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AIEEE 2010 Solved Paper

Section: Physics

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Question : 46 of 89

Marks: +1, -0

In a series

L C R

R=200 \Omega

and the voltage and the frequency of the main supply is

220 \text{ V}

50 \text{ Hz}

respectively. On taking out the capacitance from the circuit the current lags behind the voltage by

30^{\circ}

. On taking out the inductor from the circuit the current leads the voltage by

30^{\circ}

. The power dissipated in the

L C R

305 W
210 W
zero W
242 W

Solution:

When capacitance is taken out, the circular is

L R

.

\therefore \tan\varphi = \frac{\omega L}{R}

\Rightarrow \omega L = R \tan\varphi

=200 \times \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}}

Again, when inductor is taken out, the circuit is

C R

.

\therefore \tan\varphi = \frac{1}{\omega C R}

\Rightarrow \frac{1}{\omega c} = R \tan\varphi

=200 \times \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}}

Now,

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

=\sqrt{(200)^2 + \left( \frac{200}{\sqrt{3}} - \frac{200}{\sqrt{3}} \right)^2} = 200 \Omega

Power dissipated

= V_{rms} I_{rms} \cos\varphi

\; = V_{rms} \cdot \frac{V_{rms}}{Z} \cdot \frac{R}{Z}

\; \left( \; \text{ as } \; \cos\varphi = \frac{R}{Z} \right)

\; = \frac{V_{rms}^2 R}{Z^2} = \frac{(220)^2 \times 200}{(200)^2}

\; = \frac{220 \times 220}{200} = 242 \text{ W}

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