Question

a light bulb has resistance of 1000 current in it when it connected to 400v source will be (a)4A(b)3A(c)1.5A(d)0.3A​

Answer 1

Answer:

ANSWER

The frequency of AC source =f=50Hz

Thus the impedance of the inductor, X

L

=ωL=(2πf)L=20πΩ

Resistance of the resistor =20Ω

Thus, the net impedance of the circuit =

X

L

2

+R

2

=65.94Ω

Thus, the current in the circuit =

X

V

=

65.94Ω

220V

=3.33A

Answer 2

Answer:

V = IR

I = V/R

= 400/1000

I = 0.4A ~ 0.3 A

d) 0.3 A