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: Get the resistance from the DC test.
With DC there is no reactance, so $R=\frac{200}{1}=200\ \Omega$.

Step 2: Get the impedance from the AC test.
With AC, $Z=\frac{200}{0.5}=400\ \Omega$.

Step 3: Find the reactance.
$Z^2=R^2+X_L^2$, so $X_L^2=400^2-200^2=120000$, giving $X_L=200\sqrt3\ \Omega$.

Step 4: Solve for frequency.
$X_L=2\pi fL$ with $L=\frac{2\sqrt3}{\pi}$ gives $200\sqrt3=4\sqrt3\,f$, so $f=50$ Hz. \[ \boxed{50\text{ Hz}} \]