Question

For series LCR circuit, the current is observed to lead the applied voltage. An additional capacitor \(C_{1}\) when joined with capacitor C, makes the power factor of the circuit unity. The capacitor \(C_{1}\) must be connected in

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

Hint:

A leading current means capacitive reactance dominates; the extra capacitor must make the circuit resonant by joining in parallel.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Current leads voltage when \(X_C>X_L\). Power factor becomes one at resonance, \(X_L=X_C\).

Step 2: Choose the Connection
To reduce the net capacitive reactance, the capacitance must be increased. Capacitors in parallel add up, so \(C_1\) is joined in parallel.

Step 3: Resonance Condition
\(\omega L=\frac{1}{\omega(C+C_1)}\), so \(C+C_1=\frac{1}{\omega^{2}L}\).

Step 4: Solve
\(C_1=\frac{1}{\omega^{2}L}-C=\frac{1-\omega^{2}LC}{\omega^{2}L}\), parallel with C, option (C).

Final Answer:
The capacitor is joined in parallel and has capacitance (1 - w squared L C) over w squared L. \[ \boxed{\text{(C) }\text{parallel with C and has magnitude }\dfrac{(1-\omega^{2}LC)}{\omega^{2}L}} \]