A series LCR circuit is connected to an ac source of 220 V, 50 Hz. The circuit contains a resistance R = 100 and an indu

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A series LCR circuit is connected to an ac source of 220 V, 50 Hz. The circuit contains a resistance R = 100 and an inductor of inductive reactance XL = 79.6. The capacitance of the capacitor needed…

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Given: A series LCR circuit is connected to an ac source of $220 \, \text{V}$ , $50 \, \text{Hz}$ . The inductive reactance is $X_L = 79.6 \, \Omega$ .

Find: The capacitance $C$ for which the average power supplied is maximum.

To maximize the average rate at which energy is supplied, the power must be maximum. In a series LCR circuit, power is maximum at resonance.

At resonance,

X_L = X_C

Using

X_C = \frac{1}{\omega C}

we get

79.6 = \frac{1}{\omega C}

Since

\omega = 2\pi f = 2\pi \times 50

therefore

C = \frac{1}{2\pi \times 50 \times 79.6}

On calculating,

C = 40 \, \mu \text{F}

Therefore, the correct option is B.

Given: Maximum average power is required in a series LCR circuit.

Find: The capacitor value that makes the circuit resonant.

In a series LCR circuit, the impedance is minimum at resonance because the reactive parts cancel each other. That is why current becomes maximum, and hence average power is maximum.

So the condition is

X_L = X_C

with

X_C = \frac{1}{\omega C}

Substitute the given values:

79.6 = \frac{1}{2\pi \times 50 \times C}

Hence,

C = \frac{1}{2\pi \times 50 \times 79.6}

This gives

C = 40 \, \mu \text{F}

So the capacitance needed is $40 \, \mu \text{F}$ .

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