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
(C) 1 joule
(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
(C) Potential difference across 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
(C) Unit charge and unit time
(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
(C) 2 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
(C) 3.6 × 10⁶ 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
(C) 0.5 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
(C) An unlimited number of electrons
(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
(C) All resistors become short-circuited
(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
(C) 300 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
(C) 250 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.