Question
Physics Question on Electromagnetic induction
An inductor 20mH, a capacitor 100μF and a resistor 50Ω are connected in series across a source of emf, V=10sin314t. The power loss in the circuit is
A
1.13 W
B
0.79 W
C
2.74 W
D
0.43 W
Answer
0.79 W
Explanation
Solution
Pav=(ZVRMS)2R
Z=R2+(ωL−ωC1)2=56Ω
∴Pav=((2)5610)2×50=0.79W