Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A coil has $L=0.04\, H$ and $R=12\, \Omega$. When it is connected to $220 \,V , 50 \,Hz$ supply, the current flowing through the coil, in amperes, is

Alternating Current

Solution:

Impedance, $Z=\sqrt{R^{2}+4 \pi^{2} v^{2} L^{2}}$
$=\sqrt{(12)^{2}+4 \times(3.14)^{2} \times(50)^{2} \times(0.04)}=17.37 A$
Now, current, $i=\frac{V}{Z}$
$=\frac{220}{17.37}=12.7 \,\Omega$