A series LCR circuit is connected to an ac source of $220 \mathrm{~V}, 50 \mathrm{~Hz}$. The circuit contain a resistance $\mathrm{R}=100 ~\Omega$ and an inductor of inductive reactance $\mathrm{X}_{\mathrm{L}}=79.6 ~\Omega$. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be _________ $\mu \mathrm{F}$.
Answer (integer)
40
Solution
To maximize the average rate at which energy
supplied i.e. power will be maximum.
<br/><br/>So in LCR circuit power will be maximum at the
condition of resonance and in resonance condition
<br/><br/>$$
\begin{aligned}
& \therefore X_{L}=X_{C} \\\\
& 79.6=\frac{1}{2 \pi(50) \times C} \\\\
& C=\frac{1}{79.6 \times 2 \pi(50)} \\\\
& \approx 40 \mu \mathrm{F}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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