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,

Question

Write the principle of working of an ac generator. Draw its labelled diagram and explain its working.

Answer 1

Step 1: Angular frequency from $V_s=V_m\sin 2t$ is $\omega=2$ rad/s.
Step 2: Admittance of the fixed branch: $Y_Z = \dfrac{1}{1-j} = 0.5+j0.5$, so its reactive part has magnitude $0.5$.
Step 3: The capacitor's susceptance is $\omega C = 2C$; for the current to be in phase with $V_s$, this must balance the reactive part of $Y_Z$: $2C = 0.5$.
\[ \boxed{C = \frac{1}{4} \text{ F}} \]

Answer 2

Working directly in terms of currents rather than admittances, take $V_s = 1\angle 0^\circ$ as a reference phasor. The current drawn by the fixed branch is \[ I_Z = \frac{V_s}{Z} = \frac{1}{1-j} = 0.5+j0.5 \text{ A} \] which has an in-phase component of $0.5$ A and a quadrature component of $0.5$ A. The capacitor draws a purely quadrature current $I_C = j\omega C V_s = j2C$ A. For the total current $I_Z+I_C$ to line up exactly with $V_s$ (zero net phase angle), the capacitor's contribution must match $0.5$ A in magnitude at $\omega=2$ rad/s: \[ 2C = 0.5 \]

$1/4$ F: Gives a capacitor current of $2\times 0.25 = 0.5$ A, matching the branch's quadrature current exactly.
$1/2$ F: Gives $2\times0.5=1$ A of capacitor current, twice what is needed.
$1/8$ F: Gives $2\times0.125=0.25$ A, only half of what is needed.
$1/6$ F: Gives $2\times(1/6)=0.33$ A, still short of the required $0.5$ A.

Therefore, the correct answer is $1/4$ F.

Answer 3

Working directly in terms of currents rather than admittances, take $V_s = 1\angle 0^\circ$ as a reference phasor. The current drawn by the fixed branch is \[ I_Z = \frac{V_s}{Z} = \frac{1}{1-j} = 0.5+j0.5 \text{ A} \] which has an in-phase component of $0.5$ A and a quadrature component of $0.5$ A. The capacitor draws a purely quadrature current $I_C = j\omega C V_s = j2C$ A. For the total current $I_Z+I_C$ to line up exactly with $V_s$ (zero net phase angle), the capacitor's contribution must match $0.5$ A in magnitude at $\omega=2$ rad/s: \[ 2C = 0.5 \]

$1/4$ F: Gives a capacitor current of $2\times 0.25 = 0.5$ A, matching the branch's quadrature current exactly.
$1/2$ F: Gives $2\times0.5=1$ A of capacitor current, twice what is needed.
$1/8$ F: Gives $2\times0.125=0.25$ A, only half of what is needed.
$1/6$ F: Gives $2\times(1/6)=0.33$ A, still short of the required $0.5$ A.

Therefore, the correct answer is $1/4$ F.

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,

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The Correct Option is A

Approach Solution - 1

Approach Solution -2

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