Question

In the circuit shown in the figure, $V_s = V_m \sin 2t$ and $Z = 1 - j$. The value of $C$ is chosen such that the current $I$ is in phase with $V_s$. The value of $C$ in farad is, 

• A. $1/4 \, \text{F}$
• B. $1/2 \, \text{F}$
• C. $1/8 \, \text{F}$
• D. $1/6 \, \text{F}$

Hint:

When current and voltage are in phase in an AC circuit, the net reactance (or net susceptance) must be zero. Always enforce this condition using admittance for parallel circuits.

Answer 1

The Correct Option is A

Step 1: Identify the operating condition.
The current $I$ is stated to be in phase with the source voltage $V_s$. This condition implies that the overall impedance (or equivalently, the overall admittance) of the circuit must be purely resistive.
Hence, the net imaginary part of the total admittance must be zero.
Step 2: Determine the angular frequency.
Given
\[ V_s = V_m \sin 2t \]
Therefore, the angular frequency is
\[ \omega = 2 \, \text{rad/s} \]
Step 3: Find the admittance of impedance $Z$.
The given impedance is
\[ Z = 1 - j \]
Admittance is the reciprocal of impedance:
\[ Y_Z = \frac{1}{1 - j} \]
\[ Y_Z = \frac{1 + j}{(1)^2 + (1)^2} = \frac{1 + j}{2} \]
\[ Y_Z = 0.5 + j0.5 \]
Step 4: Find the admittance of the capacitor branch.
The admittance of a capacitor is given by
\[ Y_C = j \omega C \]
Substituting $\omega = 2$:
\[ Y_C = j (2C) \]
Step 5: Apply the in-phase condition.
For current to be in phase with voltage, the imaginary part of total admittance must be zero:
\[ \text{Im}(Y_Z + Y_C) = 0 \]
\[ 0.5 - 2C = 0 \]
Step 6: Solve for $C$.
\[ 2C = 0.5 \]
\[ C = 0.25 \, \text{F} \]
Step 7: Conclusion.
The value of capacitance required so that the current is in phase with the source voltage is
\[ \boxed{C = \frac{1}{4} \, \text{F}} \]

Answer 2

For the source current to be exactly in phase with \(V_s\), the net susceptance of the parallel combination made up of the fixed impedance \(Z\) and the capacitor \(C\) must vanish, i.e. the quadrature (reactive) part of \(Z\)'s admittance must be exactly balanced by the susceptance the capacitor contributes at the signal's angular frequency.

Since \(V_s = V_m \sin 2t\), the angular frequency is \(\omega = 2\) rad/s. The admittance of the fixed branch is \[ Y_Z = \frac{1}{Z} = \frac{1}{1-j} = \frac{1+j}{2} = 0.5+j0.5 \] so its reactive (imaginary) component has magnitude \(0.5\). The capacitor contributes a susceptance of magnitude \(\omega C = 2C\). Balancing the two quadrature contributions so that the total admittance is purely real requires \[ 2C = 0.5 \]

1. \(1/4\) F: Substituting gives \(2 \times \tfrac{1}{4} = 0.5\), which exactly balances the reactive component of \(Y_Z\).
2. \(1/2\) F: This gives \(2 \times \tfrac{1}{2} = 1\), overshooting the required \(0.5\) balance by double.
3. \(1/8\) F: This gives \(2 \times \tfrac{1}{8} = 0.25\), only half of what is needed to balance \(Y_Z\)'s reactive part.
4. \(1/6\) F: This gives \(2 \times \tfrac{1}{6} = 0.33\), still short of the required \(0.5\).

Only \(C = 1/4\) F makes the capacitor's contribution exactly cancel the reactive part of \(Y_Z\), leaving the total admittance purely real and the current in phase with the source voltage.

Therefore, the correct answer is \(1/4\) F.