Question

A coil has self inductance $L=0.04 \, H$ and resistance $R=12 \, \Omega$ . When it is connected to $220 \, V$ , $50 \, Hz$ supply; what will be the current flowing through the coil?

• A. $12.7 \, A$
• B. $14.7A$
• C. $11.7A$
• D. $10.7A$

Answer 1

Answer: (A)

$L=0.04 \, H$
$R=12 \, \Omega$
$V=220 \, V, \, 50 \, Hz$
Impedance $z= \, \sqrt{R^{2} + w^{2} L^{2}}$
$w=2\pi f=2\pi \times 50=100\pi $
$z= \, \sqrt{144 + \left(100 \pi \right)^{2} \times \left(0.04\right)^{2}}$
$=17.38$
$i=\frac{v}{z}=\frac{220}{17.38}=12.7 \, A.$