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  • In a series LCR circuit R = 200 and the voltage and the frequency of the main supply is 220 V and 50 Hz respe

In a series LCR circuit R = 200\Omega and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°.On taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the LCR circuit is:

Option: 1

210 W


Option: 2

0 W


Option: 3

242 W


Option: 4

305 W


Answers (1)

best_answer

Here,\mathrm{ \mathrm{R}=200 \Omega, \mathrm{V}_{\mathrm{rms}}=220 \mathrm{~V}, } \mathrm{ \mathrm{v}=50 \mathrm{~Hz} }
When only the capacitance is removed, \mathrm{\tan \phi=\frac{X_L}{R} or \tan 30^{\circ}=\frac{X_L}{R} or X_L=\frac{1}{\sqrt{3}} R }
When only the inductance is removed, \mathrm{\tan \phi^{\prime}=\frac{X_C}{R} or \tan 30^{\circ}=\frac{X_C}{R} or X_C=\frac{1}{\sqrt{3}} R }
As \mathrm{\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} }, therefore the given series LCR is in resonance.
\mathrm{\therefore \quad } Impedance of the circuit is zero \mathrm{ \mathrm{Z}=\mathrm{R}=200 \Omega }
The power dissipated in the circuit is \mathrm{\mathrm{P}=\mathrm{V}_{\mathrm{rms}} }
\mathrm{\mathrm{I}_{\mathrm{rms}} \cos \phi=\frac{\mathrm{V}_{\mathrm{rms}}^2}{\mathrm{Z}} \cos \phi \quad\left(\because \mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{Z}}\right) }
At resonance, power factor \mathrm{ \cos \phi=1 }
\mathrm{ \therefore \quad P=\frac{V_{\mathrm{rms}}^2}{Z}=\frac{(220 \mathrm{~V})^2}{(200 \Omega)}=242 \mathrm{~W} }
 

Posted by

Irshad Anwar

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