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An LCR series circuit with 100 $\Omega$ resistance is connected to an AC source of 200 V and angular frequency 300 radians per second. When only the capacitance is removed, the current lags the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60º. Then the current and power dissipated in LCR circuit are respectively
Option: 1
1A, 200 watt

Option: 2
1A, 400 watt

Option: 3
2A, 200 watt

Option: 4
2A, 400 watt

Answers (1)

best_answer

R.L Circuit Voltage -

$V_{R}= IR$

$V_{L}= {I}X_{L}$

- wherein

Phase difference -

$\tan \phi = \frac{X_{L}}{R}= \tan ^{-1}\left ( \frac{X_{L}}{R} \right )$

$\phi = \tan ^{-1}\left ( \frac{\omega L}{R} \right )$

- wherein

X_L = inductive reactance

R = resistance

R C Circuit Voltage -

$V_{R}=IR$

$V_{C}={I}X_{C}$

- wherein

Phase difference -

$\tan \phi = \frac{X_{c}}{R}$

$\phi = \tan ^{-1}\left ( \frac{X_{c}}{R} \right )= \tan ^{-1}\left ( \frac{1}{\omega cR} \right )$

-

At resonance (series resonant and circuit) If $X_{L}=X_{c}$ -

$Z_{min}= R$

- wherein

Circuit behaves as resistive circuit.

When capacitance is removed

$\tan \theta = \frac{\omega L}{R}$ or $\omega L = 100 \tan 60^{\circ}$ ..................(1)

when inductance is removed

$\tan \phi = \frac{1}{(\omega C)(R)}$ or $=100 \tan 60^{\circ}$ ....................(2)

From equation (1) and (2) $\omega L = \frac{1}{\omega C}$

So it is a condition of resonance.

so z = R = 100 $\Omega$

I = v/R = 200/100 = 2A

Power P = I² R = 4 × 100 = 400 W

Posted by

manish

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