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
(C) P = R/I
(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
(C) 3.6 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
(C) Elapsed time
(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
(C) 36 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
(C) 0.9 A and 0.9 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
(C) 2.25 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
(C) Power divided by voltage
(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
(C) Salt solution
(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
(C) Cross-sectional area of that 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
(C) J = ρE²
(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.
