Question

An inductor of 1 H is connected across a 220 V, 50 Hz supply. The peak value of the current is approximately

• A. 0.5 A
• B. 0.7 A
• C. 1 A
• D. 14 A

Answer 1

Answer: (C)

1 A

Given : $L=1H$ $\nu =50$ Hz $V_{rms}=220$ volts

Inductive reactance $X_L=wL=2\pi \nu L$

$\therefore$ $X_L=2\pi \times 50\times 1$

We get $X_L=314\Omega$

Rms value of Current $I_{rms}=\frac{V_{rms}}{X_L}$

$\therefore$ $I=\frac{220}{314}=0.7$ A

Peak value of current $I_o=\sqrt{2}{I}_{rms}$

$I_o=1.414\times 0.7=1$ A