 2.5 A for 12 seconds, the heat produced is _____ joules.

  1. (a) 900
  2. (b) 1125
  3. (c) 1800
  4. (d) 450
Answer & solution

Answer: (b) 1125

Formula — Joule's law of heating.
Heat produced = (current)² × resistance × time
H = I² × R × t
Substitute the values.
I = 2.5 A, R = 15 Ω, t = 12 s
I² = 2.5 × 2.5 = 6.25
H = 6.25 × 15 × 12
H = 6.25 × 180 = 1125 J
Answer: 1125 joules
Revision note — Heating effect of electric current
• Joule's law: H = I²Rt — heat ∝ square of current, ∝ resistance, ∝ time.
• Using V = IR, the same heat is H = VIt = V²t ÷ R.
• Doubling the current makes the heat four times.
• Heating elements use nichrome — high resistivity and high melting point.
• An electric fuse works on the heating effect; its wire has a low melting point.
Also asked from this topic: the change in heat when the current is doubled (four times); the relation for Joule's law of heating (H = I²Rt); the SI unit of the heat produced in a resistor (joule).
Square the current first — writing 2.5 × 15 × 12 = 450 is the trap option.
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 2 · Resistance, resistivity and their factors

Q179. A wire has a resistance of R. If its length is increased by a factor of 8 and its cross-sectional area is decreased by a factor of 4, what is its new resistance (R')?

  1. (a) R' = 32 R
  2. (b) R' = 8 R
  3. (c) R' = 2 R
  4. (d) R' = 16 R
Answer & solution

Answer: (a) R' = 32 R

Formula — resistance of a wire.
Resistance = resistivity × length ÷ area
R = ρ × L ÷ A
Resistivity ρ depends only on the material, so it stays the same.
Apply the changes.
New length = 8L
New area = A ÷ 4
R' = ρ × 8L ÷ (A ÷ 4)
R' = ρ × 8L × 4 ÷ A
R' = 32 × (ρL ÷ A) = 32R
Answer: R' = 32R
Revision note — Factors on which resistance depends
• R is directly proportional to length and inversely proportional to area of cross-section.
• Resistivity (ρ) is a property of the material; SI unit ohm metre (Ω m).
• Resistivity changes with temperature, not with the length or thickness of the wire.
• A wire stretched to n times its length (volume fixed) has resistance n²R — its area falls to A ÷ n.
• Metals and alloys: nichrome has a much higher resistivity than copper.
Also asked from this topic: the new resistance of a 4 Ω wire stretched to double length (16 Ω); the effect of doubling the length on resistance (it becomes four times when stretched).
Area is in the denominator — dividing it by 4 multiplies R by 4, so 8 × 4 = 32, not 8 ÷ 4 = 2.
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 2 · Ohm's law

Q180. The potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. This is known as:

  1. (a) Newton's law
  2. (b) Ohm's law
  3. (c) Ampere's law
  4. (d) Resistance law
Answer & solution

Answer: (b) Ohm's law

Answer: Option 2 — Ohm's law
Ohm's law: at constant temperature, V ∝ I, so V = IR, where R is the resistance of the wire.
Revision note — Ohm's law
• Given by Georg Simon Ohm (German physicist) in 1827.
• V = IR; R = V/I; SI unit of resistance = ohm (Ω) = 1 V/A.
• The V–I graph of an ohmic conductor is a straight line through the origin; its slope gives R.
• The law holds only if temperature and other physical conditions stay the same.
• Resistance depends on length (R ∝ l), area (R ∝ 1/A), material (resistivity ρ) and temperature: R = ρl/A.
• Non-ohmic devices (diodes, filament lamps at high temperature) do not give a straight V–I line.
Do not mix these up
• Ohm's law vs Ampere's law — Ohm links V, I and R in a circuit; Ampere's law relates the magnetic field around a closed loop to the current through it.
Also asked from this topic: what happens to current when resistance is doubled at constant voltage (it halves); the quantity measured in volt per ampere (resistance, in ohm).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 2 · Numericals on electricity

Q181. A conductor has a current of 150 mA flowing through it with a resistance of 0.8 Ω. Find the potential difference across the conductor.

  1. (a) 5.3 V
  2. (b) 120 V
  3. (c) 18.75 V
  4. (d) 0.12 V
Answer & solution

Answer: (d) 0.12 V

Formula — Ohm's law.
Potential difference = current × resistance
V = I × R
Convert the current to amperes.
1 A = 1000 mA
I = 150 mA = 150 ÷ 1000 = 0.15 A
Substitute.
R = 0.8 Ω
V = 0.15 × 0.8 = 0.12 V
Answer: 0.12 V
Revision note — Ohm's law
• At constant temperature, V ∝ I, so V = IR; the V–I graph is a straight line through the origin.
• Units: V in volt, I in ampere, R in ohm (Ω); 1 Ω = 1 V/A.
• 1 mA = 10⁻³ A; 1 kΩ = 1000 Ω.
• Rearranged: I = V/R and R = V/I.
Also asked from this topic: the current when a 12 V battery is joined to a 6 Ω resistor (2 A); the quantity whose unit is volt per ampere (ohm).
Trap: using 150 instead of 0.15 A gives 120 V — always convert mA to A first.
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 2 · Resistance, resistivity and their factors

Q182. If a wire is stretched to quadruple its length without changing its area, what will happen to its resistivity?

  1. (a) It quadruples.
  2. (b) It remains the same.
  3. (c) It doubles.
  4. (d) It halves.
Answer & solution

Answer: (b) It remains the same.

Answer: Option 2 — It remains the same.
Resistivity (ρ) is a property of the material (and its temperature), not of the wire's shape.
Stretching changes the resistance, never the resistivity.
Revision note — Resistance and resistivity
• R = ρL/A — resistance is directly proportional to length and inversely proportional to area of cross-section.
• Resistivity depends only on the nature of the material and temperature; SI unit ohm metre (Ω m).
• A wire stretched to n times its length (volume fixed) has n² times the resistance — doubling gives 4R.
• Metals have low resistivity (10⁻⁸ Ω m); alloys like nichrome have higher resistivity and are used in heaters.
• Silver is the best conductor; copper and aluminium are used for transmission wires.
Do not mix these up
• Resistance vs resistivity — resistance changes with length, area and material; resistivity changes only with material and temperature.
Also asked from this topic: the new resistance when a 4 Ω wire is stretched to double its length (16 Ω); why nichrome is used in heaters (high resistivity and high melting point).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 3 · Resistors in series and parallel

Q183. If two resistors in series are equal, say R each, what is the total resistance?

  1. (a) 2R
  2. (b) R/2
  3. (c) R
  4. (d) R^(2)
Answer & solution

Answer: (a) 2R

Answer: Option 1 — 2R
In series, resistances simply add: R_(s) = R_(1) + R_(2).
So two equal resistors give R + R = 2R.
Revision note — Series and parallel combinations
• Series: R_(s) = R_(1) + R_(2) + R_(3); same current through each; voltage divides in the ratio of resistances.
• n equal resistors R in series: R_(s) = nR.
• Parallel: 1/R_(p) = 1/R_(1) + 1/R_(2) + 1/R_(3); same voltage across each.
• n equal resistors R in parallel: R_(p) = R/n — two equal ones give R/2.
• Series maximum : parallel minimum for n equal resistors = nR : R/n = n² : 1.
• R_(s) is more than the largest resistor; R_(p) is less than the smallest.
Do not mix these up
• 2R vs R/2 — 2R is two equal resistors in series; R/2 is the same two in parallel.
Also asked from this topic: why resistors are joined in series (to increase total resistance and limit current); current when 3 Ω, 2 Ω and 1 Ω in series are connected to 9 V (1.5 A).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 3 · Electric power and electrical energy

Q184. Which factor does NOT affect the electric power consumed by an electric device?

  1. (a) Current passing through the device
  2. (b) Voltage applied to the device
  3. (c) Frequency of the device
  4. (d) Resistance of the device
Answer & solution

Answer: (c) Frequency of the device

Answer: Option 3 — Frequency of the device
Electric power is P = V × I = I²R = V²/R.
Only voltage, current and resistance appear, so frequency does not decide the power consumed.
Revision note — Electric power
• Power = rate of using electrical energy; SI unit watt (W) = 1 J/s = 1 V × 1 A.
• Three forms: P = VI, P = I²R (same current, as in series), P = V²/R (same voltage, as in parallel).
• Electrical energy = power × time; commercial unit 1 kWh = 1 unit = 3.6 × 10⁶ J.
• A device rated "100 W, 220 V" uses 100 J every second when run on 220 V.
• In household (parallel) wiring every appliance gets 220 V, so a device with lower resistance draws more power.
Do not mix these up
• Power vs energy — power is in watt; energy is in joule or kWh (the "unit" on the electricity bill).
Also asked from this topic: how to cut the power of a device at a fixed current (decrease its resistance, since P = I²R); the formula for electrical energy (Energy = Power × Time).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 3 · Resistance, resistivity and their factors

Q185. Which of the following factors directly affects the resistance of a conductor?

  1. (a) Current through the conductor
  2. (b) Length of the conductor
  3. (c) Power of the conductor
  4. (d) Voltage across the conductor
Answer & solution

Answer: (b) Length of the conductor

Answer: Option 2 — Length of the conductor
Resistance is directly proportional to length: R = ρ × L ÷ A.
A longer wire gives the electrons a longer path, so it resists more.
Revision note — Factors that affect resistance
• R ∝ L — doubling the length doubles the resistance.
• R ∝ 1/A — a thicker wire (larger area of cross-section) has less resistance.
• Material — through its resistivity ρ; SI unit of ρ is ohm metre (Ω m).
• Temperature — the resistance of a metal increases as it gets hotter.
• Ohm's law: V = IR; unit of resistance is the ohm (Ω) = 1 volt per ampere.
• Stretching a wire to double its length halves its area, so its resistance becomes 4 times.
Do not mix these up
• Resistance vs V and I — changing the voltage changes the current, but their ratio R = V/I stays the same; V and I do not decide R.
Also asked from this topic: the new resistance when a 4 Ω wire is stretched to double length (16 Ω); the effect of a rise in temperature on a metal's resistance (it increases).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 3 · Resistance, resistivity and their factors

Q186. Resistance is the property of a material that _______.

  1. (a) converts electrical energy to mechanical energy
  2. (b) allows the flow of current
  3. (c) generates electricity
  4. (d) opposes the flow of current
Answer & solution

Answer: (d) opposes the flow of current

Answer: Option 4 — opposes the flow of current
Resistance is the property of a conductor that opposes the flow of electric current through it.
By Ohm's law, R = V / I; its SI unit is the ohm (Ω).
Revision note — Resistance
• 1 ohm — resistance when 1 V across a conductor drives 1 A through it.
• Resistance depends on length (R ∝ l), area of cross-section (R ∝ 1/A), material and temperature.
• R = ρ l / A; ρ (resistivity) is a property of the material, SI unit Ω m.
• Conductors (copper, silver, aluminium) have low resistivity; alloys such as nichrome have high resistivity.
• Resistance of a metal increases with temperature.
• Series: R = R_(1) + R_(2) + …; parallel: 1/R = 1/R_(1) + 1/R_(2) + …
Also asked from this topic: SI unit of resistance (ohm); resistance of a wire stretched to double its length (4 times); property of good conductors (low resistivity).
RRB JE CBT-1 2024-25 – 16 December 2024 Shift 3 · Ohm's law

Q187. The current through a resistor is:

  1. (a) proportional to the square root of the resistance
  2. (b) directly proportional to the resistance
  3. (c) inversely proportional to the resistance
  4. (d) proportional to the square of the resistance
Answer & solution

Answer: (c) inversely proportional to the resistance

Answer: Option 3 — inversely proportional to the resistance
By Ohm's law, V = IR, so I = V/R.
For a fixed potential difference, a larger resistance lets less current flow — I ∝ 1/R.
Revision note — Ohm's law and resistance
• Ohm's law (Georg Simon Ohm): at constant temperature, current through a conductor is directly proportional to the potential difference across it — V ∝ I.
• V = IR; the constant R is the resistance, SI unit ohm (Ω) = 1 V/A.
• The V–I graph of an ohmic conductor is a straight line through the origin; its slope gives R.
• Resistance depends on length (R ∝ l), area of cross-section (R ∝ 1/A), material (resistivity ρ) and temperature: R = ρl/A.
• Doubling R at constant V halves I; doubling V at constant R doubles I.
• Current is measured by an ammeter (in series); potential difference by a voltmeter (in parallel).
Do not mix these up
• Resistance vs resistivity — resistance depends on the size and shape of the wire; resistivity (Ω m) depends only on the material and temperature.
Also asked from this topic: what happens to the current when resistance is doubled at constant voltage (it halves); the unit volt per ampere (ohm).
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