DC Circuits MCQ [Free PDF] – 3 Objective Questions with Answers

At fixed supply voltage, doubling resistance halves current and halves power because P=V²/R. Given V=20 V and R=10 Ω: I=V/R=2 A and P=VI=40 W. If R doubles to 20 Ω at the same 20 V, I=1 A and P=20 W. Key relation: I=V/R; P=VI=I²R=V²/R.

Exam focus: At fixed V, power varies inversely with R. Common trap: Do not assume power stays constant when resistance changes on a fixed-voltage source.

Equivalent resistance is found from reciprocal addition, total current is the sum of branch currents, and total real power is the sum of branch powers. For 6 Ω || 3 Ω, Req=(6×3)/(6+3)=2 Ω. Total current=12/2=6 A. Key relation: 1/R_eq=Σ(1/R_i); I_i=V/R_i; P_total=V I_total.

Exam focus: Solve branch currents first when the same supply voltage appears across every resistor. Common trap: Adding parallel resistances directly is incorrect.

B — 1 A. The 2 Ω and 6 Ω resistors are in series, so the 8 V source drives 8/(2 + 6) = 1 A.