Question

A pure inductor of $25 \,mH$ is connected to a source of $220\, V$. Given the frequency of the source as $50\, Hz$, the $rms$ current in the circuit is

• A. 7 A
• B. 14 A
• C. 28 A
• D. 42 A

Answer 1

Answer: (C)

Given, $L =25\, mH$
$=25 \times 10^{-3} H$
$f =50\, Hz$
$V_{ rms }=220\, V$ and $f_{2} =50\, Hz$
$X_{L} =2 \pi f L$
$=2 \times \frac{22}{7} \times 50 \times 25 \times 10^{-3} \Omega$
The rms current in the circuit is
$I_{\text {rms }} =\frac{V_{\text {rms }}}{X_{L}}$
$=\frac{220}{2 \times \frac{22}{7} \times 50 \times 25 \times 10^{-3}}$
$=\frac{7 \times 1000}{2 \times 5 \times 25}$
$I_{\text {rms }} =28\, A$