Question

An inductor of 1H 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. 1.4 A

Answer 1

The correct option is C 1 A

Peak value of current $\left(i_0\right)=\frac{E_0}{X_L}$
or $i_0=\frac{E_0}{L\omega }$
$\therefore$ $i_0=\frac{220\times \sqrt{2}}{1\times 2\pi \times 50}$
$i_0=\frac{220\times \sqrt{2}}{100\pi }=\frac{200\times \sqrt{2}}{100\times 314}$
$i_0=1A$