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AnLCR circuit contains resistance of 100\, \Omega and a supply of 200 \mathrm{~V} at 300 \, \mathrm{rad} angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by 60^{\circ}. If, on the other hand, only inductor is taken out the current leads by 60^{\circ} with the applied voltage. The current flowing in the circuit is

Option: 1

1 \mathrm{~A}


Option: 2

1.5 \mathrm{~A}


Option: 3

2 \mathrm{~A}


Option: 4

2.5 \mathrm{~A}


Answers (1)

best_answer

According to the given question,

\tan \tan 60^{\circ}=\frac{\omega L}{R} and  \tan \tan 60^{\circ}=\frac{1 / \omega C}{R}

\therefore \omega L=(1 / \omega C) \text { (case of resonance) }

Now   Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}=100 \Omega

\therefore I_{r m s}=\frac{E_{r m s}}{z}=\frac{200 \mathrm{~V}}{100 \Omega}=2 \mathrm{~A}

Posted by

Devendra Khairwa

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