JEE Mains · Physics · STD 12 -7. Alternating current
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 \(\mu {F}\))
- A \(4\)
- B \(25\)
- C \(400\)
- D \(40\)
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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