JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2021
- Medium
- Numerical
A series \(LCR\) circuit driven by \(300\, {V}\) at a frequency of \(50 \,{Hz}\) contains a resistance \({R}=3 \,{k} \Omega\), 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 : (Take \(\pi^{2}=10\) ) (in \(\mu {F}\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
For maximum average power
\(X_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\)
\(250 \pi=\frac{1}{2 \pi(50) \mathrm{C}}\)
\(\mathrm{C}=4 \times 10^{-6}\)
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