JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2023
- Medium
- Numerical
A series \(LCR\) circuit is connected to an ac source of \(220\,V , 50 Hz\). The circuit contain a resistance \(R=100 \Omega\) and an inductor of inductive reactance \(X_L=79.6 \Omega\). The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be \(..........\mu F\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(40\)
Step-by-step Solution
Detailed explanation
To maximize the average rate at which energy supplied i.e. power will be maximum.
So in \(LCR\) circuit power will be maximum at the condition of resonance and in resonance condition
\(X_L =X_C\)
\(79.6 =\frac{1}{\omega C}\)
\(\therefore C =\frac{1}{2 \pi \times 50 \times 79.6}\)
\(\therefore C =40 \mu F\)
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