Question

When a d.c. voltage of 200 V is applied to a coil of self-inductance $\left(\frac{2\sqrt{3}}{\pi}\right)$ H, a current of 1 A flows through it. But by replacing d.c. source with a.c. source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of a.c. supply is

• A. 100 Hz
• B. 60 Hz
• C. 75 Hz
• D. 50 Hz

Hint:

Notice that the current drops exactly in half, meaning the impedance is twice the pure resistance ($Z = 2R$). In an $RL$ circuit, if $Z = 2R$, then $X_L = \sqrt{3}R$ holds by definition! This lets you equate $2\pi f L = 200\sqrt{3}$ instantly on scratch paper.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A coil has both resistance ($R$) and self-inductance ($L$). We are given its behavior under a D.C. voltage and an A.C. voltage source, and we need to calculate the frequency ($f$) of the A.C. supply.

Step 2: Detailed Explanation:
With a D.C. source, inductive reactance is zero ($X_L = 0$), so only resistance opposes the current: $$ R = \frac{V_{\text{dc}}}{I_{\text{dc}}} = \frac{200\ \text{V}}{1\ \text{A}} = 200\ \Omega $$ With an A.C. source, total impedance ($Z$) opposes the current: $$ Z = \frac{V_{\text{ac}}}{I_{\text{ac}}} = \frac{200\ \text{V}}{0.5\ \text{A}} = 400\ \Omega $$ Using the impedance relation $Z^2 = R^2 + X_L^2$: $$ 400^2 = 200^2 + X_L^2 \implies X_L^2 = 160000 - 40000 = 120000 $$ $$ X_L = \sqrt{120000} = 200\sqrt{3}\ \Omega $$ The inductive reactance formula is $X_L = 2\pi f L$. Substituting the given value $L = \frac{2\sqrt{3}}{\pi}$: $$ 200\sqrt{3} = 2\pi f \left(\frac{2\sqrt{3}}{\pi}\right) $$ $$ 200\sqrt{3} = 4\sqrt{3}f \implies f = \frac{200}{4} = 50\ \text{Hz} $$

Step 3: Final Answer:
The frequency of the a.c. supply is 50 Hz, which corresponds to option (D).