Question

In the given circuit below, voltage \(V_C(t)\) is:

• A. \( 0.89 \cos(1000t - 6(3)43^\circ) \) V
• B. \( 0.89 \cos(1000t + 6(3)43^\circ) \) V
• C. \( 0.45 \cos(1000t + 26.57^\circ) \) V
• D. \( 0.45 \cos(1000t - 26.57^\circ) \) V

Hint:

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.

Answer 1

The Correct Option is A

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)