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 X L =79.6 Ω. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be μ F.

Question

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 X L =79.6 Ω. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be μ F. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

To maximize the average rate at which energy supplied i.e. power will be maximum.
So in LCR circuit power will be maximum at the condition of resonance and in resonance condition

X_{L} = X_{C}

79.6 = \frac{1}{ω C}

∴ C = \frac{1}{2 π \times 50 \times 79.6}

∴ C = 40 μ F

Q. A series LCR circuit is connected to an ac source of 220V,50Hz. The circuit contain a resistance R=100Ω and an inductor of inductive reactance XL​=79.6Ω. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be ________μF.

To maximize the average rate at which energy supplied i.e. power will be maximum. So in LCR circuit power will be maximum at the condition of resonance and in resonance condition XL​=XC​ 79.6=ωC1​ ∴C=2π×50×79.61​ ∴C=40μF

To maximize the average rate at which energy supplied i.e. power will be maximum.
So in LCR circuit power will be maximum at the condition of resonance and in resonance condition
$X_{L} = X_{C}$
$79.6 = \frac{1}{ω C}$
$∴ C = \frac{1}{2 π \times 50 \times 79.6}$
$∴ C = 40 μ F$