Question

A coil has self-inductance $L=0.04H$ and resistance $R=12\Omega$ , connected to $220V$, 50 Hz supply, what will be the current flow in the coil ?

• A. 11.7 A
• B. 12.7 A
• C. 10.7 A
• D. 14.7 A

Answer 1

The correct option is B 12.7 A

Given, $L=0.04H,R=12\Omega$
$V=220$ volt and $f=50Hz$
The value of current
$I=\frac{V}{Z}$
or or $I=\frac{V}{\sqrt{R^2+(\omega L{)}^2}}$
or $I=\frac{V}{\sqrt{R^2+(2\pi fL{)}^2}}$
or $I=\frac{220}{\sqrt{144+(2\pi 50\times 0.04{)}^2}}$
$\Rightarrow I=12.7A$