JEE MAIN - Physics (2010 - No. 15)
In a series $$LCR$$ circuit $$R = 200\Omega $$ and the voltage and the frequency of the main supply is $$220V$$ and $$50$$ $$Hz$$ 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 $$LCR$$ circuit is
$$305$$ $$W$$
$$210$$ $$W$$
$$zero$$ $$W$$
$$242$$ $$W$$
Explanation
When capacitance is taken out, the circular is $$LR.$$
$$\therefore$$ $$\tan \phi = {{\omega L} \over R}$$
$$ \Rightarrow \omega L = R\,\tan \phi $$
$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$
Again, when inductor is taken out, the circuit is $$CR.$$
$$\therefore$$ $$\tan \phi = {1 \over {\omega CR}}$$
$$ \Rightarrow {1 \over {\omega c}} = R\tan \phi $$
$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$
Now, $$Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}} - \omega L} \right)}^2}} $$
$$ = \sqrt {{{\left( {200} \right)}^2} + {{\left( {{{200} \over {\sqrt 3 }} - {{200} \over {\sqrt 3 }}} \right)}^2}} = 200\Omega $$
Power dissipated $$ = {V_{rms}}{I_{rms}}\cos \phi $$
$$ = {V_{rms}}.{{{V_{rms}}} \over Z}.{R \over Z}$$
$$\left( {\,\,\,} \right.$$ as $$\left. {\,\,\cos \phi = {R \over Z}\,\,\,} \right)$$
$$ = {{{V^2}rmsR} \over {{Z^2}}} = {{{{\left( {220} \right)}^2} \times 200} \over {{{\left( {200} \right)}^2}}}$$
$$ = {{220 \times 220} \over {200}} = 242\,W$$
$$\therefore$$ $$\tan \phi = {{\omega L} \over R}$$
$$ \Rightarrow \omega L = R\,\tan \phi $$
$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$
Again, when inductor is taken out, the circuit is $$CR.$$
$$\therefore$$ $$\tan \phi = {1 \over {\omega CR}}$$
$$ \Rightarrow {1 \over {\omega c}} = R\tan \phi $$
$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$
Now, $$Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}} - \omega L} \right)}^2}} $$
$$ = \sqrt {{{\left( {200} \right)}^2} + {{\left( {{{200} \over {\sqrt 3 }} - {{200} \over {\sqrt 3 }}} \right)}^2}} = 200\Omega $$
Power dissipated $$ = {V_{rms}}{I_{rms}}\cos \phi $$
$$ = {V_{rms}}.{{{V_{rms}}} \over Z}.{R \over Z}$$
$$\left( {\,\,\,} \right.$$ as $$\left. {\,\,\cos \phi = {R \over Z}\,\,\,} \right)$$
$$ = {{{V^2}rmsR} \over {{Z^2}}} = {{{{\left( {220} \right)}^2} \times 200} \over {{{\left( {200} \right)}^2}}}$$
$$ = {{220 \times 220} \over {200}} = 242\,W$$
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