A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance $\mathrm{R}=3 \mathrm{k} \Omega$, an inductor of inductive reactances $\mathrm{X}_{\mathrm{L}}=250 \pi \Omega$ and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take $\pi^{2}=10$ )

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Alternating Current

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance

R = 3 k Ω

, an inductor of inductive reactances

X_{L} = 250 π Ω

and an unknown capacitor. The value of capacitance to maximize the average power should be :
(Take

π^{2} = 10

)

Answer: (B)

The average power

P = \frac{V _{rms}^{2} R}{R ^{2} + ( X _{L} - X _{C} ) ^{2}}

is largest when

X_{C} = X_{L}

, the resonance condition. With

X_{C} = \frac{1}{2 π f C}

C = \frac{1}{2 π f X _{L}} = \frac{1}{2 π \times 50 \times 250 π} = \frac{1}{25000 π ^{2}}

Taking

π^{2} = 10

C = \frac{1}{2.5 \times 1 0 ^{5}} = 4 \times 1 0^{- 6} F = 4 μ F

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option