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A series LCR circuit is connected to an ac source of voltage $150 \sin (80 \pi t)$ volt. If the resistance of the resistor in the circuit is $25 \Omega$ and the impedance in the circuit is $75 \Omega$, the average power dissipated per cycle in the circuit is
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The correct answer is:
$50 \mathrm{~W}$
Resistance of the circuit, $\mathrm{R}=25 \Omega$
Impedance, $\mathrm{Z}=75 \Omega$
The average power dissipates per cycle in the circuit is
$$
\begin{aligned}
& P_{\text {avg }}=I_{r m s} V_{r m s} \operatorname{Cos} \theta \\
& =\frac{V_0}{\sqrt{.}} \frac{V_0}{\sqrt{ }} \times \frac{R}{Z}=\frac{V_0^2 R}{2 Z^2}=\frac{150^2 \times 25}{2 \times 75^2} \\
& =50 W
\end{aligned}
$$
Impedance, $\mathrm{Z}=75 \Omega$
The average power dissipates per cycle in the circuit is
$$
\begin{aligned}
& P_{\text {avg }}=I_{r m s} V_{r m s} \operatorname{Cos} \theta \\
& =\frac{V_0}{\sqrt{.}} \frac{V_0}{\sqrt{ }} \times \frac{R}{Z}=\frac{V_0^2 R}{2 Z^2}=\frac{150^2 \times 25}{2 \times 75^2} \\
& =50 W
\end{aligned}
$$
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