Question

In a series LCR circuit, the inductive reactance (\(X_L\)) is 10 \(\Omega\) and the capacitive reactance (\(X_C\)) is 4 \(\Omega\). The resistance (R) in the circuit is 6 \(\Omega\). The power factor of the circuit is:

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\frac{\sqrt{3}}{2}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{1}{2\sqrt{2}}\)

Hint:

The power factor ranges from 0 to 1. If \(R = X_{net}\), the phase angle is 45° and the power factor is always \(1/\sqrt{2} \approx 0.707\).

Answer 1

The Correct Option is A

Step 1: Calculate the net reactance \(X\): \[X = X_L - X_C = 10 - 4 = 6 \ \Omega\]
Step 2: Calculate the impedance \(Z\): \[Z = \sqrt{R^2 + X^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = 6\sqrt{2} \ \Omega\] [Image of the impedance triangle for a series LCR circuit]
Step 3: Calculate the power factor (\(\cos \phi\)): \[\cos \phi = \frac{R}{Z} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}\]