Question

A pure inductive coil of $30\,mH$ is connected to an $AC$ source of $220 \,V$ , $50 \,Hz$ . The $rms$ current in the coil is

• A. $50.35 \,A$
• B. $23.4 \,A$
• C. $30.5 \,A$
• D. $12.3 \,A$

Answer 1

Answer: (B)

$23.4 \,A$

Answer 2

Given $L = 30 \,mH$
$V_{rms}=220\,V$
$f =50\,Hz$
Now, $X_{L}=\omega_{L}=2\pi fL$
$=2\times3.14\times50\times30\times10^{-3}$
$=9.42 \Omega$
The rms current in the coil is
$i_{rms}=\frac{V_{rms}}{X_{L}}=\frac{220\,V}{9.42\Omega}$
$=23.4 \,A$