Q.A pure inductor of inductance $0.2\ \text{H}$ is connected across a $230\ \text{V}$ (rms), $50\ \text{Hz}$ AC supply. Calculate

Given: $L=0.2\ \text{H}$, $V_{rms}=230\ \text{V}$, $f=50\ \text{Hz}$.

1. Inductive reactance.

   $X_L = 2\pi f L = 2\pi(50)(0.2) \approx 62.83\ \Omega$

2. RMS current.

   $I_{rms} = \frac{V_{rms}}{X_L} = \frac{230}{62.83} \approx 3.66\ \text{A}$

3. Peak voltage. Since the supply's own peak voltage appears entirely across the inductor (the only element in the circuit),

   $V_0 = V_{rms}\sqrt2 = 230\times1.414 \approx 325.3\ \text{V}$

   ✓Final answer

   $X_L\approx62.8\ \Omega$, $I_{rms}\approx3.66\ \text{A}$, peak voltage $\approx325.3\ \text{V}$.

Compute $X_L=2\pi fL$, then apply $I_{rms}=V_{rms}/X_L$, and separately convert the given rms voltage to its peak value.

• Using $f$ directly in place of $\omega=2\pi f$ inside the reactance formula.
• Forgetting part (c) is independent of the inductor calculation -- it is simply the supply's own peak voltage, since the inductor is the only element present.