Question

A sinusoidal voltage of 200V ,50Hz is applied to a series LCR circuit . If R= 60ohm ,L=50.96mH , C= 398 microF , then

(a) what is the frequency of source at resonance.
(b) calculate impedance, current and dissipated power for resonance situation.

Answer 1

Answer:

Part i)

Resonance frequency is f = 35.33 Hz

Part ii)

At resonance condition net impedence is purely resistive in nature

Z = R = 60 ohm

current in the circuit is given as

i = 3.33 A

Power consumed in resonance condition is given as

P = 666.67 W

Explanation:

Part i)

When resonance occurs in series LCR circuit then power dissipated across the resistor is maximum

so it is purely resistive circuit

so we have

$f = \frac{1}{2\pi}\sqrt{\frac{1}{LC}}$

now we know that

L = 50.96 mH

$C = 398 \mu F$

now we have

$f = \frac{1}{2\pi}\sqrt{\frac{1}{(50.96 \times 10^{-3})(398\times 10^{-6})}$

$f = 35.33 Hz$

Part ii)

At resonance condition net impedence is purely resistive in nature

so we have

$Z = R = 60 ohm$

current in the circuit is given as

$i = \frac{V}{Z}$

$i = \frac{200}{60}$

$i = 3.33 A$

Power consumed in resonance condition is given as

$P = i V$

$P = 3.33(200)$

$P = 666.67 W$

#Learn

Topic : AC Resonance circuit

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