AC Circuits MCQ [Free PDF] – 14 Objective Questions with Answers

The paper uses the signed impedance angle: Z = R − jX_C, so tan φ = −X_C/R. (As a pure magnitude, the current leads by tan⁻¹(X_C/R).)

In a parallel RLC circuit the source current is the phasor sum of I_R, I_L and I_C. The phase angle is the angle between the common voltage and that resultant current.

B — R + j0. A pure resistance has only a real part and zero imaginary (reactive) part; equivalently R∠0°.

An ideal inductor stores energy in its magnetic field and returns it to the source. Voltage and current are 90° apart, so the average real power over a complete cycle is zero.

The equivalent resistance is not simply R; it is obtained after adding the two admittances. Complex-conjugate multiplication gives |4+j3|²=25. Adding admittances cancels ±j3/25 exactly. Key relation: Yeq = 1/(4+j3)+1/(4−j3)=8/25=0.32 S; Zeq=1/Yeq=3.125 Ω; I=V/Zeq=32 A; PF=cos0°=1.

Exam focus: For parallel complex branches, convert Z to Y first; it prevents sign and denominator mistakes. Common trap: Directly adding parallel impedances or assuming the equivalent resistance equals the real part of each branch.

D — 1/Ztotal = 1/Z1 + 1/Z2. In a parallel circuit admittances add, so the reciprocal relation gives Ztotal = Z1Z2/(Z1 + Z2).

C — 4 A. The resistor and capacitor currents are 90° apart, so I² = IR² + IC² → IC = √(25 − 9) = 4 A.

C — The applied voltage. All branches of a parallel circuit are connected between the same two nodes, so they share the same voltage; branch currents differ.

D — Zero. In a pure capacitor current leads voltage by 90°, so P_avg = VI cos 90° = 0; energy is only stored and returned.

A — Zero. In a pure inductor the current lags the voltage by 90°, so power factor = cos 90° = 0 (zero lagging).

B — An inductive circuit. Net reactance X = X_L − X_C is positive, so current lags voltage.

For a coil Z = R + jωL. As ω → ∞, ωL dominates R, so the impedance becomes nearly purely inductive and the voltage leads current by 90°.

B — Z = R + j(X_L − X_C) Ω. In series, R, +jX_L and −jX_C add as complex quantities.

The phase difference identifies the element before any calculation: voltage is at +45° and current at −45°, so voltage leads current by 90°, which is the ideal-inductor relation. Then Xₗ = 628/10 = 62.8 Ω and L = Xₗ/ω = 62.8/314 = 0.2 H = 200 mH.