An inductor 20mH, a capacitor `50muF` and a resistor `40Omega` are connected in series across a source of emf V=10sin340t. The power loss in ac circuit is
A
0.67W
B
0.76W
C
0.89W
D
0.51W
Text Solution
Verified by Experts
The correct Answer is:
D
Here, `omega=340Hz` Inductive reactance, `X_L=omegaL=340times20times10^-3=6.8Omega` Capacitive reactance, `X_C=1/(omegaC)=1/(340times50times10^-6)=58.8Omega` Resistance, R=40`Omega` Therefore, impedance of the circuit, `Z=sqrt(R^2+(X_C-X_L)^2)` `=sqrt(40^2+(58.8-6.8)^2)=65.6Omega` Now, rms value of current , `I_(rms)=V_(rms)/Z=10/(sqrt2times65.6)` Power dissipated in the circuit, `P=V_(rms)I_(rms)costheta=V_(rms)I_(rms)R/Z` `=10/sqrt2times10/(sqrt2times65.6)=0.46Wapprox0.51W`
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