Question

For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor \(C'\), when joined with the capacitor \(C\) present in the circuit, makes the power factor of the circuit unity. The capacitor \(C'\), must have been connected in :

• A. series with \(C\) and has a magnitude \(\frac{1-\omega^2 LC}{\omega^2 L}\).
• B. series with \(C\) and has a magnitude \(\frac{C}{(\omega^2 LC-1)}\).
• C. parallel with \(C\) and has a magnitude \(\frac{C}{(\omega^2 LC-1)}\).
• D. parallel with \(C\) and has a magnitude \(\frac{1-\omega^2 LC}{\omega^2 L}\).

Hint:

Current leading voltage means \(X_C \gt X_L\); unity power factor needs \(X_L = X_C\), so the effective capacitance must be increased to \(1/(\omega^2 L)\), which needs a parallel capacitor.

Answer 1

The Correct Option is D

Current leads the voltage, so the circuit is capacitive: \(\frac{1}{\omega C} \gt \omega L\).
Unity power factor requires net reactance zero: \(\frac{1}{\omega C_{eq}} = \omega L\), i.e. \(C_{eq} = \frac{1}{\omega^2 L}\).
Since \(\frac{1}{\omega C} \gt \omega L\), we have \(C \lt \frac{1}{\omega^2 L}\), so the capacitance must be increased. Capacitors in parallel add, so \(C'\) is in parallel with \(C\).
\(C + C' = \frac{1}{\omega^2 L}\) gives \(C' = \frac{1}{\omega^2 L} - C = \frac{1-\omega^2 LC}{\omega^2 L}\).
Answer: (4) parallel with C and has a magnitude \(\frac{1-\omega^2 LC}{\omega^2 L}\)