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 circui

Question

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 circui

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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - 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`