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← JEE Main 2022 · 30 Jun · Shift 1
PhysicsInteger+4 / -1Question 57 of 90

JEE Main 2022 · 30 Jun · Shift 1, Physics question 57

A series LCR circuit with R=25011R=\dfrac{250}{11} Ω\Omega and XL=7011X_L=\dfrac{70}{11} Ω\Omega is connected across a 220 V, 50 Hz supply. The value of capacitance needed to maximize the average power of the circuit will be ______ μ\muF. (Take : π=227\pi=\dfrac{22}7)

Answer and worked solution

Official answer: 500

Resonance. XC=XL=7011X_C=X_L=\frac{70}{11}: C=112π⋅50⋅70=11×7100×22×70=5×10−4C=\dfrac{11}{2\pi\cdot50\cdot70}=\dfrac{11\times7}{100\times22\times70}=5\times10^{-4} F =500=500 μF.

Answer: 500