Electric Current

VidyaarOctober 19, 20260146 views

Q100. How much electrical energy is consumed when a 1 kW appliance operates for one hour?

[RRB Group D — 18 August 2026, Shift 3]

(A) 1 watt-hour

(B) 1 kilowatt-hour

(D) 1 kilowatt

Answer: (B) 1 kilowatt-hour

Explanation: Electrical energy equals power multiplied by the operating time. A 1,000-watt device running for one hour consumes 1,000 watt-hours, or 1 kilowatt-hour. The kilowatt-hour measures energy, not power, and is commonly used for household billing.

Q99. Which electrical quantity is identical across resistors connected in parallel?

[RRB Group D — 17 August 2026, Shift 3]

(A) Current through each branch

(B) Resistance of each branch

(D) Power used by every branch

Answer: (C) Potential difference across each branch

Explanation: Each parallel branch connects directly between the same two circuit nodes. Consequently, every branch has the same potential difference. Currents can differ because their resistances differ; the total supply current equals the sum of branch currents.

Q98. What physical dimensions are used when defining a material’s resistivity from its resistance?

[RRB Group D — 17 August 2026, Shift 2]

(A) Unit length and unit cross-sectional area

(B) Unit mass and unit volume

(D) Unit temperature and unit pressure

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

Explanation: Resistivity is defined using R = ρL/A, giving ρ = RA/L. Conceptually it is the resistance of a specimen of unit length and unit cross-sectional area. Unlike resistance, resistivity does not depend on the specimen dimensions at a fixed temperature.

Q97. Resistors of 4 Ω and 12 Ω are connected in parallel across 6 V. What is the total current?

[RRB Group D — 14 August 2026, Shift 3]

(A) 0.5 A

(B) 1.5 A

(D) 3 A

Answer: (C) 2 A

Explanation: For a parallel pair, 1/R = 1/4 + 1/12 = 1/3, so equivalent resistance is 3 Ω. By Ohm’s law the source current is 6/3 = 2 A. Branch currents are 1.5 A and 0.5 A.

Q96. One kilowatt-hour is equivalent to how many joules?

[RRB Group D — 13 August 2026, Shift 2]

(A) 3.6 × 10³ J

(B) 3.6 × 10⁵ J

(D) 3.6 × 10⁹ J

Answer: (C) 3.6 × 10⁶ J

Explanation: One kilowatt is 1,000 joules per second, while one hour contains 3,600 seconds. Multiplying gives 1,000 × 3,600 = 3,600,000 joules. Thus one electricity-billing unit corresponds to 3.6 megajoules of energy.

Q95. A bulb uses 60 W when supplied with 120 V. Find the resistance of its filament.

[RRB Group D — 13 August 2026, Shift 1]

(A) 2 Ω

(B) 60 Ω

(D) 240 Ω

Answer: (D) 240 Ω

Explanation: The electrical power equation is P = V²/R. Rearranging gives R = V²/P = 120²/60 = 240 Ω. The value applies at the bulb’s operating temperature; a cold filament usually has substantially lower resistance.

Q94. What does a cell or battery provide to drive charge around a circuit?

[RRB Group D — 11 August 2026, Shift 3]

(A) A fixed electrical resistance

(B) A potential difference

(D) A magnetic field only

Answer: (B) A potential difference

Explanation: Chemical reactions within a cell separate charge and establish a potential difference between its terminals. Once a conducting loop is closed, that potential difference drives current. The battery transfers energy to moving charges rather than creating charge from nothing.

Q93. What is the effect of connecting additional resistors in series with a fixed-voltage source?

[RRB Group D — 10 August 2026, Shift 2]

(A) Total resistance decreases

(B) Total resistance increases, reducing current

(D) Supply voltage necessarily rises

Answer: (B) Total resistance increases, reducing current

Explanation: Resistances in series add arithmetically because the same current passes through each element. If the battery voltage is unchanged, Ohm’s law gives a smaller current as the total resistance rises. This principle is used to limit circuit current.

Q92. Calculate the charge crossing a conductor carrying 0.5 A continuously for 10 minutes.

[RRB Group D — 9 August 2026, Shift 3]

(A) 5 C

(B) 30 C

(D) 600 C

Answer: (C) 300 C

Explanation: Charge transported is Q = It. Convert ten minutes into 600 seconds; then Q = 0.5 × 600 = 300 coulombs. The ampere is a coulomb per second, so using seconds rather than minutes is essential.

Q91. A 220 V, 1000 W resistive heater is operated on 110 V. What power does it dissipate?

[RRB Group D — 9 August 2026, Shift 2]

(A) 1000 W

(B) 500 W

(D) 125 W

Answer: (C) 250 W

Explanation: Assuming the heater resistance remains approximately constant, P is proportional to V². Halving the supply voltage therefore lowers power to one-quarter of its rated value: 1,000/4 = 250 W. This is different from merely halving the current.

Q90. A 9 V cell is connected to 3 Ω, 2 Ω and 1 Ω resistors in series. What current flows?

[RRB Group D — 6 August 2026, Shift 3]

(A) 9 A

(B) 1.5 A

(D) 6 A

Answer: (B) 1.5 A

Explanation: Series resistance is the sum 3 + 2 + 1 = 6 Ω. Applying Ohm’s law to the complete loop gives I = V/R = 9/6 = 1.5 A. Every resistor carries this identical current.

Q89. What is the main function of an electrical fuse in household wiring?

[RRB Group D — 6 August 2026, Shift 2]

(A) Increase source voltage

(B) Interrupt an excessive current

(D) Prevent all temperature changes

Answer: (B) Interrupt an excessive current

Explanation: The fuse contains a metal link chosen to melt when current becomes dangerously large. Its melting opens the circuit, helping prevent overheating and fire. It does not stabilize voltage or store charge and must have a suitable current rating.

Q88. On which effect of current does an ordinary fuse primarily operate?

[RRB Group D — 5 February 2026]

(A) Magnetic effect

(B) Chemical effect

(D) Photoelectric effect

Answer: (C) Heating effect

Explanation: A fuse responds to excess current because the resistive element produces heat according to Joule’s law, H = I²Rt. If its temperature reaches the melting point, the connection breaks. The rapid increase of heating with current is vital.

Q87. Which expression gives the dimensional nature of electric charge in terms of current and time?

[RRB NTPC Undergraduate CBT — 9 May 2026]

(A) Current divided by time

(B) Current multiplied by time

(D) Power divided by time

Answer: (B) Current multiplied by time

Explanation: Electric current is defined as charge passing per unit time, I = Q/t. Rearranging yields Q = It, so the dimensions of charge are those of electric current multiplied by time. Its SI unit is the coulomb.

Q86. In a metallic conductor, the conventional current direction is ____ the direction of electron drift.

[RRB Group D — 29 December 2025]

(A) the same as

(B) opposite to

(D) unrelated to

Answer: (B) opposite to

Explanation: Electrons have negative charge and drift toward higher electric potential in a simple metal circuit. Conventional current is defined in the direction a positive charge would move, so it is opposite to electron drift. The current convention predates electron discovery.

Q85. What is produced around a wire when electric current passes through it?

[RRB Group D — 14 December 2025]

(A) Only an electric charge

(B) A magnetic field

(D) A permanent temperature decrease

Answer: (B) A magnetic field

Explanation: A current-carrying conductor generates a magnetic field around itself. Its direction can be found with the right-hand thumb rule. The field forms closed curves around a straight wire and provides the basic operating principle of electromagnets.

Q84. What energy conversion occurs in a purely resistive conductor carrying current?

[RRB Group D — 8 December 2025]

(A) Electrical energy into heat

(B) Heat into gravitational energy

(D) Sound into chemical energy

Answer: (A) Electrical energy into heat

Explanation: Moving charge carriers transfer energy to the lattice through collisions, increasing its thermal energy. For a resistance R carrying current I, the heating rate is I²R. This is the intended process in kettles and other resistive heating appliances.

Q83. A 110 W electrical appliance is operated at 220 V. Calculate the current it draws.

[RRB Group D — 5 December 2025]

(A) 0.25 A

(B) 0.5 A

(D) 5 A

Answer: (B) 0.5 A

Explanation: Power is the product of potential difference and current, P = VI. Therefore I = 110/220 = 0.5 A. The current is one-half ampere under the stated operating conditions, without requiring the appliance’s resistance to be given.

Q82. Electric current is defined as the rate of flow of which physical quantity?

[RRB Group D — 2 December 2025]

(A) Electrical charge

(B) Electrical resistance

(D) Potential difference

Answer: (A) Electrical charge

Explanation: A current of one ampere means one coulomb of charge crosses a section every second. Mathematically, average current equals charge divided by elapsed time. Electrons carry current in metals; ions also transport charge in electrolytes.

Q81. Why is nichrome preferred for electric heating elements?

[SSC CGL — 21 September 2025, Shift 3]

(A) Low resistance and low melting point

(B) High resistivity and high melting point

(D) Rapid oxidation and low strength

Answer: (B) High resistivity and high melting point

Explanation: Nichrome has relatively high resistivity, so a useful length generates substantial heat, and its high melting point allows it to tolerate operating temperatures. It also resists oxidation well. These properties distinguish it from an ordinary fuse wire.

Q80. An electric bell mainly operates by which effect of electric current?

[SSC CGL — 16 September 2025, Shift 1]

(A) Chemical effect

(B) Magnetic effect

(D) Cooling effect

Answer: (B) Magnetic effect

Explanation: When current passes through the bell’s coil, the soft-iron core becomes an electromagnet. It attracts the armature and moves the striker; the contact repeatedly breaks and reconnects. The resulting rapid vibration produces the ringing sound.

Q79. A circuit needs which arrangement for a steady electric current?

[RRB NTPC Undergraduate — 29 August 2025, Shift 1]

(A) An open wire without a source

(B) A closed conducting path with a source

(D) Only an insulating switch

Answer: (B) A closed conducting path with a source

Explanation: Current persists when there is a complete conducting loop and an energy source maintaining a potential difference. An open switch interrupts the path, stopping ordinary conduction. A resistor alone cannot sustain a continuous current without an applied source.

Q78. An electric motor primarily uses which effect of current, rather than the heating effect?

[RRB NTPC Undergraduate — 29 August 2025, Shift 1]

(A) Magnetic effect

(B) Evaporation effect

(D) Capillary effect

Answer: (A) Magnetic effect

Explanation: Electric motors exploit the forces experienced by current-carrying conductors in magnetic fields. These forces produce torque and rotate the shaft. Some resistive heat is unavoidably produced in the windings, but heating is not the motor’s useful primary action.

Q77. A charge of 100 C passes a cross-section in 50 s. What is the mean current?

[RRB NTPC Undergraduate — 29 August 2025, Shift 1]

(A) 0.5 A

(B) 2 A

(D) 50 A

Answer: (B) 2 A

Explanation: Average current is the quantity of charge transported per second. Dividing Q = 100 C by t = 50 s gives I = 2 A. Larger charge transferred in the same time corresponds to greater current.

Q76. How much potential difference corresponds to 24 joules of work transferring 2 coulombs of charge?

[UPSC NDA — 2025-I]

(A) 6 V

(B) 12 V

(D) 48 V

Answer: (B) 12 V

Explanation: Potential difference is work done per unit charge, V = W/Q. Dividing 24 joules by 2 coulombs gives 12 volts. This is the energy transferred to each coulomb as it moves between the two circuit points.

Q75. Three identical 1 Ω resistors are connected in parallel. Find their equivalent resistance.

[UPSC NDA — 2025-I]

(A) 3 Ω

(B) 1 Ω

(D) 1/2 Ω

Answer: (C) 1/3 Ω

Explanation: Reciprocals of resistances add for parallel branches. Here 1/R = 1 + 1 + 1 = 3 per ohm, so R = 1/3 Ω. Adding parallel pathways always decreases the equivalent resistance.

Q74. What is the SI unit used to express specific electrical resistance or resistivity?

[SSC CGL — 26 September 2024]

(A) Ohm

(B) Ohm metre

(D) Volt per ampere

Answer: (B) Ohm metre

Explanation: The resistivity relation is ρ = RA/L. Resistance is measured in ohms, area in square metres and length in metres; consequently the derived SI unit is ohm metre, Ω m. Resistivity is not measured in plain ohms.

Q73. Which is the SI base unit of electric current?

[SSC CGL — 13 September 2024]

(A) Coulomb

(B) Volt

(D) Watt

Answer: (C) Ampere

Explanation: The ampere is an SI base unit and measures electric current. A coulomb is the derived unit of charge, equal to one ampere-second. The volt measures potential difference, whereas the watt measures the rate of energy transfer.

Q72. Which rule determines the direction of the magnetic field surrounding a straight current-carrying wire?

[SSC JE — 7 June 2024]

(A) Right-hand thumb rule

(B) Pascal’s law

(D) Archimedes’ principle

Answer: (A) Right-hand thumb rule

Explanation: Point the right thumb in the conventional current direction; the curled fingers indicate the magnetic field lines surrounding the conductor. This is the right-hand thumb rule, frequently used to establish field orientation in electromagnetism.

Q71. Approximately how many electrons pass a point in a wire each second when the current is 1 A?

[SSC JE — 5 June 2024]

(A) 6.25 × 10¹²

(B) 6.25 × 10¹⁵

(D) 6.25 × 10²¹

Answer: (C) 6.25 × 10¹⁸

Explanation: One ampere transports one coulomb of charge per second. The magnitude of the elementary charge is approximately 1.6 × 10⁻¹⁹ C, so 1/(1.6 × 10⁻¹⁹) yields about 6.25 × 10¹⁸ electrons each second.

Q70. A magnetic field can be produced by which of the following?

[SSC GD Constable — 1 March 2024]

(A) A moving electric charge

(B) An isolated stationary neutron

(D) A stationary neutral molecule only

Answer: (A) A moving electric charge

Explanation: Moving electrical charges constitute current and produce magnetic fields. A stationary electric charge creates an electric field, whereas motion contributes a magnetic field. This connection underlies the magnetic force around current-carrying wires and the action of electromagnets.

Q69. A circuit contains 5 Ω and 7 Ω in series, connected in parallel with 36 Ω. What is the equivalent resistance?

[UPSC NDA — 2024-II]

(A) 9 Ω

(B) 12 Ω

(D) 48 Ω

Answer: (A) 9 Ω

Explanation: The first branch has series resistance 5 + 7 = 12 Ω. Combining it in parallel with 36 Ω gives R = (12 × 36)/(12 + 36) = 9 Ω. The result is lower than either branch resistance.

Q68. In an alternating-current supply of frequency 50 Hz, how often does the current reverse direction?

[UPSC NDA — 2024-II]

(A) Every 0.02 s

(B) Every 0.01 s

(D) Every 1 s

Answer: (B) Every 0.01 s

Explanation: A 50 Hz waveform completes fifty cycles per second, so one full period lasts 1/50 = 0.02 second. Polarity and current direction reverse after each half-cycle, meaning reversals occur every 0.01 second.

Q67. Which among electric current, current density, power and wattless current is a vector?

[WBCS Mains (Science) — 2023]

(A) Electric current

(B) Current density

(D) Wattless current

Answer: (B) Current density

Explanation: Current density includes the direction of charge transport through a surface and is therefore a vector quantity. Circuit current is treated as a signed scalar, while electrical power has magnitude without direction. Current density is measured in amperes per square metre.

Q66. Who experimentally discovered the magnetic effect of electric current?

[WBCS Mains (Science) — 2023]

(A) Michael Faraday

(B) Hans Christian Oersted

(D) Niels Bohr

Answer: (B) Hans Christian Oersted

Explanation: In 1820 Oersted observed that a nearby magnetic compass needle deflected when current flowed through a wire. This demonstrated a connection between electricity and magnetism. Faraday later discovered electromagnetic induction, while Ampère developed mathematical descriptions of magnetic forces.

Q65. Which is NOT an application of the chemical effect of an electric current?

[SSC CHSL — 16 August 2023, Shift 1 (memory-based)]

(A) Electrolysis

(B) Electroplating

(D) Galvanization

Answer: (C) Electromagnetism

Explanation: Electromagnetism concerns magnetic fields generated by moving charges, so it is the magnetic rather than chemical effect of current. Electrolysis and electroplating involve reactions driven by electrical energy. Some industrial galvanizing methods also use electrochemical zinc deposition.

Q64. Which description correctly identifies an electrical conductor?

[SSC MTS — 19 June 2023]

(A) A substance that allows electric charge to flow

(B) A material that prohibits all electron motion

(D) A substance that creates electricity without energy

Answer: (A) A substance that allows electric charge to flow

Explanation: Conductors contain mobile charged particles and allow electrical current when a potential difference is applied. Metals possess mobile electrons, while electrolyte solutions conduct through ions. Insulators have relatively few freely mobile carriers under normal conditions.

Q63. An electric heater rated 220 V draws 0.5 A. What current would it draw at 120 V if its resistance stays unchanged?

[UPSC NDA — 2023-II]

(A) 0.27 A

(B) 0.50 A

(D) 1.10 A

Answer: (A) 0.27 A

Explanation: The heater resistance at rated voltage is R = 220/0.5 = 440 Ω. At the reduced voltage, Ohm’s law predicts I = 120/440 ≈ 0.273 A. The approximation assumes resistance does not change with temperature.

Q62. Which alloy is commonly chosen as a heating element in an electric iron?

[UPSC NDA — 2023-II]

(A) Solder

(B) Nichrome

(D) Soft iron

Answer: (B) Nichrome

Explanation: Nichrome combines high electrical resistivity with a high melting point and resistance to oxidation. Consequently it can produce heat without melting rapidly. Soft iron is useful for electromagnet cores, whereas low-melting solder is unsuited for a heating coil.

Q61. A wire is stretched to twice its original length without loss of material. How does its resistance change?

[UPSC NDA — 2023-I]

(A) Becomes one-fourth

(B) Becomes one-half

(D) Becomes four times

Answer: (D) Becomes four times

Explanation: If the wire’s volume stays constant, doubling length halves cross-sectional area. Since R = ρL/A, both changes multiply resistance, making it four times the original. With constant applied voltage, the resulting current falls to one-quarter.

Q60. A lamp rated 80 W at 220 V is connected to a 110 V supply. Assuming constant resistance, its power becomes:

[UPSC NDA — 2023-I]

(A) 10 W

(B) 20 W

(D) 80 W

Answer: (B) 20 W

Explanation: Power at constant resistance obeys P = V²/R. Reducing the supply voltage to half its rated value gives one-quarter the original power: 80/4 = 20 W. In a real incandescent bulb, resistance varies with filament temperature.

Q59. A high-resistivity material is chosen for a lamp filament primarily to achieve which result?

[RRB Group D — 11 October 2022]

(A) Rapid heat generation

(B) Zero energy loss

(D) Cold operation

Answer: (A) Rapid heat generation

Explanation: A thin, relatively high-resistance filament releases electrical energy as heat, reaching temperatures at which it emits visible light. Tungsten is preferred because it also survives very high temperatures. The filament’s glow is an example of incandescence.

Q58. A lamp works at 220 V while drawing 2 A. What is its electrical power?

[RRB Group D — 7 October 2022]

(A) 110 W

(B) 220 W

(D) 880 W

Answer: (C) 440 W

Explanation: Electrical power at a device is the potential difference across it multiplied by its current. Substituting the values gives P = 220 × 2 = 440 watts. This is the instantaneous conversion rate under the stated operating conditions.

Q57. Two resistors have combined resistance 120 Ω in series and 30 Ω in parallel. Find their separate resistance values.

[RRB Group D — 7 October 2022, Shift 1]

(A) 70 Ω and 50 Ω

(B) 80 Ω and 40 Ω

(D) 60 Ω and 60 Ω

Answer: (D) 60 Ω and 60 Ω

Explanation: Let the resistances be R₁ and R₂. Series connection gives R₁ + R₂ = 120 Ω; their parallel equivalent is R₁R₂/(R₁ + R₂) = 30 Ω. Therefore the product is 3,600, satisfied by 60 Ω each.

Q56. Which equation correctly gives Joule heat produced in a resistor during time t?

[RRB Group D — 29 September 2022]

(A) H = I²Rt

(B) H = IRt²

(D) H = I²R/t

Answer: (A) H = I²Rt

Explanation: Joule’s heating law is H = I²Rt. For constant resistance and elapsed time, doubling the current causes four times as much heat to be generated. It explains both normal resistance heating and why overloads can damage electrical wiring.

Q55. A conductor obeys Ohm’s law. How is current determined from voltage V and resistance R?

[RRB Group D — 28 September 2022]

(A) I = VR

(B) I = V/R

(D) I = V + R

Answer: (B) I = V/R

Explanation: Ohm’s law states V = IR when temperature and other physical conditions are constant. Rearranging for current gives I = V/R. At unchanged potential difference, raising the resistance lowers current in inverse proportion.

Q54. Two groups each have six identical resistors connected in parallel across 20 V. Group A draws 6 A and B draws 8 A. What are the individual resistance values?

[RRB Group D — 19 September 2022]

(A) 20 Ω and 15 Ω

(B) 15 Ω and 20 Ω

(D) 6 Ω and 8 Ω

Answer: (A) 20 Ω and 15 Ω

Explanation: Six identical resistors of resistance r in parallel have equivalent resistance r/6. Hence r = 6V/I. For group A, r = 120/6 = 20 Ω; for B, r = 120/8 = 15 Ω.

Q53. Each of two networks uses twelve equal resistors in parallel across 30 V. They draw 10 A and 18 A. What are the individual resistor values?

[RRB Group D — 16 September 2022, Shift 3]

(A) 26 Ω and 25 Ω

(B) 36 Ω and 15 Ω

(D) 36 Ω and 20 Ω

Answer: (D) 36 Ω and 20 Ω

Explanation: With twelve equal parallel elements, total resistance equals individual resistance divided by twelve. Therefore r = 12V/I. Network A gives 12 × 30/10 = 36 Ω; network B gives 12 × 30/18 = 20 Ω.

Q52. On which principle does an ordinary electrical transformer operate?

[RRB Group D — 14 September 2022]

(A) Mutual electromagnetic induction

(B) Chemical electrolysis

(D) Thermal expansion

Answer: (A) Mutual electromagnetic induction

Explanation: A changing current in the primary winding produces a changing magnetic flux through the transformer core. That changing flux induces an electromotive force in the secondary winding. For this reason transformers ordinarily require changing current, such as alternating current.

Q51. In a parallel arrangement, how does the equivalent resistance compare with the lowest individual resistance?

[RRB Group D — 13 September 2022]

(A) It is always greater

(B) It equals the highest resistance

(D) It is the arithmetic sum

Answer: (C) It is smaller than the smallest resistance

Explanation: Connecting resistors in parallel provides multiple pathways for charge. Their conductances add, so the total resistance is less than the resistance of the easiest single path. This remains true for any finite positive branch resistances.

Q50. Which relation expresses the electric power used by a circuit element?

[RRB Group D — 12 September 2022]

(A) P = V/I

(B) P = VI

(D) P = I/V

Answer: (B) P = VI

Explanation: Power represents electrical work transferred each second. Since each coulomb transfers energy equal to the potential difference V, and I coulombs flow each second, their product VI equals watts. For ohmic resistors this is also I²R.

Q49. Two resistors of 10 Ω each are in parallel, and the pair is in series with another 10 Ω resistor across 6 V. Find the circuit power.

[RRB Group D — 9 September 2022]

(A) 1.2 W

(B) 2.4 W

(D) 6 W

Answer: (B) 2.4 W

Explanation: The parallel pair contributes 5 Ω because two identical 10 Ω resistors halve their resistance. Adding the third resistor in series gives 15 Ω. Total power is V²/R = 36/15 = 2.4 W.

Q48. When temperature stays fixed, the potential difference across an ohmic conductor is proportional to what?

[RRB Group D — 6 September 2022]

(A) Current flowing through it

(B) Square of the length only

(D) Magnetic field strength only

Answer: (A) Current flowing through it

Explanation: Ohm’s law states that voltage and current are directly proportional in an ohmic conductor when physical conditions remain unchanged. Their ratio is its resistance. If the voltage doubles at constant resistance, the current doubles as well.

Q47. A resistor of 4 Ω is connected across 12 V. What electrical power does it dissipate?

[RRB Group D — 5 September 2022]

(A) 3 W

(B) 16 W

(D) 48 W

Answer: (C) 36 W

Explanation: For a resistor, P = V²/R. Substituting 12 V and 4 Ω gives P = 144/4 = 36 W. The same result follows because the current is 12/4 = 3 A and power equals 12 × 3.

Q46. A 10 Ω and a 20 Ω resistor are connected separately in parallel across 6 V. What currents pass through them?

[RRB Group D — 2 September 2022]

(A) 0.6 A and 0.3 A

(B) 0.3 A and 0.6 A

(D) 0.2 A and 0.4 A

Answer: (A) 0.6 A and 0.3 A

Explanation: Each parallel resistor experiences the complete 6 V supply. Current through the 10 Ω branch is 6/10 = 0.6 A, while the 20 Ω branch carries 6/20 = 0.3 A. Greater resistance gives smaller current.

Q45. Resistors 6 Ω and 12 Ω are connected in parallel across 3 V. How much heat is produced in one second?

[RRB Group D — 26 August 2022]

(A) 0.75 J

(B) 1.5 J

(D) 4.5 J

Answer: (C) 2.25 J

Explanation: The equivalent resistance is (6 × 12)/(6 + 12) = 4 Ω. Power is V²/R = 9/4 = 2.25 W. In one second the electrical energy converted to heat equals 2.25 joules.

Q44. Which quotient gives a conductor’s electrical resistance?

[RRB Group D — 25 August 2022]

(A) Current divided by voltage

(B) Voltage divided by current

(D) Voltage multiplied by charge

Answer: (B) Voltage divided by current

Explanation: Ohm’s law identifies resistance as the ratio of potential difference to current for an ohmic conductor. Thus R = V/I, measured in ohms. One ohm permits one ampere to flow under an applied potential difference of one volt.

Q43. Which substance is least likely to conduct an ordinary electric current?

[RRB Group D — 22 August 2022]

(A) Copper

(B) Graphite

(D) Octane

Answer: (D) Octane

Explanation: Octane is a nonpolar hydrocarbon without appreciable mobile ions or freely moving electrons under normal conditions. Metals conduct through electrons, graphite also has delocalized electrons, and dissolved salts conduct through positive and negative ions.

Q42. Which material property chiefly determines resistivity at a given temperature?

[RRB Group D — 17 August 2022]

(A) The nature of the material

(B) Length of a specific wire

(D) Total number of wire loops

Answer: (A) The nature of the material

Explanation: Resistivity is an intrinsic material property, largely determined by its charge carriers and microscopic structure at a particular temperature. Making the wire longer changes its resistance, but not its resistivity. Temperature changes can alter resistivity significantly.

Q41. What is the point-form version of Ohm’s law using current density J and electric field E?

[UPPSC Assistant Engineer (Electrical) — 29 May 2022]

(A) J = E/σ

(B) J = σE

(D) J = E + σ

Answer: (B) J = σE

Explanation: The microscopic form of Ohm’s law is J = σE, where σ is electrical conductivity. It relates the local current density vector to the electric field vector. In isotropic ohmic materials, the two vectors point in the same direction.

Q40. Why does the current suddenly become very large when a short circuit occurs?

[UPSC CDS — 10 April 2022]

(A) Circuit resistance becomes very small

(B) Battery charge becomes infinite

(D) The wire becomes an insulator

Answer: (A) Circuit resistance becomes very small

Explanation: In a short circuit an unintended, low-resistance route bypasses the load. For a fixed supply voltage, Ohm’s law predicts a large current as resistance becomes very small. Fuses or circuit breakers are installed to interrupt such dangerous faults.

Q39. A 50 Ω wire is cut into five equal pieces, all connected in parallel. What is their combined resistance?

[UPSC NDA — 2022-I]

(A) 2 Ω

(B) 5 Ω

(D) 250 Ω

Answer: (A) 2 Ω

Explanation: Each equal piece has resistance 50/5 = 10 Ω. Placing five 10 Ω pieces in parallel gives 10/5 = 2 Ω. More generally, cutting a uniform wire into n pieces and paralleling them yields R/n².

Q38. Which intrinsic property of a material describes opposition to the flow of electric charge?

[UPPSC Polytechnic Lecturer (Electrical) — 22 November 2021]

(A) Conductivity

(B) Thermoelectricity

(D) Resistivity

Answer: (D) Resistivity

Explanation: Resistivity quantifies how strongly a material opposes electric current, independently of the dimensions of any particular specimen. Its unit is ohm metre. Conductivity expresses the reciprocal tendency to carry current; the product of the two properties is one.

Q37. To change a moving-coil galvanometer into a DC ammeter, what should be connected across it?

[UPPSC Polytechnic Lecturer (Electrical) — 22 November 2021]

(A) High resistance in series

(B) Low shunt resistance in parallel

(D) Capacitor in series

Answer: (B) Low shunt resistance in parallel

Explanation: An ammeter must offer little resistance to the measured circuit. A low-resistance shunt in parallel with the galvanometer carries most of the current, protecting the delicate coil while allowing the instrument to measure a larger current.

Q36. Which device–principle pair is incorrectly matched: potentiometer–Wheatstone bridge, voltameter–chemical effect, induction coil–electromagnetic induction, transformer–mutual induction?

[UPPSC GIC Lecturer (Physics) — 2021]

(A) Potentiometer–Wheatstone bridge

(B) Voltameter–chemical effect

(D) Transformer–mutual induction

Answer: (A) Potentiometer–Wheatstone bridge

Explanation: A potentiometer uses the proportional relation between potential drop and length along a uniform current-carrying wire. It is not based on Wheatstone-bridge balance. The other listed principles correctly describe electrolysis-based charge measurement and electromagnetic induction devices.

Q35. A nichrome wire carrying current is shortened to half its length without changing its cross-section or applied voltage. How does the current change?

[UPSC CDS — 14 November 2021]

(A) It becomes one-fourth

(B) It becomes half

(D) It remains unchanged

Answer: (C) It doubles

Explanation: For an unchanged material and cross-sectional area, resistance is proportional to length. Halving the wire length therefore halves resistance. Since supply voltage is unchanged, Ohm’s law predicts twice the current, although resistive heating may change the final temperature.

Q34. Which welding method is especially extensively carried out by industrial robots?

[MPPSC Assistant Engineer (Mechanical) — 2021]

(A) Spot welding

(B) MIG welding

(D) Thermit welding

Answer: (A) Spot welding

Explanation: Resistance spot welding passes a large electric current between electrodes pressing metal sheets together. Joule heating at the contact forms a weld. Industrial robots repeat the placement, pressure and current cycle rapidly and precisely, particularly in automobile assembly.

Q33. Which instrument is used to measure electric current?

[SSC MTS — 5 October 2021]

(A) Barometer

(B) Thermometer

(D) Hygrometer

Answer: (C) Ammeter

Explanation: An ammeter measures the current flowing in a circuit and is connected in series with the part being tested. It has low internal resistance to minimize interference with the current. A voltmeter instead measures the potential difference across two points.

Q32. A refrigerator rated at 5 kW operates 10 hours daily for 30 days. At ₹4 per unit, what is the cost?

[UPSC NDA and Naval Academy — 2020 (Combined Examination)]

(A) ₹600

(B) ₹3,000

(D) ₹12,000

Answer: (C) ₹6,000

Explanation: Energy consumed is 5 kW × 10 h × 30 = 1,500 kWh. One electricity unit is one kilowatt-hour. At ₹4 per unit, the bill for this usage alone is 1,500 × 4 = ₹6,000.

Q31. Which metal is commonly used for the filament of an incandescent electric bulb?

[UPSC NDA and Naval Academy — 2020 (Combined Examination)]

(A) Copper

(B) Tungsten

(D) Aluminium

Answer: (B) Tungsten

Explanation: Tungsten has an exceptionally high melting point, allowing the filament to glow intensely without immediately melting. Its electrical resistance converts energy into heat and light. The sealed bulb also limits oxidation that would otherwise destroy the hot filament.

Q30. A 20 Ω resistance wire is cut into two equal lengths and the parts are placed in parallel. Calculate the resulting resistance.

[UPSC NDA and Naval Academy — 2020 (Combined Examination)]

(A) 5 Ω

(B) 10 Ω

(D) 40 Ω

Answer: (A) 5 Ω

Explanation: Cutting the uniform wire in half produces two 10 Ω pieces. Connecting these two equal resistances in parallel yields 10/2 = 5 Ω. Because both shortening and adding a branch reduce resistance, the total becomes one-quarter of the original.

Q29. What occurs in the electric current when a circuit is shorted?

[UPSC NDA and Naval Academy — 2020 (Combined Examination)]

(A) Current decreases greatly

(B) Current suddenly becomes very large

(D) Current becomes exactly zero

Answer: (B) Current suddenly becomes very large

Explanation: An accidental low-resistance path bypasses the intended load, producing an excessive current from the supply. The resulting I²R heating may damage cables or start a fire. This is why distribution circuits include protective fuses or breakers.

Q28. Which electrical instrument detects the presence of small currents?

[UPSC NDA and Naval Academy — 2020 (Combined Examination)]

(A) Galvanometer

(B) Transformer

(D) Dynamo

Answer: (A) Galvanometer

Explanation: A sensitive galvanometer detects small electric currents through the deflection of a moving coil or other indicator. With suitable added resistances, its operating range can be adapted to measure larger currents or potential differences as an ammeter or voltmeter.

Q27. What is the name of the mechanism by which moving conducting fluid in Earth’s outer core generates electric currents and magnetism?

[RPSC Senior Teacher Grade II (Science) — 4 July 2019]

(A) Joule effect

(B) Seebeck effect

(D) Dynamo effect

Answer: (D) Dynamo effect

Explanation: The geodynamo operates as molten, electrically conducting iron moves through magnetic fields in Earth’s outer core. Induced currents reinforce the magnetic field, allowing it to persist. Joule heating and thermoelectric effects describe different physical processes.

Q26. An electrical safety fuse wire should have which combination of properties?

[UPSC NDA — 21 April 2019]

(A) Low melting point and suitable relatively high resistivity

(B) High melting point and zero resistivity

(D) No conductivity whatsoever

Answer: (A) Low melting point and suitable relatively high resistivity

Explanation: When a fault causes excessive current, the fuse element must heat and melt before the protected wiring becomes unsafe. Suitable resistivity allows sufficient resistive heating, while a low melting point ensures prompt interruption. A conductor that never melts cannot serve this purpose.

Q25. Does Ohm’s law apply to every conducting material under every condition?

[UPSC NDA — 2019-II]

(A) Yes, at all temperatures and voltages

(B) No, only materials with linear voltage-current behaviour obey it

(D) Only magnetic materials obey it

Answer: (B) No, only materials with linear voltage-current behaviour obey it

Explanation: Ohm’s law describes conductors whose current varies linearly with applied voltage under fixed physical conditions. Diodes, filament bulbs across changing temperatures, and many electrolytes can show nonlinear characteristics. Hence the voltage-to-current ratio need not remain constant in all materials.

Q24. Current produced by convection of metallic liquid inside Earth helps create its magnetic field. What is this called?

[RPSC Senior Teacher Grade II (Science) — 1 November 2018]

(A) Seebeck effect

(B) Thomson effect

(D) Meissner effect

Answer: (C) Dynamo effect

Explanation: Earth’s flowing molten core is conductive. Its motion through an existing magnetic field induces electric currents which in turn generate magnetic fields. This self-sustaining feedback is the dynamo mechanism that explains much of the planet’s global magnetism.

Q23. Which SI unit measures electrical resistance?

[RRB Group D — 31 October 2018]

(A) Ampere

(B) Ohm

(D) Farad

Answer: (B) Ohm

Explanation: The ohm is the SI derived unit of electrical resistance. One ohm is the resistance through which one ampere flows under one volt. Amperes measure current, coulombs measure charge, and farads are units of capacitance.

Q22. Which is the SI unit of electrical potential difference?

[RRB Group D — 26 October 2018]

(A) Watt

(B) Volt

(D) Ohm

Answer: (B) Volt

Explanation: Potential difference is energy transferred per unit charge and has the SI unit volt. One volt equals one joule per coulomb. Unlike the ampere, it does not express the rate of flow of charge through a conductor.

Q21. Which law relates heat generated in an electrical conductor to I²Rt?

[RRB Group D — 11 October 2018]

(A) Joule’s law of heating

(B) Newton’s law of cooling

(D) Faraday’s second law only

Answer: (A) Joule’s law of heating

Explanation: Joule established that heat released in a resistor grows with the square of current, its resistance and the duration of current flow. The equation H = I²Rt is crucial when sizing heaters, fuses, and wiring against thermal damage.

Q20. What physical phenomenon is responsible for the operation of a basic electric heater?

[RRB Group D — 10 October 2018]

(A) Magnetic attraction

(B) Chemical reaction

(D) Photoelectric emission

Answer: (C) Heating effect of current

Explanation: A heater contains a resistive element that converts electrical energy to internal energy through interactions between moving electrons and the atomic lattice. The heat developed depends on current squared, element resistance, and operating time.

Q19. For a uniform wire made of one material, resistance is inversely proportional to which quantity?

[RRB Group D — 4 October 2018]

(A) Its length

(B) Its cross-sectional area

(D) Applied voltage only

Answer: (B) Its cross-sectional area

Explanation: The resistance of a uniform wire is R = ρL/A. With material and length unchanged, increasing cross-sectional area gives charges more conducting pathways and reduces resistance. Doubling the area halves resistance if temperature remains fixed.

Q18. What is the usual commercial unit of electrical energy?

[RRB Group D — 20 September 2018]

(A) Ampere

(B) Watt

(D) Ohm

Answer: (C) Kilowatt-hour

Explanation: Electricity meters record energy in kilowatt-hours, often called units. Multiply appliance power in kilowatts by the number of operating hours to obtain consumption. Watts measure power, whereas kilowatt-hours measure the total amount of energy delivered.

Q17. A wire carrying an electric current exhibits which magnetic behaviour?

[SSC MTS — 10 October 2017]

(A) It acts as a magnet

(B) It loses all magnetic effects

(D) It stops interacting with a compass

Answer: (A) It acts as a magnet

Explanation: The movement of charge through a conductor creates a magnetic field around it. A compass placed near the wire may deflect. The strength of the field increases with current, a principle exploited by electromagnets.

Q16. Current in a coil changes from +2 A to −2 A in 0.05 s, inducing 8 V. What is its self-inductance?

[RPSC Senior Teacher Grade II (Science) — 1 July 2017]

(A) 0.05 H

(B) 0.1 H

(D) 0.8 H

Answer: (B) 0.1 H

Explanation: The magnitude of induced voltage is L|ΔI/Δt|. Here the total current change is 4 A, occurring in 0.05 second. Consequently L = 8 × 0.05/4 = 0.1 henry. The voltage opposes the change in current.

Q15. What shunt resistance converts a 0.1 mA meter of internal resistance 100 Ω into an ammeter capable of 100 mA?

[CGPSC Assistant Engineer (Electrical) — 2017]

(A) 0.1 Ω

(B) 1 Ω

(D) 100 Ω

(E) 1000 Ω

Answer: (A) 0.1 Ω

Explanation: At full scale the meter carries 0.1 mA while the remainder, 99.9 mA, bypasses it through the shunt. Equating their voltage drops gives Rsh = (0.0001 × 100)/0.0999 ≈ 0.1001 Ω, approximately 0.1 Ω.

Q14. Three equal resistors each of resistance R are first connected in parallel and then in series. What is the series-to-parallel resistance ratio?

[UPSC NDA — 2016-II]

(A) 1:3

(B) 3:1

(D) 1:9

Answer: (C) 9:1

Explanation: In series three identical resistors have total resistance 3R. In parallel their equivalent resistance is R/3. Dividing series by parallel resistance gives (3R)/(R/3) = 9. The comparison demonstrates the large impact of circuit arrangement.

Q13. What is the protective purpose of a fuse in a household electric circuit?

[UPSC NDA — 18 September 2016]

(A) To prevent damage from excessive current

(B) To boost electric power

(D) To measure the resistance of appliances

Answer: (A) To prevent damage from excessive current

Explanation: The fuse acts as a sacrificial safety link. Too much current raises its temperature by resistive heating until it melts, disconnecting power. It can help protect against overloads and short circuits but is not a voltage regulator or measuring instrument.

Q12. When does the resistivity of a uniform conductor change even if only its geometric shape changes?

[UPSC NDA — 2016-II]

(A) It always changes with wire length

(B) It changes with cross-section only

(D) It becomes zero upon bending

Answer: (C) It does not change solely because of the dimensions

Explanation: Resistivity is an intrinsic property of the material at a given temperature. Changing length or area modifies resistance through R = ρL/A but leaves ρ unchanged. Only conditions such as composition and temperature alter the material’s resistivity.

Q11. What device converts chemical energy into electrical energy?

[RPSC RAS Prelims — 2015 (re-examination of 2013)]

(A) Electric motor

(B) Generator

(D) Battery

Answer: (D) Battery

Explanation: A battery contains electrochemical cells in which spontaneous redox reactions establish a potential difference between terminals. When connected to a circuit, chemical energy is converted to electrical energy. Generators instead convert mechanical energy, while motors consume electrical energy.

Q10. Three identical resistors have a series equivalent of 90 Ω. What is their parallel equivalent?

[UPSC NDA — 2015-II]

(A) 10 Ω

(B) 30 Ω

(D) 270 Ω

Answer: (A) 10 Ω

Explanation: Each resistor must be 90/3 = 30 Ω because three resistances add in series. When these three 30 Ω resistors are put in parallel, their combined resistance is 30/3 = 10 Ω.

Q9. Two same-material wires A and B have length ratio 1:2 and radii ratio 2:1. What is their resistance ratio R_A:R_B?

[UPSC NDA — 2014-I]

(A) 2:1

(B) 1:2

(D) 1:4

Answer: (C) 1:8

Explanation: Wire resistance is ρL/(πr²). Wire B is twice as long as A and has half its radius. These changes increase its resistance by 2 × 4 = 8, making the ratio of A to B equal to 1:8.

Q8. If reversing a battery makes the current through a semiconductor device nearly zero, what might the device be?

[UPSC NDA — 2014-I]

(A) Pure p-type bulk semiconductor

(B) Pure n-type bulk semiconductor

(D) A p-n junction diode

Answer: (D) A p-n junction diode

Explanation: A p-n junction diode conducts readily when forward-biased, but blocks most current when reverse-biased. Reversing the battery switches between those two conditions. The directional behaviour is called rectification and enables conversion of alternating signals to unidirectional current.

Q7. Which energy-conversion pairs are correct?
(i) Electric generator: mechanical to electrical;
(ii) Electric motor: electrical to mechanical;
(iii) Diesel engine: light to electrical;
(iv) Solar cell: chemical to mechanical.

[RPSC RAS Prelims — 2013]

(A) (i) and (ii)

(B) (i) and (iii)

(D) (iii) and (iv)

Answer: (A) (i) and (ii)

Explanation: A generator uses electromagnetic induction to convert mechanical work into electric energy. A motor converts electrical input to useful mechanical motion. Diesel engines derive mechanical output from fuel, and photovoltaic cells convert incident light into electrical energy.

Q6. A resistor carries current I under potential difference V. What does the product VI represent?

[UPSC NDA — 2013-I]

(A) Electrical resistance

(B) Total heat in any duration

(D) Rate of change of resistance

Answer: (C) Rate of electrical energy transfer

Explanation: Power is energy transferred per second, and its electrical expression is VI. When the resistor turns all supplied energy into thermal energy, this product is the heating rate. Total heat over a duration t is VIt, not merely VI.

Q5. A safety fuse marked 16 A is intended to interrupt the circuit under what condition?

[UPSC NDA — 2012-II]

(A) Current stays below 16 A

(B) The resistance becomes 16 Ω

(D) Current exceeds its rated 16 A value

Answer: (D) Current exceeds its rated 16 A value

Explanation: The number on a fuse indicates the current it is designed to carry safely under specified conditions. Sustained excessive current produces enough heat to melt the link and break the circuit. It is not a resistance or temperature rating.

Q4. An electrical wire initially has resistance 10 Ω. If stretched to ten times its original length at constant volume, what resistance results?

[UPSC NDA — 2012-I]

(A) 1 Ω

(B) 10 Ω

(D) 1000 Ω

Answer: (D) 1000 Ω

Explanation: If a wire is stretched to ten times its length without changing volume, its area becomes one-tenth. Resistance then rises by 10 × 10 = 100 times, since R = ρL/A. Therefore the new resistance is 1,000 Ω.

Q3. Two copper wires with lengths L and 2L have equal cross-sectional areas. What is the ratio of their resistivities?

[UPSC NDA — 2011-II]

(A) 4:1

(B) 2:1

(D) 1:2

Answer: (C) 1:1

Explanation: Resistivity depends on the conducting material and temperature rather than length. Because both wires are copper and are assumed to be at the same temperature, their resistivities are equal, despite the second wire having twice the electrical resistance.

Q2. For which type of current is a conventional clamp-on ammeter particularly useful?

[RPSC Technical Education Lecturer — 2011]

(A) Low alternating current only

(B) High alternating current

(D) High direct current only

Answer: (B) High alternating current

Explanation: Traditional clamp-on ammeters sense the alternating magnetic field surrounding a conductor using transformer action. They are useful for measuring substantial AC currents without breaking the circuit. Basic transformer-type clamps cannot directly measure steady DC current without additional sensor technology.

Q1. What is the greatest number of 60 W lamps usable simultaneously on a 240 V circuit protected by a 4 A fuse?

[UPSC NDA — 2010-I]

(A) 4

(B) 8

(D) 16

Answer: (D) 16

Explanation: Each lamp draws I = P/V = 60/240 = 0.25 A. Assuming identical lamps connected in parallel, a 4 A supply can carry 4/0.25 = 16 lamps at their rated current. The calculation assumes no additional connected loads.