Electric Circuits Practice Exam — Free

The Thevenin voltage is the open-circuit voltage at a, so write KCL at node a with nothing connected to the terminals (currents leaving the node, $V_a$ unknown):

Multiplying through by 12 gives $V_a - 36 + 2V_a - 24 = 0$, so $3V_a = 60$ and $V_{th} = V_a = 20\,\text{V}$.

For $R_{th}$, deactivate the independent sources: the voltage source becomes a short circuit and the current source an open circuit. Looking in from a–b, the 12 Ω and 6 Ω resistors are then in parallel:

Superposition confirms $V_{th}$. The voltage source alone forms a divider, giving $36\times\tfrac{6}{18} = 12\,\text{V}$ at a; the current source alone pushes 2 A through $12\parallel 6 = 4\,\Omega$, giving 8 V; the two contributions add to 20 V. The matching Norton equivalent is $I_N = 20/4 = 5\,\text{A}$ in parallel with 4 Ω.

Option A (12 V) is the tempting mistake: it is the voltage-source contribution alone, forgetting that the current source also lifts node a. Option C adds the resistors in series instead of in parallel when finding $R_{th}$, and option D computes the current-source contribution with the 6 Ω resistor alone instead of $12\parallel 6$.

$V_{th} = 20\,\text{V}$, $R_{th} = 4\,\Omega$ (option B).

The time constant of a series RC circuit is $\tau = RC = (4000\,\Omega)(50\times 10^{-6}\,\text{F}) = 0.20\,\text{s}$.

A capacitor charging from rest toward a DC source follows $v_C(t) = V_s\left(1 - e^{-t/\tau}\right)$, because the capacitor voltage cannot jump and approaches $V_s$ exponentially. At $t = 0.30$ s, $t/\tau = 1.5$ and $e^{-1.5} = 0.2231$:

The current is set by the voltage left across the resistor (KVL around the loop):

The same value comes from the decaying form $i(t) = \frac{V_s}{R}\,e^{-t/\tau} = (5\,\text{mA})(0.2231) = 1.12\,\text{mA}$, which confirms the arithmetic.

Sanity check: after one time constant the capacitor reaches 63 % of its final voltage and after two it reaches 86 %; at 1.5 time constants it sits at 78 %, matching $15.5/20$.

$v_C(0.30\,\text{s}) \approx 15.5\,\text{V}$ and $i \approx 1.12\,\text{mA}$.

(a) Apparent power follows from the definition of power factor, $\text{pf} = P/S$:

Reactive power comes from the power triangle: $Q_1 = \sqrt{S^2 - P^2} = \sqrt{12.5^2 - 10^2} = 7.5\,\text{kvar}$ (equivalently $P\tan\theta_1$ with $\theta_1 = \cos^{-1}0.80 = 36.87^\circ$ and $\tan\theta_1 = 0.75$). The line current is $I_1 = S/V = 12\,500/240 = 52.1\,\text{A}$.

(b) A parallel capacitor leaves the real power $P$ unchanged and only reduces the net reactive power. The target angle is $\theta_2 = \cos^{-1}0.95 = 18.19^\circ$, so the reactive power that may remain is

The capacitor must supply $Q_C = Q_1 - Q_2 = 7.50 - 3.29 = 4.21\,\text{kvar}$. For a capacitor across voltage $V$, $Q_C = \omega C V^2$ with $\omega = 2\pi(60) = 377\,\text{rad/s}$:

The new apparent power is $S_2 = P/0.95 = 10.53\,\text{kVA}$, so $I_2 = 10\,530/240 = 43.9\,\text{A}$, a 16 % reduction in line current for the same real power. Check: $Q_2 = \sqrt{10.53^2 - 10^2} = 3.29\,\text{kvar}$, consistent with the value used above.

(a) $S = 12.5\,\text{kVA}$, $Q = 7.5\,\text{kvar}$, $I = 52.1\,\text{A}$; (b) $C \approx 194\,\mu\text{F}$, $I \approx 43.9\,\text{A}$.