-->

Laws Of Motion

Question
CBSEENPH11026252

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

Solution

D.

1.1 H

Power P = I × V

             = V R × V

           P = V2 R

Resistance of bulb

          R = V2P

           R = 100250

               = 200 Ω 

Current through bulb

            (I) = VR

                 = 100200

              I  =  0.5 A

In a circuit containing inductive reactance (XL ) and resistance (R ), impedance (Z) of the circuit is

             Z = R2 + ω2 L2              ...... (i)

             Z = 2000.5

             Z = 400 Ω

Now,      XL2 = Z2 - R2

             XL2  = ( 400 )2  - ( 200 )2

           2 πf L2 = 12 × 104

             L = 23 × 1002π × 50

                  = 23 π

              L = 1.1 H