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

Question

A coil has L=0.04 H and R=12 Ω. When it is connected to 220 V , 50 Hz supply, the current flowing through the coil, in amperes, is (A) 10.7 (B) 11.7 (C) 14.7 (D) 12.7

Tardigrade Admin · Accepted Answer

Impedance, Z=√R2+4 π2 v2 L2 =√(12)2+4 ×(3.14)2 ×(50)2 ×(0.04)=17.37 A Now, current, i=(V/Z) =(220/17.37)=12.7 Ω

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

10.7

11.7

14.7

12.7

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

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