An electric bulb has a rated power of 50 W at 100 V. If it is used on an AC source 200 V, 50 Hz, a choke has to be used in series with it. This choke should have an inductance of
0.1 mH
1 mH
0.1 mH
1.1 H
D.
1.1 H
Power P = I × V
= V R × V
P = V2 R
Resistance of bulb
R =
R =
= 200 Ω
Current through bulb
(I) =
=
I = 0.5 A
In a circuit containing inductive reactance (XL ) and resistance (R ), impedance (Z) of the circuit is
Z = ...... (i)
Z =
Z = 400 Ω
Now,
= ( 400 )2 - ( 200 )2
=
L = 1.1 H