Question

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

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

Answer 1

Answer: (C)

Given, $L=25mH$
$=25\times 10^{-3}H$
$f=50Hz$
$V_{rms}=220V$ and $f_2=50Hz$
$X_L=2\pi fL$
$=2\times \frac{22}{7}\times 50\times 25\times 10^{-3}\Omega$
The rms current in the circuit is
$I_{\text{rms }}=\frac{V_{\text{rms }}}{X_L}$
$=\frac{220}{2\times \frac{22}{7}\times 50\times 25\times 10^{-3}}$
$=\frac{7\times 1000}{2\times 5\times 25}$
$I_{\text{rms }}=28A$