Questions
A bulb is rated at 100V, 100W, 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 Hz.
detailed solution
Correct option is D
R=Vrms2p=(100)2100=100ΩFor the bulb to be operated at its rated value the rms current through it should be 1A I=2001002+[2π(50L)]2⇒L=3πHTalk to our academic expert!
Similar Questions
If a current flows in an AC circuit across which an AC potential of has been applied, then the power consumption P in the circuit will be
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
