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A series LCR circuit driven by 300 V at 50 Hz has R=3 kΩ, inductive reactance X_L=250π Ω and an unknown capacitor. The capacitance that maximises the average power is (Takeπ²=10)

Asked in JEE Main 26th Aug 1st Shift 2021 · Power and bandwidth at resonance

Answer: (2) 4 μF

Step-by-step solution

Maximum power at resonance: X_C=X_L=250π, and X_C=1/(ω C)=1/(100π C).

1/(100π C)=250π⇒ C=1/(25000π²)=1/(25000×10)=4×10⁻⁶ F=4 μF.

Why the other options are wrong

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