Question

An inductor of 1 henry is connected across a $220v,50Hz$ supply. The peak value of the current is approximate:

• A. $0.5A$
• B. $0.7A$
• C. $1A$
• D. $1.4A$

Answer 1

The correct option is C $1A$

Peak voltage = rms voltage $\times \sqrt{2}$

$=220\times \sqrt{2}$

$=311V$

peak current= $\frac{peakvoltage}{inductivereactance}$

$=\frac{311}{\omega L}$

$=\frac{311}{2\pi \times 50\times 1}$ $(\because \omega =2\pi f)$

$=1A.$