NTA Abhyas JEE Main2020PhysicsAlternating CurrentPractice
A bulb is rated at 100 V , 100 W , it can be treated as a resistor. Find out the inductance of an inductor (called choke coil) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of 200 V and 50 H z .
Options
- Aπ 3   H
- B100 H
- C2 π   H
- D3 π H
Correct answer
D. 3 π H
Step-by-step solution
From the rating of the bulb, the resistance of the bulb can be calculated. R = V r m s 2 P = ( 100 ) 2 100 = 100 Ω For the bulb to be operated at its rated value the rms current through it should be 1 A Also, I r m s = V r m s Z ∴ 1 = 200 100 2 + ( 2 π 5 0 L ) 2 ⇒ L = 3 π H