Circuit Analysis (DC and AC Steady State) practice questions

Answer: A) 3 A

The Norton current equals the Thevenin short-circuit current, I_N = V_th/R_th = 12 V / 4 Ω = 3 A, and R_N = R_th.

• ACorrect. The Norton current is the short-circuit current: I_N = V_th/R_th = 12 V / 4 Ω = 3 A, with R_N = R_th = 4 Ω in parallel.
• BIncorrect. 12 × 4 = 48 multiplies voltage by resistance, which gives V·Ω, not a current. Ohm's law gives I = V/R.
• CIncorrect. 4/12 ≈ 0.33 is R_th/V_th, the inverse of the correct ratio. Its units are 1/A, not amperes. The short-circuit current is V_th/R_th = 3 A.
• DIncorrect. This uses the Thevenin voltage value as the current and ignores R_th. Shorting the output terminals drives V_th through R_th, giving 3 A.

Answer: B) 3 μF

Capacitors in series combine like resistors in parallel: 1/C = 1/C1 + 1/C2, which gives 3 μF here. A series equivalent is always smaller than the smallest capacitor.

• AIncorrect. Adding capacitances directly gives the equivalent for capacitors in parallel, not in series.
• BCorrect. For series capacitors, 1/C = 1/4 + 1/12 = 1/3, so C = 3 μF (equivalently C1C2/(C1 + C2) = 48/16 = 3 μF).
• CIncorrect. This is the arithmetic mean of the two values, which isn't how capacitances combine.
• DIncorrect. This is the sum of reciprocals, 1/4 + 1/12 = 1/3, without the final inversion. The equivalent is the reciprocal of that sum, 3 μF.

Answer: D) 4 A

• AIncorrect. The current divides.
• BIncorrect. That's the 4-Ω branch.
• CIncorrect. The branches aren't equal.
• DCorrect. \(6 \times 4/(2 + 4)\).

Answer: B) 5 μF

• AIncorrect. That's parallel.
• BCorrect. Series capacitors combine like parallel resistors.
• CIncorrect. Series capacitance is smaller.
• DIncorrect. Use product over sum.

Answer: B) 2 A

• AIncorrect. The sources oppose.
• BCorrect. (12 − 6)/3.
• CIncorrect. Include the opposing source.
• DIncorrect. That multiplies instead.

Answer: A) Zero

• ACorrect. The bridge ratios are equal.
• BIncorrect. Balance means zero detector current.
• CIncorrect. It's zero.
• DIncorrect. It's zero.

Answer: A) 20 W

For a resistor driven by a sinusoid, average power is V_rms²/R = 100/5 = 20 W. Equivalently, it is V_m²/(2R) using the peak value.

• ACorrect. P = V_rms²/R = (10)²/5 = 20 W.
• BIncorrect. This answer converts to the peak value (V_m ≈ 14.14 V) and then uses V_m²/R without the factor of ½. The correct average power is V_m²/(2R) = 20 W.
• CIncorrect. This answer treats 10 V as the peak value, giving (10/√2)²/5 = 10 W. The stem says the 10 V is already an rms value.
• DIncorrect. This answer multiplies the peak voltage (14.14 V) by the rms current (2 A). Average power uses V_rms × I_rms = 10 × 2 = 20 W.

Answer: D) 600 VAR

Reactive power is Q = Vrms·Irms·sinθ = 120 × 10 × 0.5 = 600 VAR. VI cosθ gives real power, and VI alone gives apparent power.

• AIncorrect. This is half the correct value, as if VI sinθ were divided by 2. Rms quantities need no factor of 1/2.
• BIncorrect. 120 × 10 × cos 30° ≈ 1,039 is the real power P in watts, not the reactive power.
• CIncorrect. 120 × 10 = 1,200 is the apparent power S in VA, with no phase-angle factor applied.
• DCorrect. Q = VI sinθ = 120 × 10 × sin 30° = 1,200 × 0.5 = 600 VAR.

Answer: A) 2 mH

Uncoupled inductors in parallel combine like resistors in parallel: L = L1L2/(L1 + L2) = 2 mH. A parallel equivalent is always smaller than the smallest inductor.

• ACorrect. Uncoupled parallel inductors combine like parallel resistors: 1/L = 1/6 + 1/3 = 1/2, so L = 2 mH.
• BIncorrect. Adding inductances directly gives the equivalent for uncoupled inductors in series, not in parallel.
• CIncorrect. This is the arithmetic mean of the two values, which isn't how inductances combine.
• DIncorrect. This is the sum of reciprocals, 1/6 + 1/3 = 1/2, without the final inversion. The equivalent is the reciprocal of that sum, 2 mH.

Answer: B) 10 V in series with 5 Ω

• AIncorrect. Multiply the current by the resistance.
• BCorrect. \(V_{th} = I_N R_N\).
• CIncorrect. The resistance stays 5 Ω.
• DIncorrect. That divides instead.