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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.

📚 187 questionsSSC 17 · Railway 170
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SSC Stenographer 2026 – 10 September 2026 Shift 3 · Ohm's law

Q1. Which law quantifies the relationship between resistance, voltage, and current?

  1. (a) Boyle's Law
  2. (b) Ohm's Law
  3. (c) Hooke's Law
  4. (d) Newton's Third Law
Answer & solution

Answer: (b) Ohm's Law

Answer: Option 2 — Ohm's Law
Ohm's law states that at constant temperature the current through a conductor is directly proportional to the potential difference across its ends.
That proportionality is written V = I × R, which is exactly the link between voltage, current and resistance.
Revision note — Ohm's law and resistance
• Statement: under constant physical conditions V ∝ I, so V = IR, where R is the resistance measured in ohm (Ω).
• One ohm is one volt per ampere; the law is named after Georg Simon Ohm, who published it in 1827.
• R = ρL/A — resistance grows with length and falls with area of cross-section; ρ is resistivity, measured in ohm-metre.
• Stretching a wire so that its length doubles makes its resistance four times, because the volume stays the same and the area halves.
• Series: R = R₁ + R₂ + … ; Parallel: 1/R = 1/R₁ + 1/R₂ + … , so the parallel resistance is always less than the smallest branch.
• Electric power P = VI = I²R = V²/R; a 100 W bulb working at 200 V has a filament resistance of 400 Ω.
• Boyle's law — at constant temperature P ∝ 1/V for a fixed mass of gas; Hooke's law — within the elastic limit stress ∝ strain.
• Newton's third law — to every action there is an equal and opposite reaction.
Do not mix these up
• Resistance vs resistivity — resistance depends on the size and shape of the sample; resistivity is a property of the material alone.
• Ohm's law vs Kirchhoff's laws — Ohm's law links V, I and R in a single conductor; Kirchhoff's junction and loop rules handle a network.
• Ohmic vs non-ohmic conductors — metals obey Ohm's law; a diode or a filament lamp does not.
Also asked from this topic: the effect on current of doubling the resistance at constant voltage (it halves); the quantity measured in volt per ampere (ohm).
RRB Group D 2025-26 – 25 August 2026 Shift 1 · Electric cells and batteries

Q2. What role does the chemical energy in a cell play in an electric circuit?

  1. (a) It changes the resistance of the circuit
  2. (b) It directly moves electrons along the wire
  3. (c) It creates and maintains the potential difference
  4. (d) It measures the flow of current
Answer & solution

Answer: (c) It creates and maintains the potential difference

Correct Answer:
Correct Answer: Option 3 – It creates and maintains the potential difference
Explanation:
A cell (battery) converts storedchemical energyinto electrical energy through internal chemical reactions. This process creates and maintains apotential difference (voltage)between its two terminals, which is what drives the flow of electrons (current) through the external circuit.
Important Facts:
• Potential difference is the 'push' that causes current to flow in a closed circuit.
• The cell does not directly move electrons through the wire; it is the electric field set up by the potential difference that does so.
• Current is measured by an ammeter, not created by chemical energy directly.
RRB Group D 2025-26 – 21 August 2026 Shift 1 · Resistance, resistivity and their factors

Q3. Which of the following materials has the highest resistivity?

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

Answer: (d) Manganin

Explanation:
Approximate resistivity values: Silver ≈ 1.6×10⁻⁸ Ωm, Tungsten ≈ 5.6×10⁻⁸ Ωm, Iron ≈ 10×10⁻⁸ Ωm, Manganin ≈ 44–48×10⁻⁸ Ωm. Manganin (an alloy of Cu, Mn, Ni) is deliberately made with a high, nearly temperature-independent resistivity, which is why it is used in standard resistance coils and resistance boxes.
Correct Answer: Option 4 – Manganin
RRB Group D 2025-26 – 20 August 2026 Shift 1 · Resistors in series and parallel

Q4. Two identical bulbs are connected in parallel to a battery. If one bulb burns out, what happens to the brightness of the other bulb?

  1. (a) The brightness of the other bulb decreases.
  2. (b) Both bulbs turn off.
  3. (c) The brightness of the other bulb increases.
  4. (d) The brightness of the other bulb remains unchanged.
Answer & solution

Answer: (d) The brightness of the other bulb remains unchanged.

Correct Answer:
Correct Answer: Option 4 – The brightness of the other bulb remains unchanged.
Explanation:
In aparallel circuit, each component is connected across the same voltage source independently, on its own separate branch. If one bulb burns out (its branch becomes open), the voltage across the remaining bulb and the current through it are unaffected, so its brightnessremains unchanged.
Important Facts:
• In a series circuit, if one bulb burns out, the circuit breaks and all bulbs go off.
• Household wiring uses parallel circuits so that appliances operate independently.
RRB Group D 2025-26 – 20 August 2026 Shift 1 · Heating effect of current and Joule's law

Q5. What is the main purpose of filling electric bulbs with nitrogen and argon gases?

  1. (a) To increase the brightness of the bulb
  2. (b) To cool the filament rapidly
  3. (c) To prolong the life of the filament
  4. (d) To make the bulb transparent
Answer & solution

Answer: (c) To prolong the life of the filament

Correct Answer:
Correct Answer: Option 3 – To prolong the life of the filament
Explanation:
Electric bulbs are filled with chemically inert gases likenitrogen and argonto prevent oxidation (burning) of the hot tungsten filament, therebyprolonging its life. These gases also help reduce evaporation of the filament material.
Important Facts:
• Tungsten is used as the filament due to its very high melting point (~3400°C).
• A vacuum was used in earlier bulbs, but inert gas filling further reduces filament evaporation.
RRB Group D 2025-26 – 20 August 2026 Shift 3 · Heating effect of current and Joule's law

Q6. If the resistance in a circuit is doubled and the current remains constant, what happens to the amount of heat produced in the resistor over the same time period?

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

Answer: (b) It doubles

Correct Answer:
Option 2 – It doubles
Explanation:
Heat produced in a resistor, by Joule's law, is H = I²Rt. Since current (I) and time (t) remain constant, H is directly proportional to R. If resistance Rdoubles, the heat produced alsodoubles.
RRB Group D 2025-26 – 19 August 2026 Shift 1 · Resistance, resistivity and their factors

Q7. If a copper wire and an iron wire of the same length and cross-sectional area are compared, which wire will have higher resistance?

  1. (a) Equal in both
  2. (b) Cannot be determined
  3. (c) Iron wire
  4. (d) Copper wire
Answer & solution

Answer: (c) Iron wire

Correct Answer:
Correct Answer: Option 3 – Iron wire
Explanation:
Resistance R = ρL/A, where ρ is resistivity. For the same length (L) and area (A), resistance depends only onresistivity.Iron has a much higher electrical resistivity (~10 × 10⁻⁸ Ωm) than copper (~1.7 × 10⁻⁸ Ωm), so theiron wire has higher resistance.
Important Facts:
• Copper is a better electrical conductor than iron, which is why copper is used in electrical wiring.
• Resistivity is a material property independent of the wire's dimensions.
RRB Group D 2025-26 – 19 August 2026 Shift 2 · Resistance, resistivity and their factors

Q8. A student uses two wires of the same material and length but different thickness in the same circuit. Which statement best explains why the thicker wire causes a higher ammeter reading?

  1. (a) The thicker wire has a larger area, resulting in higher resistance and lower current
  2. (b) The thicker wire has more resistivity, resulting in higher current
  3. (c) The thicker wire has a larger area, resulting in lower resistance and higher current
  4. (d) The thicker wire has a different material, resulting in higher current
Answer & solution

Answer: (c) The thicker wire has a larger area, resulting in lower resistance and higher current

Correct Answer:
Correct Answer: Option 3 – The thicker wire has a larger area, resulting in lower resistance and higher current
Explanation:
Resistance of a wire is given by R = ρL/A, where A is the cross-sectional area. For wires of the same material (same ρ) and length (same L), athicker wire has a larger cross-sectional area (A), so its resistance islower. By Ohm's Law (I = V/R), a lower resistance at the same voltage produces ahigher current, giving a higher ammeter reading.
Important Facts:
• Resistance is inversely proportional to cross-sectional area.
• Resistivity (ρ) depends only on the material, not thickness, so it stays the same for both wires here.
RRB Group D 2025-26 – 19 August 2026 Shift 3 · Numericals on electricity

Q9. A heater is rated at 1000 W, 250 V. If it is connected to a 250 V supply, what will be the current drawn by the heater and what is the resistance of the heater?

  1. (a) 5.0 A, 50 Ω
  2. (b) 3.5 A, 71.4 Ω
  3. (c) 4 A, 62.5 Ω
  4. (d) 2.5 A, 100 Ω
Answer & solution

Answer: (c) 4 A, 62.5 Ω

Correct Answer:
Option 3 – 4 A, 62.5 Ω
Explanation:
Since the heater is connected to its rated voltage (250 V), it draws its rated power (1000 W). Current: I = P/V = 1000/250 =4 A. Resistance: R = V/I = 250/4 =62.5 Ω.
Important Facts:
• Power formula: P = VI = I²R = V²/R.
• Resistance can also be found directly as R = V²/P = 250²/1000 = 62.5 Ω.
Correct Answer: Option 3 – 4 A, 62.5 Ω
RRB Group D 2025-26 – 18 August 2026 Shift 1 · Resistance, resistivity and their factors

Q10. When the length of a wire is doubled while keeping material and area of cross-section constant, what happens to its resistance?

  1. (a) The resistance becomes half
  2. (b) The resistance becomes twice as much
  3. (c) The resistance becomes four times as much
  4. (d) The resistance remains unchanged
Answer & solution

Answer: (b) The resistance becomes twice as much

Resistance and wire length
Resistance of a wire is given byR = ρL/A, where ρ is resistivity, L is length, and A is the cross-sectional area.
Since R is directly proportional to L, doubling the length (keeping material and area constant) doubles the resistance.
Correct Answer: The resistance becomes twice as much
RRB Group D 2025-26 – 18 August 2026 Shift 3 · Ohm's law

Q11. In a laboratory, a student wants to verify Ohm's Law for a new material. Which experimental approach will most accurately determine if the material obeys Ohm's Law?

  1. (a) Measure current for several applied voltages and plot voltage versus current
  2. (b) Observe the heat produced at a single applied voltage
  3. (c) Measure resistance at one voltage and compare to tabulated value
  4. (d) Calculate voltage for a fixed current
Answer & solution

Answer: (a) Measure current for several applied voltages and plot voltage versus current

Concept: Verifying Ohm's Law Experimentally
Ohm's Law states that V = IR (voltage is directly proportional to current, at constant temperature) for an ohmic conductor. To experimentally verify this for a new material, one must vary the applied voltage across several values, measure the corresponding current for each, and then plot aV vs I graph. If the material obeys Ohm's law, the graph will be a straight line passing through the origin (constant slope = resistance). A single-point measurement (options 2, 3, 4) cannot establish a proportional/linear relationship — multiple data points are required.
Correct Answer: Measure current for several applied voltages and plot voltage versus current
RRB Group D 2025-26 – 17 August 2026 Shift 1 · Potential difference and EMF

Q12. The potential difference between two points in a circuit is 1 volt if:

  1. (a) 1 ampere current flows for 1 second
  2. (b) 1 coulomb of charge flows through a resistance of 1 ohm
  3. (c) 1 joule of work is done in moving 1 coulomb of charge
  4. (d) 1 joule of energy is consumed in 1 second
Answer & solution

Answer: (c) 1 joule of work is done in moving 1 coulomb of charge

Definition of Potential Difference (Volt)
Potential difference (V) between two points is defined as the work done (W) in moving a unit positive charge (Q) from one point to the other:V = W/Q.
The potential difference is1 voltwhen1 joule of work is done in moving 1 coulomb of chargebetween the two points (1 V = 1 J/C).
Correct Answer: 1 joule of work is done in moving 1 coulomb of charge
RRB Group D 2025-26 – 17 August 2026 Shift 2 · Resistance, resistivity and their factors

Q13. The resistivity of a material is defined as the resistance offered by a conductor of that material having:

  1. (a) Double the length and unit cross-sectional area
  2. (b) Half the length and double the cross-sectional area
  3. (c) Unit length and unit cross-sectional area
  4. (d) Unit length and half the cross-sectional area
Answer & solution

Answer: (c) Unit length and unit cross-sectional area

Definition of Resistivity
Resistance of a conductor is given by R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. Resistivity (ρ) is therefore numerically equal to the resistance of a conductor of that material havingunit length and unit cross-sectional area(ρ = R when L = 1 and A = 1).
Correct Answer: Unit length and unit cross-sectional area
RRB Group D 2025-26 – 17 August 2026 Shift 3 · Resistors in series and parallel

Q14. Which of the following properties is unique to the parallel combination of resistors compared to the series combination?

  1. (a) Total resistance equals sum of resistors
  2. (b) Current through each resistor is the same
  3. (c) Each resistor receives the same voltage
  4. (d) All resistors must be identical
Answer & solution

Answer: (c) Each resistor receives the same voltage

Series vs Parallel Combination of Resistors
In a parallel combination, all resistors are connected across the same two points, so:
• Each resistor gets thesame voltage(equal to the source voltage), while the current divides among them based on individual resistance — the opposite of what happens in series.
• In series, current is the same through all resistors and voltage divides, and total resistance in series equals the sum (not parallel).
Correct Answer: Each resistor receives the same voltage
RRB Group D 2025-26 – 17 August 2026 Shift 3 · Electric cells and batteries

Q15. Which component is best represented by a symbol with multiple alternating long and short lines in a circuit diagram?

  1. (a) A resistor of resistance R
  2. (b) An electric cell
  3. (c) Electric bulb
  4. (d) A battery or a combination of cells
Answer & solution

Answer: (d) A battery or a combination of cells

Circuit Symbol Identification
In standard circuit diagrams, a single electric cell is represented by one long line (positive terminal) and one short line (negative terminal).
• Whenmultiple alternating long and short linesappear together, this represents abattery— a combination/series of multiple cells.
• Resistors and bulbs have different distinct symbols (zig-zag line, circle with X, respectively).
Correct Answer: A battery or a combination of cells
RRB Group D 2025-26 – 14 August 2026 Shift 1 · Ohm's law

Q16. Which of the following graphs best represents Ohm’s Law for a resistor at constant temperature?

  1. (a) A straight line graph passing through the origin when current is plotted against voltage
  2. (b) A straight line graph not passing through the origin when current is plotted against voltage
  3. (c) A horizontal line graph when current is plotted against voltage
  4. (d) A curved line graph passing through the origin when current is plotted against voltage
Answer & solution

Answer: (a) A straight line graph passing through the origin when current is plotted against voltage

Answer: Option 1 – A straight line through the origin (current vs voltage).
• Ohm's law (V = IR) gives astraight line through the originfor I vs V at constant temperature.
RRB Group D 2025-26 – 14 August 2026 Shift 3 · Numericals on electricity

Q17. In a circuit, two resistors R₁ and R₂ are connected in parallel. If R₁ = 4 Ω and R₂ = 12 Ω, and the total voltage across the parallel combination is 6V, what is the total current flowing in the circuit?

  1. (a) 2 A
  2. (b) 10 A
  3. (c) 1.5 A
  4. (d) 8 A
Answer & solution

Answer: (a) 2 A

Answer: Option 1 – 2 A.
• Parallel: R = (4×12)/(4+12) = 3 Ω. I = V/R = 6/3 =2 A.
RRB Group D 2025-26 – 14 August 2026 Shift 3 · Heating effect of current and Joule's law

Q18. If a resistor is connected across a battery and used for a longer time period, which of the following will increase, assuming current and resistance remain constant?

  1. (a) The total heat produced in the resistor
  2. (b) The resistance of the resistor
  3. (c) The charge per unit time
  4. (d) The potential difference across the resistor
Answer & solution

Answer: (a) The total heat produced in the resistor

Answer: Option 1 – The total heat produced in the resistor.
• Heat H = I²Rt. With I and R constant, more time ⇒ moretotal heat.
RRB Group D 2025-26 – 13 August 2026 Shift 1 · Numericals on electricity

Q19. If the power consumed by a bulb is 60 W and the voltage across it is 120 V, what is the resistance of the bulb?

  1. (a) 2 Ω
  2. (b) 240 Ω
  3. (c) 60 Ω
  4. (d) 0.5 Ω
Answer & solution

Answer: (b) 240 Ω

Answer: Option 2 – 240 Ω.
• R = V²/P = 120²/60 = 14400/60 =240 Ω.
RRB Group D 2025-26 – 13 August 2026 Shift 1 · Numericals on electricity

Q20. A student constructs a circuit with a battery of 12 V and two resistors of 6 ohms and 4 ohms connected in series. What is the total current flowing through the circuit according to Ohm's Law?

  1. (a) 0.6 A
  2. (b) 1.0 A
  3. (c) 1.2 A
  4. (d) 2.0 A
Answer & solution

Answer: (c) 1.2 A

Answer: Option 3 – 1.2 A.
• Series resistance = 6 + 4 = 10 Ω. I = V/R = 12/10 =1.2 A.
RRB Group D 2025-26 – 13 August 2026 Shift 2 · Resistors in series and parallel

Q21. A technician needs to design a circuit where removing one device does not interrupt the operation of others. Which resistor arrangement should the technician choose?

  1. (a) The technician should use a series combination of resistors.
  2. (b) The technician should connect resistors in a random arrangement.
  3. (c) The technician should use a series-parallel mixed combination of resistors.
  4. (d) The technician should use a parallel combination of resistors.
Answer & solution

Answer: (d) The technician should use a parallel combination of resistors.

Answer: Option 4 – Parallel combination of resistors.
• In aparallelcircuit, removing one device doesn't break the others' paths.
RRB Group D 2025-26 – 12 August 2026 Shift 1 · Electric power and electrical energy

Q22. A student wants to reduce the power consumed by a device without changing the current. Which action will achieve this?

  1. (a) Decrease the resistance in the circuit
  2. (b) Increase the resistance in the circuit
  3. (c) increase the temperature in the circuit
  4. (d) Increase the voltage applied
Answer & solution

Answer: (a) Decrease the resistance in the circuit

Answer: Option 1 – Decrease the resistance in the circuit.
• Power P = I²R. With current fixed, loweringRlowers the power consumed.
RRB Group D 2025-26 – 12 August 2026 Shift 2 · Numericals on electricity

Q23. An electric kettle has a resistance of 10 ohms and is connected to a 220 V supply. For 5 minutes of operation, how much charge flows through it?

  1. (a) 6600 C
  2. (b) 1100 C
  3. (c) 13200 C
  4. (d) 2200 C
Answer & solution

Answer: (a) 6600 C

Answer: Option 1 – 6600 C.
• I = V/R = 220/10 = 22 A; t = 5 min = 300 s. Charge Q = I×t = 22×300 =6600 C.
RRB Group D 2025-26 – 12 August 2026 Shift 3 · Ohm's law

Q24. What will happen to the resistance of a wire if both the current and voltage are increased by the same factor?

  1. (a) The resistance will depend on the factor increase.
  2. (b) The resistance will remain unchanged.
  3. (c) The resistance will increase.
  4. (d) The resistance will decrease.
Answer & solution

Answer: (b) The resistance will remain unchanged.

Answer: Option 2 – The resistance will remain unchanged.
• R = V/I. Increasing V and I by the same factor keeps the ratio (resistance) constant.
RRB Group D 2025-26 – 11 August 2026 Shift 1 · Heating effect of current and Joule's law

Q25. If the resistance of a wire is doubled and same current flows through it for same time, heat produced will:

  1. (a) Remain same
  2. (b) Become half
  3. (c) Become double
  4. (d) Become four times
Answer & solution

Answer: (c) Become double

Answer: Option 3 – Become double.
• Heat H = I²Rt. With I and t constant, H ∝ R. Doubling Rdoublesthe heat produced.
RRB Group D 2025-26 – 11 August 2026 Shift 2 · Numericals on electricity

Q26. A wire of resistance 10 ohms is connected to a 12 V battery. If another identical wire is connected in parallel to the first, what is the total power consumed by both wires together?

  1. (a) 14.4 W
  2. (b) 1.44 W
  3. (c) 28.8 W
  4. (d) 120 W
Answer & solution

Answer: (c) 28.8 W

Answer: Option 3 – 28.8 W.
• Two 10 Ω wires in parallel ⇒ R = 5 Ω. Power P = V²/R = 12²/5 = 144/5 =28.8 W.
RRB Group D 2025-26 – 11 August 2026 Shift 3 · Electric cells and batteries

Q27. In an electric circuit, what produces the potential difference needed for charge flow?

  1. (a) The gravitational pull on electrons
  2. (b) Connecting wires alone between two points
  3. (c) The resistance of the wire to current
  4. (d) A battery consisting of one or more electric cells
Answer & solution

Answer: (d) A battery consisting of one or more electric cells

Answer: Option 4 – A battery consisting of one or more electric cells.
• Abattery/cellprovides the potential difference that drives charge through a circuit.
RRB Group D 2025-26 – 10 August 2026 Shift 1 · Numericals on electricity

Q28. A heater is rated at 1000 W. If it is used for 2 hours, how much energy does it consume in kilowatt hours?

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

Answer: (b) 2 kWh

Answer: Option 2 — 2 kWh
Energy = 1 kW × 2 h =2 kWh→ Option 2.
RRB Group D 2025-26 – 10 August 2026 Shift 1 · Conductors, insulators and semiconductors

Q29. Why is copper commonly used for electrical wiring in homes?

  1. (a) It is magnetic
  2. (b) It is a poor conductor
  3. (c) It has a low melting point
  4. (d) It has low electrical resistance
Answer & solution

Answer: (d) It has low electrical resistance

Answer: Option 4 — It has low electrical resistance
Copper is used for wiring because of itslow electrical resistance(excellent conductor) → Option 4.
RRB Group D 2025-26 – 10 August 2026 Shift 2 · Electric current and its unit

Q30. Which combination of symbols would you use to indicate a wire joint in a schematic circuit diagram?

  1. (a) Two parallel lines
  2. (b) A rectangle
  3. (c) A dot at the intersection of two lines
  4. (d) Lines crossing without a dot
Answer & solution

Answer: (c) A dot at the intersection of two lines

Answer: Option 3 – A dot at the intersection of two lines.
• In circuit diagrams, asolid dotat a crossing shows the wires arejoined(connected).
• Lines crossingwithouta dot mean the wires are not connected.
RRB Group D 2025-26 – 10 August 2026 Shift 2 · Resistors in series and parallel

Q31. Which of the following is a practical reason for using a series combination of resistors in an electrical circuit?

  1. (a) To maintain the same voltage across each resistor
  2. (b) To increase the current in the circuit
  3. (c) To increase total resistance and limit current
  4. (d) To decrease voltage across each resistor
Answer & solution

Answer: (c) To increase total resistance and limit current

Answer: Option 3 – To increase total resistance and limit current.
• In aseriescombination, resistances add up (R = R₁+R₂+…), giving higher total resistance, whichlimits the current– useful for protecting components.
RRB Group D 2025-26 – 9 August 2026 Shift 1 · Heating effect of current and Joule's law

Q32. A student wants to design a device that maximizes heat production for a given battery and time. Which configuration should the student choose?

  1. (a) Use a resistor with the highest possible resistance value
  2. (b) Use a resistor with very low resistance value
  3. (c) Short-circuit the battery with zero resistance
  4. (d) Add more batteries without changing resistance
Answer & solution

Answer: (b) Use a resistor with very low resistance value

Answer: Option 2 — Use a resistor with very low resistance value
For a given battery and time, H = V²t/R, solower R gives more heat→ Option 2.
RRB Group D 2025-26 – 9 August 2026 Shift 1 · Resistors in series and parallel

Q33. If two resistors of 4 Ω and 6 Ω are connected in parallel, the equivalent resistance will be:

  1. (a) Between 4 Ω and 6 Ω
  2. (b) Less than 4 Ω
  3. (c) Greater than 6 Ω
  4. (d) Equal to 10 Ω
Answer & solution

Answer: (b) Less than 4 Ω

Answer: Option 2 — Less than 4 Ω
Parallel resistance is alwaysless than the smallestresistor (here < 4 Ω). (Actual = 2.4 Ω) → Option 2.
RRB Group D 2025-26 – 9 August 2026 Shift 2 · Numericals on electricity

Q34. A heater is rated at 220 V and 1000 W. If it is connected to a 110 V supply, what will be the power consumed by the heater?

  1. (a) 200 W
  2. (b) 1000 W
  3. (c) 250 W
  4. (d) 500 W
Answer & solution

Answer: (c) 250 W

Answer: Option 3 — 250 W
R = V²/P = 220²/1000 = 48.4 Ω. At 110 V: P = 110²/48.4 =250 W→ Option 3.
RRB Group D 2025-26 – 9 August 2026 Shift 3 · Numericals on electricity

Q35. If a bulb filament draws 0.5 A of current for 10 minutes, what is the total charge passed through the circuit?

  1. (a) 60 coulombs
  2. (b) 0.05 coulombs
  3. (c) 300 coulombs
  4. (d) 30 coulombs
Answer & solution

Answer: (c) 300 coulombs

Answer: Option 3 — 300 coulombs
Q = I × t = 0.5 × (10 × 60) = 0.5 × 600 =300 C→ Option 3.
RRB Group D 2025-26 – 9 August 2026 Shift 3 · Resistance, resistivity and their factors

Q36. Two uniform wires made of the same material, having the same length, and maintained at the same temperature are compared. Which factor will cause one wire to have greater electrical resistance than the other?

  1. (a) Larger electric current
  2. (b) Higher voltage
  3. (c) Smaller cross-sectional area
  4. (d) Higher cross-sectional area
Answer & solution

Answer: (c) Smaller cross-sectional area

Answer: Option 3 — Smaller cross-sectional area
R = ρL/A. For same material/length, asmaller cross-sectional areagives greater resistance → Option 3.
RRB Group D 2025-26 – 6 August 2026 Shift 1 · Heating effect of current and Joule's law

Q37. A manufacturer wants to design heating elements that generate more heat when a current passes through. Which material property should be maximized for this purpose?

  1. (a) Conductivity should be maximized to increase heat generation for a given current.
  2. (b) Resistivity should be maximized to increase heat generation for a given current.
  3. (c) A material with zero resistivity should be used for maximum heat generation.
  4. (d) Resistivity should be minimized to increase heat generation for a given current.
Answer & solution

Answer: (b) Resistivity should be maximized to increase heat generation for a given current.

Answer: Option 2 — Resistivity should be maximized
Heat H = I²R, so ahigh-resistivitymaterial generates more heat for a given current → Option 2.
RRB Group D 2025-26 – 6 August 2026 Shift 1 · Resistors in series and parallel

Q38. A circuit consists of three resistors of 4 ohms, 6 ohms, and 12 ohms connected in parallel. Which statement is correct about the total resistance?

  1. (a) The total resistance is equal to the largest individual resistance
  2. (b) The total resistance is less than the smallest individual resistance
  3. (c) The total resistance equals the sum of all resistances
  4. (d) The total resistance is equal to one of the individual resistances
Answer & solution

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

Answer: Option 2 — The total resistance is less than the smallest individual resistance
In parallel, total resistance is alwaysless than the smallestresistor → Option 2.
RRB Group D 2025-26 – 6 August 2026 Shift 2 · Resistance, resistivity and their factors

Q39. A student sets up two circuits: Circuit X uses a thin, long copper wire, and Circuit Y uses a thick, short copper wire, both connected to identical batteries. Which circuit will have a higher resistance, and why?

  1. (a) Circuit Y because the resistance decreases with length and increases with cross-sectional area
  2. (b) Circuit X because the resistance decreases with length and increases with cross-sectional area
  3. (c) Circuit X because the resistance increases with length and decreases with cross-sectional area
  4. (d) Circuit Y because the resistance increases with length and decreases with cross-sectional area
Answer & solution

Answer: (c) Circuit X because the resistance increases with length and decreases with cross-sectional area

Answer: Option 3 — Circuit X (resistance ↑ with length, ↓ with area)
R = ρL/A. The thin, long wire (Circuit X) hashigher resistance→ Option 3.
RRB Group D 2025-26 – 6 August 2026 Shift 3 · Resistance, resistivity and their factors

Q40. A copper wire and an iron wire have the same length and thickness. Which wire will have lower resistance and why, when both are kept at the same temperature?

  1. (a) Both wires, because material does not affect resistance
  2. (b) The copper wire, because copper is a better conductor
  3. (c) The iron wire, because iron is a better conductor
  4. (d) Both wires, because thickness is the only factor
Answer & solution

Answer: (b) The copper wire, because copper is a better conductor

Answer: Option 2 — Copper wire, because copper is a better conductor
Copper has lower resistivity than iron, so thecopper wire has lower resistance→ Option 2.
RRB Group D 2025-26 – 6 August 2026 Shift 3 · Numericals on electricity

Q41. If a 9V battery is connected across three resistors of 3 ohms, 2 ohms, and 1 ohm in series, what is the current flowing through the circuit?

  1. (a) 1.5 A
  2. (b) 9 A
  3. (c) 3 A
  4. (d) 6 A
Answer & solution

Answer: (a) 1.5 A

Answer: Option 1 — 1.5 A
Series resistance = 3 + 2 + 1 = 6 Ω. I = V/R = 9/6 =1.5 A→ Option 1.
RRB Group D 2025-26 – 5 August 2026 Shift 1 · Resistance, resistivity and their factors

Q42. A student wants to compare the resistivity of two different materials using wires of identical length and cross-sectional area. Which measurement must be made for both wires to determine their resistivity?

  1. (a) Measuring the resistance of each wire
  2. (b) Measuring the brightness of a bulb in the circuit
  3. (c) Measuring the mass of each wire
  4. (d) Measuring the voltage across each wire
Answer & solution

Answer: (a) Measuring the resistance of each wire

Answer: Option 1 — Measuring the resistance of each wire
With equal length and area, resistivity ρ = R·A/L depends only on R, someasure the resistanceof each wire → Option 1.
RRB Group D 2025-26 – 5 August 2026 Shift 3 · Heating effect of current and Joule's law

Q43. Why is tungsten used as the filament material in electric bulbs?

  1. (a) It produces more heat than other metals at lower temperatures.
  2. (b) It reacts with nitrogen and argon to create a brighter light.
  3. (c) It has a high melting point and can withstand high temperatures without melting.
  4. (d) It is the cheapest metal available for manufacturing filaments.
Answer & solution

Answer: (c) It has a high melting point and can withstand high temperatures without melting.

Answer: Option 3 — High melting point, withstands high temperature without melting
Tungstenis chosen for filaments because of itsvery high melting point (~3400°C)→ Option 3.
RRB Group D 2025-26 – 5 August 2026 Shift 3 · Electric current and its unit

Q44. What is the conventional direction of electric current in a metallic circuit?

  1. (a) From the negative terminal to the positive terminal
  2. (b) From the positive terminal to the negative terminal
  3. (c) Randomly between any two points
  4. (d) In the direction of electron movement
Answer & solution

Answer: (b) From the positive terminal to the negative terminal

Answer: Option 2 — From positive to negative terminal
Conventional current flows frompositive to negative(opposite to electron flow) → Option 2.
RRB Group D 2025-26 – 4 August 2026 Shift 2 · Electric current and its unit

Q45. A science project requires a variable resistor to adjust the current in a circuit. Which symbol should be used in the schematic diagram to represent this component?

  1. (a) Parallel long and short lines
  2. (b) A simple zigzag line
  3. (c) A straight line with a gap and bridge
  4. (d) A zigzag line with an arrow across it
Answer & solution

Answer: (d) A zigzag line with an arrow across it

Answer: Option 4 — A zigzag line with an arrow across it
Avariable resistor (rheostat)is shown as a resistor (zigzag) with anarrowacross it → Option 4.
RRB Group D 2025-26 – 4 August 2026 Shift 2 · Heating effect of current and Joule's law

Q46. Why does a metal filament bulb glow when electric current passes through it?

  1. (a) The filament offers resistance, converting electrical energy to heat and light
  2. (b) A chemical reaction in the filament produces light
  3. (c) The filament is magnetized by the current
  4. (d) The filament conducts current with zero resistance
Answer & solution

Answer: (a) The filament offers resistance, converting electrical energy to heat and light

Answer: Option 1 — The filament offers resistance, converting electrical energy to heat and light
The high-resistance filament heats up (I²R) and glows, converting electrical energy to heat and light → Option 1.
RRB Group D 2025-26 – 3 August 2026 Shift 1 · Numericals on electricity

Q47. If a bulb is rated at 60 W and operates at 120 V, what is the current flowing through the bulb?

  1. (a) 2.0 A
  2. (b) 1.2 A
  3. (c) 0.25 A
  4. (d) 0.5 A
Answer & solution

Answer: (d) 0.5 A

Answer: Option 4 — 0.5 A
Power:P = VI ⟹ I = P/V.
I = 60 / 120 =0.5 A→ Option 4.
RRB Group D 2025-26 – 3 August 2026 Shift 1 · Resistance, resistivity and their factors

Q48. Why are alloys preferred for the heating elements in devices like toasters?

  1. (a) They have a low melting point
  2. (b) They are insulators
  3. (c) They are good electrical conductors with low resistivity
  4. (d) They have higher resistivity and do not oxidise readily at high temperatures
Answer & solution

Answer: (d) They have higher resistivity and do not oxidise readily at high temperatures

Answer: Option 4 — They have higher resistivity and do not oxidise readily at high temperatures
Concept:Heating elements (e.g., nichrome) are alloys because they havehigh resistivity(so they get hot when current flows), ahigh melting point, anddo not oxidise/burneven when red-hot → Option 4.
RRB Group D 2025-26 – 3 August 2026 Shift 2 · Heating effect of current and Joule's law

Q49. If a resistor is connected to a voltage source and the resistance is halved while the voltage remains constant, what happens to the heat produced in a fixed time?

  1. (a) The heat produced remains unchanged
  2. (b) The heat produced increases to twice its previous value
  3. (c) The heat produced increases to four times its previous value
  4. (d) The heat produced decreases to half its previous value
Answer & solution

Answer: (b) The heat produced increases to twice its previous value

Answer: Option 2 — Increases to twice its previous value
Heat H = V²t/R (constant V and t). If R ishalved, H = V²t/(R/2) =2 × (V²t/R)→ heatdoubles→ Option 2.
RRB Group D 2025-26 – 3 August 2026 Shift 3 · Conductors, insulators and semiconductors

Q50. Which material would be the best choice for electrical transmission lines?

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

Answer: (a) Aluminium

Answer: Option 1 — Aluminium
Transmission lines need agood conductor that is light and cheap:aluminium→ Option 1. (Glass/rubber are insulators; tungsten is used in bulb filaments.)
RRB Group D 2025-26 – 3 August 2026 Shift 3 · Numericals on electricity

Q51. An electric heater is rated 2000 W and is used for 2 hours. How much energy is consumed in joules?

  1. (a) 14,400,000 joules
  2. (b) 7,200,000 joules
  3. (c) 3,600,000 joules
  4. (d) 720,000 joules
Answer & solution

Answer: (a) 14,400,000 joules

Answer: Option 1 — 14,400,000 joules
Energy = Power × time = 2000 W × (2 × 3600 s) = 2000 × 7200 =1,44,00,000 J→ Option 1.
RRB NTPC UG CBT-1 2025-26 – 18 June 2026 Shift 1 · Resistance, resistivity and their factors

Q52. The quantity represented by the unit volt per ampere corresponds to which electrical unit?

  1. (a) Coulomb
  2. (b) Ohm
  3. (c) Joule
  4. (d) Watt
Answer & solution

Answer: (b) Ohm

Ohm's Law: V = I × R
Rearranging: R = V/I
Unit of Resistance (R) = Volt/Ampere
Volt/Ampere =Ohm (Ω)
Answer: Option 2 → Ohm
RRB NTPC UG CBT-1 2025-26 – 16 June 2026 Shift 2 · Heating effect of current and Joule's law

Q53. Heat generated in an electrical resistor is measured in which SI unit?

  1. (a) Volt
  2. (b) Joule
  3. (c) Coulomb
  4. (d) Watt
Answer & solution

Answer: (b) Joule

Topic: Heat and Electrical Energy
Heat is a form of energy
SI unit of energy =Joule (J)
Heat generated in resistor = I²Rt (in Joules)
Note: Watt is unit of power (energy/time)
Answer: Option 2 → Joule
RRB NTPC UG CBT-1 2025-26 – 16 June 2026 Shift 3 · Electric power and electrical energy

Q54. Electrical energy can be calculated using which of the following formulas?

  1. (a) Energy = Power × Time
  2. (b) Energy = Voltage × Current
  3. (c) Energy = Current × Resistance
  4. (d) Energy = Power ÷ Time
Answer & solution

Answer: (a) Energy = Power × Time

Electrical Energy Formula
Energy (E) =Power (P) × Time (t)
Unit: Watt-hour (Wh) or Joule (J)
Where Power = Voltage × Current
Other formulas:
Power = V × I (not energy)
Power = I² × R (not energy)
Answer: Option 1 → Energy = Power × Time
RRB NTPC UG CBT-1 2025-26 – 16 June 2026 Shift 3 · Electric charge and Coulomb's law

Q55. If you measure the amount of electric charge flowing in a circuit, which SI unit would you use to record the result?

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

Answer: (b) Coulomb

Topic: Electric Charge and SI Units
Electric Chargeis measured inCoulomb (C).
Key SI Units in Electricity:
•Coulomb (C)→ Amount of charge
•Ampere (A)→ Rate of flow of charge (current)
•Volt (V)→ Electric potential difference
•Ohm (Ω)→ Electrical resistance
1 Ampere = 1 Coulomb per second (1 A = 1 C/s)
Answer: Option 2 → Coulomb
RRB NTPC UG CBT-1 2025-26 – 13 June 2026 Shift 3 · Heating effect of current and Joule's law

Q56. What change occurs in the amount of heat produced when the electric current in a conductor is doubled?

  1. (a) Half
  2. (b) Same as before
  3. (c) Double
  4. (d) Four times
Answer & solution

Answer: (d) Four times

Joule's Law of Heating:
Heat produced:H = I²Rt
Where I = current, R = resistance, t = time
When current is doubled: I' = 2I
New heat: H' = (2I)²Rt = 4I²Rt = 4H
Heat becomesfour timesthe original value
Answer: Option 4 → Four times
SSC GD 2026 – 31 May 2026 Shift 3 · Resistance, resistivity and their factors

Q57. A wire of resistance 4 Ω is stretched until its length is doubled. What will be its new resistance?

  1. (a) 4 Ω
  2. (b) 8 Ω
  3. (c) 12 Ω
  4. (d) 16 Ω
Answer & solution

Answer: (d) 16 Ω

Correct Answer: 16 Ω (Option 4)
When a wire is stretched to double its length, the volume remains constant:
Original: Length = L, Area = A, Resistance R = ρL /A = 4 Ω
After stretching: New length = 2L
Since Volume is constant: A × L = A' × 2L ⟹ A' = A/2 (area halves)
New Resistance R' = ρ × (2L) / (A/2) = ρ × 2L × 2/A = 4ρL/A = 4R = 4 × 4 = 16 Ω
Key rule: When a wire is stretched to n times its original length, resistance becomes n² times the original.
Here n = 2, so R' = 2² × 4 = 4 × 4 = 16 Ω
Why others are wrong:
- 4 Ω: This would be the case if nothing changed.
- 8 Ω: This would result only if resistance doubled, not quadrupled.
- 12 Ω: Not a physically derived value for this scenario.
RRB NTPC UG CBT-1 2025-26 – 9 May 2026 Shift 2 · Conductors, insulators and semiconductors

Q58. Electrical conductors are materials that possess which of the following properties?

  1. (a) No resistivity
  2. (b) High resistivity
  3. (c) Low resistivity
  4. (d) Constant resistivity
Answer & solution

Answer: (c) Low resistivity

Electrical Conductors - Properties
Conductors allow electric current to flow easily through them.
Key Property:Low Resistivity (ρ)
Examples: Copper (ρ ≈ 1.7×10⁻⁸ Ωm), Silver, Aluminum
Low resistivity → Less opposition to current flow → Good conductor
Note: No material has zero resistivity (except superconductors at very low temperatures)
Answer: Option 3 → Low resistivity
SSC GD 2026 – 8 May 2026 Shift 2 · Heating effect of current and Joule's law

Q59. What is the primary reason for filling chemically inert gases
like nitrogen and argon in electric bulbs?

  1. (a) To make the bulb heavier
  2. (b) To increase brightness of light
  3. (c) To increase the life of the filament
  4. (d) To reduce production cost
Answer & solution

Answer: (c) To increase the life of the filament

Correct Answer: To increase the life of the filament — Option (c)
Chemically inert gases like nitrogen and argon are filled in electric bulbs primarily to increase the life of the tungsten filament. When the bulb is operational, the filament reaches very high temperatures (~2500°C). In vacuum, the tungsten evaporates rapidly. Inert gases like argon suppress the evaporation of tungsten from the filament, slow down the rate of filament deterioration, and thereby significantly increase the bulb's operational life. Argon is preferred as it reduces heat loss by convection less than nitrogen.
Option (a) To make the bulb heavier — Incorrect: The weight of the gas inside the bulb is negligible and has no practical purpose in making it heavier.
Option (b) To increase brightness of light — Incorrect: The inert gas does not increase brightness; the brightness depends on the filament's temperature and wattage.
Option (d) To reduce production cost — Incorrect: Using inert gases increases cost slightly; the purpose is to extend filament life, not reduce cost.
RRB NTPC UG CBT-1 2025-26 – 8 May 2026 Shift 2 · Heating effect of current and Joule's law

Q60. According to Joule’s law of heating, heat produced in a resistor is expressed by which relation?

  1. (a) H = I²Rt
  2. (b) H = Vt
  3. (c) H = V/I
  4. (d) H = IR
Answer & solution

Answer: (a) H = I²Rt

Joule's Law of Heating
When current flows through a resistor, electrical energy converts to heat energy.
Formula: H = I²Rt
Where:
H = Heat produced (Joules)
I = Current (Amperes)
R = Resistance (Ohms)
t = Time (seconds)
Heat produced is directly proportional to:
• Square of current (I²)
Answer: Option 1 → H = I²Rt
RRB NTPC GL CBT-1 2025-26 – 27 March 2026 Shift 2 · Heating effect of current and Joule's law

Q61. According to Joule’s law of heating, the heat produced in a resistor is _____________________.

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

Answer: (d) directly proportional to the square of current

Joule's Law of Heating
Formula: H = I²Rt
Where:
H = Heat produced
I = Current through resistor
R = Resistance
t = Time
From the formula, heat H is directly proportional to:
• Square of current (I²)
• Resistance (R)
• Time (t)
Answer: Option 4 → directly proportional to the square of current
RRB NTPC GL CBT-1 2025-26 – 27 March 2026 Shift 2 · Conductors, insulators and semiconductors

Q62. Which of the following materials is a good conductor of electricity?

  1. (a) Copper
  2. (b) Rubber
  3. (c) Wood
  4. (d) Glass
Answer & solution

Answer: (a) Copper

Topic: Electrical Conductivity
Conductorsallow electric current to flow easily due to free electrons.
Copperis a metal with abundant free electrons, making it an excellent conductor.
Rubber, Wood, and Glass areinsulators— they resist current flow.
Answer: Option 1 → Copper
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Resistance, resistivity and their factors

Q63. A device often used in electric circuits to change the resistance without changing the voltage source is the ___________.

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

Answer: (a) Rheostat

The device used to change resistance without altering the voltage source is called a Rheostat.
A Rheostat controls current by varying resistance in the circuit.
Voltmeters measure voltage and do not change resistance.
Fuses protect circuits by breaking the circuit when current is too high.
Ammeters measure current and do not adjust resistance.
Hence, the correct device is Rheostat.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Numericals on electricity

Q64. A wire has a resistance of 10 Ω. If it is stretched to three times its original length (volume remaining constant), its new resistance will be _______.

  1. (a) 90 Ω
  2. (b) 10 Ω
  3. (c) 30 Ω
  4. (d) 3.33 Ω
Answer & solution

Answer: (a) 90 Ω

Original resistance, R = 10 Ω
Length is stretched to 3 times, so new length, L2 = 3L1
Volume remains constant, so cross-sectional area changes accordingly
Volume formula
V = Area × Length
Initial volume
A1 × L1 = Final volume: A2 × L2
So, A2 = A1 × (L1 / L2) = A1 × (1/3)
Resistance R is given by R = ρ × (Length / Area)
New resistance R2 = ρ × (L2 / A2) = ρ × (3L1 / (A1 /3)) = ρ × (3L1 × 3 / A1) = 9 × (ρ × L1 / A1) = 9 × R
Hence, new resistance = 9 × 10 Ω = 90 Ω
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Numericals on electricity

Q65. How much work is required to move a charge of 3 C between two points that have a potential difference of 15 V?

  1. (a) 14 J
  2. (b) 45 J
  3. (c) 21 J
  4. (d) 5 J
Answer & solution

Answer: (b) 45 J

The work done to move a charge (Q) across a potential difference (V) is given by: Work = Q × V
Here, Q = 3 C, V = 15 V
Work = 3 × 15 = 45 J
Therefore, the work required is 45 joules.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 2 · Numericals on electricity

Q66. If a conductor has a potential difference of 1 V across its ends and a current of 1 A flows through it, what is the resistance of the conductor?

  1. (a) 2 Ω
  2. (b) 10 Ω
  3. (c) 0.5 Ω
  4. (d) 1 Ω
Answer & solution

Answer: (d) 1 Ω

Use Ohm's Law which states: V = IR
Given, V = 1 volt and I = 1 ampere
Calculate resistance, R = V / I
So, R = 1 V / 1 A = 1 ohm
Therefore, resistance of the conductor is 1 ohm.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Numericals on electricity

Q67. Two resistors of 5 Ω and 15 Ω are connected in series to a battery of 10 V. How much will flow through the circuit?

  1. (a) 0.4 A
  2. (b) 0.5 A
  3. (c) 2 A
  4. (d) 200 A
Answer & solution

Answer: (b) 0.5 A

The total resistance in series is the sum of resistances.
R_total = 5 Ω + 15 Ω = 20 Ω
Current I = Voltage / Total resistance
I = 10 V / 20 Ω = 0.5 A
The current flowing through the circuit is 0.5 A.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Electric power and electrical energy

Q68. Which of the following relations is incorrect for electric power (P)?

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

Answer: (b) P=R/V2

Option 2 is incorrect: P = R / V² is not a valid formula for electric power.
Correct formulas for electric power are:
P = VI (Power = Voltage × Current)
P = I²R (Power = Current squared × Resistance)
P = V²/R (Power = Voltage squared / Resistance)
Thus, the power related to resistance and voltage is P = V²/R, not P = R/V².
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Numericals on electricity

Q69. A 90 W electric bulb does how much work in 10 seconds?

  1. (a) 9000 J
  2. (b) 9 J
  3. (c) 900 J
  4. (d) 90 J
Answer & solution

Answer: (c) 900 J

Power (P) = 90 W means 90 joules of work done per second.
Work done (W) = Power × Time
Time = 10 seconds
So, Work done = 90 × 10 = 900 joules.
Therefore, a 90 W bulb does 900 joules of work in 10 seconds.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Numericals on electricity

Q70. A device has a resistance of 75 Ω and carries a current of 4 A. The power dissipated is _______.

  1. (a) 18.75 W
  2. (b) 300 W
  3. (c) 1400 W
  4. (d) 1200 W
Answer & solution

Answer: (d) 1200 W

Correct option: 2
Power dissipated in a device (P) is given by the formula: P = I² × R
Given, current I = 4 A, resistance R = 75 Ω
Calculate power: P = 4² × 75 = 16 × 75 = 1200 W
Check calculation again: yes, 1200 W is correct
Among the given options, 1200 W is option 4, not option 2
Re-evaluated carefully: 4² = 16
16 × 75 = 1200 W
Correct power dissipated is 1200 W .
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Numericals on electricity

Q71. If 12 J of work is done to move a charge of 2 C between two points, what is the potential difference between these points?

  1. (a) 24 V
  2. (b) 14 V
  3. (c) 10 V
  4. (d) 6 V
Answer & solution

Answer: (d) 6 V

Work done (W) = 12 J
Charge (Q) = 2 C
Potential difference (V) is given by V = W / Q
Substitute the values: V = 12 / 2 = 6
Check options carefully and re-calculate if necessary:
Since V = 6 volts, correct option is 4
Re-checking calculations shows V = 6 V is correct
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Numericals on electricity

Q72. Six resistors of 3 Ω each are connected in parallel in a circuit. What will be their effective resistance?

  1. (a) 2 Ω
  2. (b) 18 Ω
  3. (c) 3 Ω
  4. (d) 0.5 Ω
Answer & solution

Answer: (d) 0.5 Ω

Six resistors each of 3 Ω are connected in parallel.
The effective resistance R in parallel is given by:
1/R = 1/R1 + 1/R2 + ... + 1/Rn
Here, all resistors have equal resistance (3 Ω), so:
1/R = 6/3 = 2
R = 1/2 = 0.5 Ω
Therefore, the effective resistance is 0.5 Ω.
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 2 · Numericals on electricity

Q73. An electric oven has a power of 1200 W when it is operated at 100 V. How much will be the current flowing through the oven?

  1. (a) 0.08 A
  2. (b) 12 A
  3. (c) 8 A
  4. (d) 1.2 A
Answer & solution

Answer: (b) 12 A

Power (P) = 1200 W
Voltage (V) = 100 V
Current (I) = Power / Voltage
I = 1200 / 100
I = 12 A
So, the current flowing through the oven is 12 amperes.
Final answer: Option 2
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Resistance, resistivity and their factors

Q74. The resistivity of an alloy is generally higher than its constituent metals, and this property makes alloys suitable for ___________.

  1. (a) Connecting wires (e.g., copper transmission lines)
  2. (b) Heating elements (e.g., nichrome)
  3. (c) Circuit fuses (e.g., lead/tin alloy)
  4. (d) Filaments of electric bulbs (e.g., tungsten)
Answer & solution

Answer: (b) Heating elements (e.g., nichrome)

The resistivity of an alloy is generally higher than that of its constituent pure metals.
Higher resistivity means the alloy resists electric current more than pure metals.
This property makes alloys ideal for use in heating elements, where controlled resistance is needed to generate heat.
Nichrome, an alloy, is commonly used for heating elements due to its high resistivity and durability.
Therefore, alloys are suitable for heating elements rather than connecting wires or filaments, which require low resistivity.
Hence, the correct option is heating elements (e.g., nichrome).
Final answer: Option 2
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 2 · Electric circuits and circuit symbols

Q75. Which of the following is represented by a circle with the letter 'V' in a circuit diagram?

  1. (a) Voltmeter
  2. (b) Variable resistance
  3. (c) Voltage regulator
  4. (d) Vacuum diode
Answer & solution

Answer: (a) Voltmeter

The letter 'V' inside a circle in a circuit diagram stands for Voltmeter.
A voltmeter is an instrument used to measure voltage across two points in an electrical circuit.
This symbol is standard in circuit diagrams to indicate where voltage measurement is taken.
Therefore, the correct representation is Voltmeter.
Final answer: Option 1
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 3 · Heating effect of current and Joule's law

Q76. Why is tungsten used almost exclusively for making filaments of electric bulbs?

  1. (a) It readily oxidizes at high temperatures.
  2. (b) It has low electrical resistance.
  3. (c) It has a very high melting point.
  4. (d) It does not allow heat to pass.
Answer & solution

Answer: (c) It has a very high melting point.

The primary reason tungsten is used for electric bulb filaments is its extremely high melting point.
This allows the filament to be heated to very high temperatures without melting.
This property makes tungsten ideal for producing light when electric current passes through the filament.
Other metals would melt or evaporate quickly at such high temperatures.
Hence, tungsten's high melting point ensures durability and efficiency of the bulb filament.
Final answer: Option 3
RRB JE CBT-1 2025-26 – 25 February 2026 Shift 3 · Numericals on electricity

Q77. If the current in a wire is 2 amperes, how much charge flows through a cross-section in 5 seconds?

  1. (a) 7 coulombs
  2. (b) 10 coulombs
  3. (c) 2.5 coulombs
  4. (d) 0.4 coulombs
Answer & solution

Answer: (b) 10 coulombs

Current (I) is the rate of flow of charge (Q) through a conductor.
The relation is Q = I × t where,
I = 2 amperes (current)
t = 5 seconds (time)
Calculate charge: Q = 2 × 5
Q = 10 coulombs
So, 10 coulombs of charge flows in 5 seconds.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 2 · Numericals on electricity

Q78. A device operates at 2 amperes current and 220 volts potential difference for 5 hours. How much electrical energy does it consume, expressed in kilowatt hours (kWh)?

  1. (a) 1.1 kWh
  2. (b) 0.22 kWh
  3. (c) 5 kWh
  4. (d) 2.2 kWh
Answer & solution

Answer: (d) 2.2 kWh

Power (P) = Voltage (V) × Current (I)
P = 220 V × 2 A = 440 watts = 0.44 kW
Energy consumed (E) = Power × Time
E = 0.44 kW × 5 hours = 2.2 kWh
The device consumes 2.2 kilowatt hours of electrical energy.
Final answer: Option 4
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 2 · Heating effect of current and Joule's law

Q79. A room heater and an electric bulb both convert electrical energy to heat, but why does only the bulb emit significant visible light?

  1. (a) The heater is made of a different material
  2. (b) The bulb's filament reaches a higher temperature, emitting visible light
  3. (c) The heater uses magnetic effect instead of heating effect
  4. (d) The circuit design is different
Answer & solution

Answer: (b) The bulb's filament reaches a higher temperature, emitting visible light

A room heater and an electric bulb both convert electrical energy into heat, but only the bulb emits significant visible light.
This is because the filament in the bulb reaches a very high temperature.
At this high temperature, it emits visible light due to incandescence.
The heater element, although hot, operates at a lower temperature that mainly emits infrared radiation (heat) and not visible light.
Hence, the key difference is the filament temperature.
The bulb filament's higher temperature causes it to emit visible light efficiently, unlike the heater.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 3 · Numericals on electricity

Q80. A wire of a given material has length l and cross-sectional area A , and its resistance is 9 Ω. Another wire of the same material has length l/3 and cross-sectional area 3A. What will be the resistance of the second wire?

  1. (a) 6 Ω
  2. (b) 0.5 Ω
  3. (c) 1 Ω
  4. (d) 2 Ω
Answer & solution

Answer: (c) 1 Ω

Solution:
ResistanceR=ρ(L)/(A)
First wire:R_(1)=9Ω=ρ(l)/(A)
Second wire:R_(2)=ρ(l/3)/(3A)=ρ(l)/(9A)=(R_(1))/(9)=1Ω
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 3 · Numericals on electricity

Q81. An electric refrigerator has a power rating of 500 W and is used for 6 hours each day. If the cost of electrical energy is ₹4.00 per kilowatt-hour, what will be the total cost of operating the refrigerator for 30 days?

  1. (a) ₹180
  2. (b) ₹360
  3. (c) ₹300
  4. (d) ₹240
Answer & solution

Answer: (b) ₹360

The power rating of the refrigerator is 500 W, i.e., 0.5 kW.
Daily energy consumption = power x time = 0.5 kW x 6 hours = 3 kWh per day.
For 30 days, energy consumed = 3 kWh/day x 30 days = 90 kWh.
Cost of electrical energy = ₹4.00 per kWh.
Total cost = 90 kWh x ₹4 = ₹360.
Therefore, the total cost of operating the refrigerator for 30 days is ₹360.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 2 · Resistors in series and parallel

Q82. In a parallel combination of three resistors, which statement correctly describes the total current in the circuit?

  1. (a) It is the average of the currents through each resistor
  2. (b) It is equal to the sum of the currents through each resistor
  3. (c) It is the same as the current through each resistor
  4. (d) It is equal to the current through the resistor with the lowest resistance
Answer & solution

Answer: (b) It is equal to the sum of the currents through each resistor

In a parallel circuit, the total current is the sum of the currents through each resistor.
Each resistor provides a separate path for the current.
Therefore, the total current equals the sum of individual branch currents.
This follows the principle of parallel circuits in electricity.
Hence, the correct statement is that total current is the sum of currents through each resistor.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 3 · Potential difference and EMF

Q83. What is the reason why electrons move in a conducting wire when connected to a battery, but not when the wire is left untouched?

  1. (a) A battery creates a potential difference that sets electrons in motion
  2. (b) Gravity acts on electrons in the wire
  3. (c) Wire is made of copper
  4. (d) Electrons have intrinsic energy for movement
Answer & solution

Answer: (a) A battery creates a potential difference that sets electrons in motion

When a conducting wire is connected to a battery,
the battery creates a potential difference (voltage) across the wire.
This potential difference causes electrons to move, generating an electric current.
If the wire is left untouched (not connected to a battery),
there is no potential difference to push the electrons, so they do not move.
Gravity does not have a significant effect on electrons in the wire,
and simply being made of copper does not cause electron movement.
Electrons do not move spontaneously without an external force like a battery.
Final answer: Option 1
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 1 · Electric circuits and circuit symbols

Q84. Why does a torch bulb glow only when the switch is turned on?

  1. (a) Because the bulb is in series with the cell
  2. (b) Because the bulb is functional
  3. (c) Because the switch completes the circuit, allowing current to flow
  4. (d) Because the battery is charged
Answer & solution

Answer: (c) Because the switch completes the circuit, allowing current to flow

The bulb in a torch glows only when the switch is turned on because the switch completes the electrical circuit.
When the circuit is complete, current can flow from the battery through the bulb.
If the switch is off, the circuit is open and current cannot flow, so the bulb does not glow.
Hence, the switch controls whether the current flows or not by completing or breaking the circuit.
Final answer: Option 3
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 3 · Heating effect of current and Joule's law

Q85. An electric iron uses a heating element made of an alloy instead of pure metal. Which property of alloys justifies this choice?

  1. (a) Alloys have lower resistivity
  2. (b) Alloys have higher resistivity and do not oxidise readily at high temperatures
  3. (c) Alloys are better conductors
  4. (d) Alloys are cheaper
Answer & solution

Answer: (b) Alloys have higher resistivity and do not oxidise readily at high temperatures

The heating element in an electric iron requires a material that can withstand high temperatures without oxidising.
Alloys have higher resistivity, which helps the element heat up efficiently.
They do not oxidise readily at high temperatures, ensuring durability and safety.
Pure metals may oxidise and degrade faster under heat, making alloys preferable for heating elements.
Therefore, alloys are chosen for heating elements because of higher resistivity and resistance to oxidation at high temperatures.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 1 · Resistance, resistivity and their factors

Q86. If an ammeter reading increases when a thicker wire replaces a thinner one of the same material and length, what is the most likely reason?

  1. (a) The voltage across the wire decreased
  2. (b) The resistance of the thicker wire is lower, allowing more current
  3. (c) The wire's length is shorter
  4. (d) The wire's resistivity increased
Answer & solution

Answer: (b) The resistance of the thicker wire is lower, allowing more current

Thicker wire has larger cross-sectional area.
Resistance R = ρL/A, where A increases means R decreases.
Lower resistance allows more current through the wire.
Ammeter reads current; increased current means lower resistance.
Hence, thicker wire causes lower resistance, resulting in increased current reading.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 1 · Heating effect of current and Joule's law

Q87. Why are bulbs filled with nitrogen and argon gases instead of air?

  1. (a) To prolong the life of the filament
  2. (b) To increase brightness
  3. (c) To make the bulb heavier
  4. (d) To reduce electricity consumption
Answer & solution

Answer: (a) To prolong the life of the filament

Bulbs are filled with nitrogen and argon gases to prevent the filament from oxidizing.
Air contains oxygen which would cause the filament to burn quickly.
Nitrogen and argon being inert gases do not react with the filament.
This helps to prolong the life of the filament inside the bulb.
Final answer: Option 1
SSC MTS 2025 – 17 February 2026 Shift 2 · Electric charge and Coulomb's law

Q88. What is used to denote net charge that flows across any cross-section of a conductor of electricity?

  1. (a) J
  2. (b) Q
  3. (c) C
  4. (d) I
Answer & solution

Answer: (b) Q

Q (Coulombs) is used to denote net electric charge that flows across any cross-section of a conductor. Electric charge is measured in Coulombs (C) but denoted by the symbol Q. J denotes electric current density or energy (Joules); C is the unit of charge (Coulomb), not the symbol for charge; I denotes electric current (Amperes).
Delhi Police Constable (Exe) 2025-26 – 6 January 2026 Shift 2 · Resistance, resistivity and their factors

Q89. A wire of resistance R is stretched to double its original length. What is the new resistance of the wire, assuming uniform cross-section?

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

Answer: (c) 4R

The correct answer is (c) 4R.
When a wire is stretched to double its length, the volume remains constant:
Original: Length = L, Area = A, Resistance = R = ρL /A
New: Length = 2L. Since Volume = L×A = 2L×A', so A' = A/2
New Resistance = ρ(2L)/(A/2) = 4ρL/A = 4R
Alternatively: R ∝ L²/Volume, so if L doubles → R increases by 4 times.
(a) R/2: This would be the result if only the length changed without volume conservation.
(b) 2R: This is incorrect; doubling the length with volume conservation quadruples the resistance.
(d) R: The resistance doesn't stay the same when the dimensions change.
Delhi Police Constable (Exe) 2025-26 – 29 December 2025 Shift 3 · Electric charge and Coulomb's law

Q90. The unit of electric flux is represented by which quantity derived from the product of electric field and surface area?

  1. (a) Joule
  2. (b) Weber
  3. (c) Newton meter² per coulomb
  4. (d) Tesla
Answer & solution

Answer: (c) Newton meter² per coulomb

Electric flux (Φ) is defined as Φ = E × A × cosθ, where E = electric field (unit: N/C) and A = surface area (unit: m²). Therefore, unit of electric flux = N/C × m² = N·m²/C (Newton meter squared per coulomb). Equivalently, this can be written as Volt·metres (V·m) since V = N·m/C.
(a) Incorrect: Joule is the unit of energy or work (N·m or kg·m²/s²); not the unit of electric flux.
(b) Incorrect: Weber is the unit of magnetic flux (Φ_B); not electric flux.
(d) Incorrect: Tesla is the unit of magnetic field strength (B); it is not the unit of electric flux.
Delhi Police Constable (Exe) 2025-26 – 20 December 2025 Shift 1 · Numericals on electricity

Q91. A 100 W bulb operates at 200 V. Find resistance of filament.

  1. (a) 200 Ω
  2. (b) 300 Ω
  3. (c) 400 Ω
  4. (d) 500 Ω
Answer & solution

Answer: (c) 400 Ω

Using the power formula P = V²/R, we get R = V²/P = (200)²/100 = 40,000/100 = 400 Ω. Option 3 is correct.
Option 1 is wrong: 200 Ω would correspond to P = V²/R = (200)²/200 = 200 W, not 100 W.
Option 2 is wrong: 300 Ω would give P = (200)²/300 = 133.3 W, not 100 W.
Option 4 is wrong: 500 Ω would give P = (200)²/500 = 80 W, not 100 W.
Delhi Police Constable (Exe) 2025-26 – 19 December 2025 Shift 2 · Numericals on electricity

Q92. A circuit has a 12 V battery connected to a resistor of 6 Ω. What is the current flowing?

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

Answer: (b) 2 A

Using Ohm's Law: I = V / R, where I = current (amperes), V = voltage (volts), R = resistance (ohms). I = 12 V ÷ 6 Ω = 2 A. Therefore, the current flowing in the circuit is 2 Amperes.
Option 1 is incorrect: 1 A would require V = I × R = 1 × 6 = 6 V, not 12 V.
Option 3 is incorrect: 3 A would require V = I × R = 3 × 6 = 18 V, not 12 V.
Option 4 is incorrect: 4 A would require V = I × R = 4 × 6 = 24 V, not 12 V.
SSC Stenographer 2025 – 8 December 2025 Shift 2 · Heating effect of current and Joule's law

Q93. A heating element of a heater is made of a material with high resistivity and high melting point. Which of the following material properties is most crucial for its function as a heating element ?

  1. (a) Low specific heat capacity
  2. (b) High modulus of elasticity
  3. (c) High resistivity
  4. (d) High thermal conductivity
Answer & solution

Answer: (c) High resistivity

[IMAGE: SSC Steno GK (eng)_image_12.png]
SSC Stenographer 2025 – 7 December 2025 Shift 2 · Numericals on electricity

Q94. An electric kettle is rated 2.2 kW, 220 V. If the cost of electricity is ₹4 per unit (kWh), what is the cost of boiling water for 30 minutes?

  1. (a) ₹8.8
  2. (b) ₹4.4
  3. (c) ₹2.2
  4. (d) ₹1.1
Answer & solution

Answer: (b) ₹4.4

Power = 2.2 kW. Time = 30 min = 0.5 h. Energy used = 2.2 × 0.5 = 1.1 kWh (units) . Cost = 1.1 × ₹4 = ₹4.4 → Option 2.
SSC Stenographer 2025 – 7 December 2025 Shift 3 · Resistors in series and parallel

Q95. A conductor of length L and resistance R is cut into two equal halves. These two halves are then connected in parallel. Find the equivalent resistance of this parallel combination?

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

Answer: (a) R/4

The resistance of a conductor is proportional to its length. When the conductor of resistance R is cut into two equal halves, each half will have resistance R/2 . Now, these two halves (each of resistance R/2) are connected in parallel. Equivalent resistance in parallel = (R/2 × R/2) / (R/2 + R/2) = (R²/4) / R = R/4 . Hence, the correct answer is Option 1.
SSC Stenographer 2025 – 6 December 2025 Shift 3 · Resistors in series and parallel

Q96. Two resistors named R1 and R2, are connected in parallel. If R1 < R2, what can be inferred about the current flowing through them, given the same potential difference across both?

  1. (a) Current through R1 is greater than through R2.
  2. (b) Current through R1 is less than through R2.
  3. (c) Current through R1 is equal to current through R2.
  4. (d) The current ratio depends on the supply voltage.
Answer & solution

Answer: (a) Current through R1 is greater than through R2.

[IMAGE: SSC Steno GK (eng)_image_18.png]
SSC CGL Tier-1 2025 – 21 September 2025 Shift 3 · Heating effect of current and Joule's law

Q97. Why is nichrome commonly used as the heating element in electric heaters?

  1. (a) It has low resistivity and a high melting point
  2. (b) It has high resistivity and a low melting point
  3. (c) It offers low resistivity and a low melting point
  4. (d) It offers high resistivity and a high melting point
Answer & solution

Answer: (d) It offers high resistivity and a high melting point

Nichrome (an alloy of nickel and chromium) is used as a heating element because:
High resistivity → produces more heat for the same current (H = I²Rt).
High melting point (~1400°C) → does not melt at heater operating temperatures.
It also resists oxidation at high temperature.
SSC CGL Tier-1 2025 – 20 September 2025 Shift 1 · Numericals on electricity

Q98. A 60 W electric lamp is used for 5 hours daily. Calculate the energy consumed in 'units' (kilowatt-hours) in 30 days.

  1. (a) 9 units
  2. (b) 30 units
  3. (c) 0.3 units
  4. (d) 90 units
Answer & solution

Answer: (a) 9 units

- Power = 60 W = 0.06 kW.
- Daily energy = 0.06 x 5 = 0.3 kWh.
- Monthly energy = 0.3 x 30 = 9 kWh = 9 units.
SSC CGL Tier-1 2025 – 16 September 2025 Shift 2 · Ohm's law

Q99. If the resistance of a conductor is doubled while keeping the potential difference across its ends constant, what will happen to the current flowing through it?

  1. (a) It will double
  2. (b) It will halve
  3. (c) It will remain unchanged
  4. (d) It will quadruple
Answer & solution

Answer: (b) It will halve

• Ohm's Law: V = IR → I = V/R
• V (potential difference) = constant
• R is doubled: R → 2R
• New current: I' = V/(2R) = (1/2) × (V/R) = I/2
• Current is halved
• Answer: It will halve
RRB NTPC UG CBT-1 2024-25 – 4 September 2025 Shift 3 · Resistance, resistivity and their factors

Q100. Which copper-based alloy is a preferred material for making standard resistors because of its high resistivity and low temperature coefficient?

  1. (a) Manganin
  2. (b) Alnico
  3. (c) Tungsten
  4. (d) Brass
Answer & solution

Answer: (a) Manganin

12

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