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
(C) 60 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
(C) 1:8
(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
(C) An intrinsic semiconductor sample
(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)
(C) (ii) and (iv)
(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
(C) Rate of electrical energy transfer
(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
(C) Ambient temperature reaches 16°C
(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
(C) 100 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
(C) 1:1
(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
(C) Low direct current only
(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
(C) 12
(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.
