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 reactance XL = 250 π Ω and an unknown capacitor. The value of capacitance to maximize the average power should be: (Take π 2 = 10) (a) 4 μ F (b) 25 μ F (c) 400 μ F (d) 40 μ F

Answer: (a). Average power is maximum at resonance: XC = XL = 250π Ω. C = (1)/(ω XC) = (1)/(100π × 250π) = (1)/(25000π²) With π² = 10: C = (1)/(2.5×10⁵) F = 4 μ F, so the answer is (a).

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains… | JEE Main 2021 (26 Aug, Shift 1)

Q59JEE Main 2021 · 26 AUG · SHIFT 1Difficult

Power, Power Factor & Wattless 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 XL = 250 $\pi\ \Omega$ and an unknown capacitor. The value of capacitance to maximize the average power should be: (Take $\pi$ 2 = 10)

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Answer: (a)

Average power is maximum at resonance: $X_C = X_L = 250\pi\ \Omega$ .
$C = \frac{1}{\omega X_C} = \frac{1}{100\pi \times 250\pi} = \frac{1}{25000\pi^2}$
With $\pi^2 = 10$ : $C = \frac{1}{2.5\times10^5}$ F $= 4\ \mu$ F, so the answer is (a).

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