Question:
A series LCR circuit driven by $300 \mathrm{~V}$ at a frequency of $50 \mathrm{~Hz}$ contains a resistance $\mathrm{R}=3 \mathrm{k} \Omega$, an inductor of inductive reactance $X_{L}=250 \pi \Omega$ and an unknown capacitor. The value of capacitance to maximize the average power should be : $\left(\right.$ Take $\left.\pi^{2}=10\right)$
Correct Option: 1
Solution:
For maximum average power
$\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$
$250 \pi=\frac{1}{2 \pi(50) \mathrm{C}}$
$\mathrm{C}=4 \times 10^{-6}$
Option (1)