Question

A resistor of \( 400\,\Omega \) and an inductance of \( 0.4\,\text{H} \) are in series with an a.c. source of e.m.f. \( E = 200\sqrt{2}\sin(1000t) \). The power factor of the circuit is

• A. $0.4$
• B. $0.5$
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( \frac{1}{\sqrt{3}} \)

Hint:

If $R = X_L$, then: \[ \cos\phi = \frac{1}{\sqrt{2}} \]

Answer 1

The Correct Option is C

Concept: For an $R$-$L$ series circuit, \[ \text{Power factor} = \cos\phi = \frac{R}{Z}, \quad Z = \sqrt{R^2 + X_L^2}, \quad X_L = \omega L \]

Step 1: Identify angular frequency.
Given: \[ E = 200\sqrt{2}\sin(1000t) \Rightarrow \omega = 1000\ \text{rad/s} \]

Step 2: Calculate inductive reactance.
\[ X_L = \omega L = 1000 \times 0.4 = 400\ \Omega \]

Step 3: Calculate impedance.
\[ Z = \sqrt{R^2 + X_L^2} = \sqrt{400^2 + 400^2} = \sqrt{2 \times 400^2} = 400\sqrt{2} \]

Step 4: Find power factor.
\[ \cos\phi = \frac{R}{Z} = \frac{400}{400\sqrt{2}} = \frac{1}{\sqrt{2}} \] Answer: \( \frac{1}{\sqrt{2}} \)