Alternating Current Circuits: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A current i(t)=(2+4sinωt) A flows through a resistor of resistance 5Ω. Here 2 A is a steady (DC) part and 4 A is the peak value of the sinusoidal part. What is the time-averaged power dissipated in the resistor? Exactly one option is correct.
A40 W
B100 W
C60 W
D117 W
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Answer: C
Average of i2: ⟨(2+4sinωt)2⟩=4+16⟨sinωt⟩+16⟨sin2ωt⟩=4+0+8=12A2. So Irms=12 A and P=12×5=60 W.
Traps: ignoring the DC part gives 40 W; using 22+42 gives 100 W; adding rms values linearly, (2+22)2×5≈117 W. The cross term averages to zero, so the mean squares add.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A resistor of resistance R is connected in series with a coil that has both resistance r and inductance L, across a sinusoidal AC supply. Ideal AC voltmeters (reading rms values) give: 30 V across the resistor, 40 V across the coil, and 60 V across the supply. What is the power factor of the coil alone (that is, r divided by the coil's impedance)? Exactly one option is correct.
A43
B3629
C125
D2411
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Answer: D
The same current flows through both, so VR is along the current phasor and Vcoil leads it by the coil angle θ, where cosθ=r/Zcoil. The phasors add: V2=VR2+Vcoil2+2VRVcoilcosθ.
3600=900+1600+2400cosθ⇒cosθ=24001100=2411.
Trap: 3629 is the power factor of the whole circuit, (VR+Vcoilcosθ)/V=(30+40⋅2411)/60. The ratio 30/40 means nothing here. The readings do not add arithmetically (30+40=60) because the phasors are not parallel.
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
A coil of resistance 30Ω and inductance 80 mH is connected to a sinusoidal source of rms voltage 100 V and angular frequency ω=500 rad/s. A capacitor of finite capacitance C is now inserted in series with the coil, and the rms current is found to be exactly the same as before. Use the convention that a phase angle compares the current with the source voltage. Which option gives C and the phase of the current after the capacitor is inserted? Exactly one option is correct.
AC=12.5μF; the current leads the source voltage by tan−1(4/3)
BC=25μF; the current leads the source voltage by tan−1(4/3)
CC=25μF; the current lags the source voltage by tan−1(4/3)
DC=50μF; the current is in phase with the source voltage
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Answer: B
XL=ωL=500×0.08=40Ω. Same rms current with the same source means the same impedance, so ∣XL−XC∣=XL.
Either XC=0 (no capacitor, excluded because C is finite) or XC=2XL=80Ω, giving C=ωXC1=500×801=25μF.
Now XL−XC=−40Ω: the circuit is capacitive and tan∣ϕ∣=40/30=4/3, so the current leads by tan−1(4/3)≈53∘ (before, it lagged by the same angle).
Insight: equal current magnitude means the reactance flips sign with equal magnitude. 50 μF is the resonance value (XC=40Ω), which would increase the current.
Difficulty: Very hard
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