A series LCR circuit is connected to an ac source of 220 V, 50 Hz. The circuit contain a resistance R = 100 Ω and an inductor of inductive reactance XL = 79.6 Ω. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be _________. Numerical answer type. Work it out, then open the answer.

Answer: 40. Average power is maximum at resonance: XC = XL = 79.6 Ω. C = (1)/(2π f XC) = (1)/(2×3.14×50×79.6) C = (1)/(25000) = 40×10⁻⁶ F = 40 μ F

A series LCR circuit is connected to an ac source of 220 V, 50 Hz… | JEE Main 2023 (1 Feb, Shift 1)

Q68JEE Main 2023 · 1 FEB · SHIFT 1Medium

Resonance, Bandwidth & Q Factor

A series LCR circuit is connected to an ac source of $220\text{ V}, 50\text{ Hz}$ . 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 _________ $\mu\text{F}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 40

Average power is maximum at resonance: $X_C = X_L = 79.6\ \Omega$ .
$C = \frac{1}{2\pi f X_C} = \frac{1}{2\times3.14\times50\times79.6}$
$C = \frac{1}{25000} = 40\times10^{-6}$ F $= 40\ \mu$ F

Report a mistake

Resonance, Bandwidth & Q Factor question from JEE Main 2023 (1 Feb, Shift 1)

More questions on Resonance, Bandwidth & Q Factor