A series LCR circuit with $R = {{250} \over {11}}\,\Omega$ and ${X_L} = {{70} \over {11}}\,\Omega$ is connected across a 220 V, 50 Hz supply. The value of capacitance needed to maximize the average power of the circuit will be _________ $\mu$F. (Take : $\pi = {{22} \over 7}$)
Answer (integer)
500
Solution
For maximum power
<br/><br/>$$
\begin{aligned}
&\text { power factor }=\cos \theta=1\\\\
& \therefore \frac{R}{Z}=1 \\\\
&R^{2}=Z^{2} \\\\
&R^{2}=\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}+\mathrm{R}^{2} \\\\
&\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \\\\
&\frac{70}{11}=\frac{1}{100 \pi \times C} \\\\
&\Rightarrow C=\frac{11}{7000 \pi}=500 \times 10^{-6} F=500 \mu F
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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