Question 239: AC fundamentals

Question

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

Accepted Answer

When capacitance is taken out, the circular is $$LR.$$
$$	herefore$$ $$	an \phi  = {{\omega L} \over R}$$
$$ \Rightarrow \omega L = R\,	an \phi $$
$$ = 200 	imes {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$
Again, when inductor is taken out, the circuit is $$CR.$$
$$	herefore$$ $$	an \phi  = {1 \over {\omega CR}}$$
$$ \Rightarrow {1 \over {\omega c}} = R	an \phi $$
$$ = 200 	imes {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$
Now, $$Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}} - \omega L} ight)}^2}} $$
$$ = \sqrt {{{\left( {200} ight)}^2} + {{\left( {{{200} \over {\sqrt 3 }} - {{200} \over {\sqrt 3 }}} ight)}^2}}  = 200\Omega $$
Power dissipated $$ = {V_{rms}}{I_{rms}}\cos \phi $$
$$ = {V_{rms}}.{{{V_{rms}}} \over Z}.{R \over Z}$$
$$\left( {\,\,\,} ight.$$ as $$\left. {\,\,\cos \phi  = {R \over Z}\,\,\,} ight)$$
$$ = {{{V^2}rmsR} \over {{Z^2}}} = {{{{\left( {220} ight)}^2} 	imes 200} \over {{{\left( {200} ight)}^2}}}$$
$$ = {{220 	imes 220} \over {200}} = 242\,W$$

Answer

$$305$$ $$W$$

Answer

$$210$$ $$W$$

Answer

$$zero$$ $$W$$

Answer

$$242$$ $$W$$

AC fundamentals

Choose the correct answer: