In the given circuit below, voltage \(V_C(t)\) is:
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AC Parallel Circuits. Use impedances: \(Z_R = R\), \(Z_C = 1/(j\omega C)\). Equivalent impedance for parallel components: \(1/Z_{eq = 1/Z_1 + 1/Z_2 + ...\). Voltage across parallel components \(V = I_{source \times Z_{eq\). Convert complex numbers to polar form (\(M\angle\phi\)) for magnitude and phase.
This is an AC circuit analysis problem The current source is \(I(t) = \cos(1000t)\) A, so the phasor current is \(I = 1 \angle 0^\circ\) A, and the angular frequency is \(\omega = 1000\) rad/s
The resistor \(R = 2 \, \Omega\)
The capacitor \(C = 1 \, \text{mF} = 1 \times 10^{-3} \, \text{F}\)
The impedance of the capacitor is \(Z_C = \frac{1}{j\omega C} = \frac{1}{j(1000)(10^{-3})} = \frac{1}{j} = -j \, \Omega\)
The resistor and capacitor are in parallel The voltage \(V_C\) across the capacitor is the same as the voltage across the parallel combination Let the equivalent impedance of the parallel combination be \(Z_{eq}\)
$$ \frac{1}{Z_{eq}} = \frac{1}{R} + \frac{1}{Z_C} = \frac{1}{2} + \frac{1}{-j} = \frac{1}{2} + j $$
$$ Z_{eq} = \frac{1}{0 5 + j} $$
Multiply numerator and denominator by the conjugate (0 5 - j):
$$ Z_{eq} = \frac{0 5 - j}{(0 5 + j)(0 5 - j)} = \frac{0 5 - j}{0 5^2 - (j)^2} = \frac{0 5 - j}{0 25 - (-1)} = \frac{0 5 - j}{(1)25} $$
$$ Z_{eq} = \frac{0 5}{(1)25} - j\frac{1}{(1)25} = 0 4 - j0 8 \, \Omega $$
The voltage across the parallel combination (which is \(V_C\)) is given by Ohm's law in phasor form: \(V_C = I \times Z_{eq}\)
$$ V_C = (1 \angle 0^\circ \, \text{A}) \times (0 4 - j0 8 \, \Omega) = 0 4 - j0 8 \, \text{V} $$
Convert the phasor voltage \(V_C\) to polar form (Magnitude \(\angle\) Angle):
Magnitude \(|V_C| = \sqrt{0 4^2 + (-0 8)^2} = \sqrt{0 16 + 0 64} = \sqrt{0 80} \approx 0 894\) V
Angle \(\phi = \arctan\left(\frac{-0 8}{0 4}\right) = \arctan(-2) \approx -6(3)43^\circ\)
So, the phasor voltage is \(V_C \approx 0 894 \angle -6(3)43^\circ\) V
Converting back to the time domain:
$$ V_C(t) = |V_C| \cos(\omega t + \phi) \approx 0 894 \cos(1000t - 6(3)43^\circ) \, \text{V} $$
This matches option (1)
