Question

A series LCR circuit is connected to an AC source of \(220\) V, \(50\) Hz. The circuit contains resistance of \(100\ \Omega\), capacitor of capacitive reactance \(X_{C}\ \Omega\) and inductor of inductive reactance \(X_{L}=150\ \Omega\). If the power factor of the circuit is \(\frac{8}{10}\) and, \(X_{L}>X_{C}\), then the value of \(X_{C}\) is

• A. \(50\ \Omega\)
• B. \(140\ \Omega\)
• C. \(75\ \Omega\)
• D. \(100\ \Omega\)

Hint:

Use power factor equal to R over Z to find Z, then find the net reactance.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Power factor \(\cos\phi=\dfrac{R}{Z}\), so \(Z=\dfrac{100}{0.8}=125\ \Omega\).

Step 2: Net reactance
\(Z^{2}=R^{2}+(X_{L}-X_{C})^{2}\), so \((X_{L}-X_{C})^{2}=125^{2}-100^{2}=5625\) and \(X_{L}-X_{C}=75\ \Omega\) (positive as \(X_{L}>X_{C}\)).

Step 3: Calculation
\(X_{C}=150-75=75\ \Omega\).

Final Answer:
The capacitive reactance is 75 ohm. \[ \boxed{\text{(C) }75\ \Omega} \]