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

Question

A coil has self inductance L=0.04 H and resistance R=12 Ω . 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

Tardigrade Admin · Accepted Answer

L=0.04 H R=12 Ω V=220 V, 50 Hz Impedance z= √R2 + w2 L2 w=2π f=2π × 50=100π z= √144 + (100 π )2 × (0.04)2 =17.38 i=(v/z)=(220/17.38)=12.7 A.

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

$12.7 A$

$14.7 A$

$11.7 A$

$10.7 A$

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

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

12.7A

14.7A

11.7A

10.7A