Question
Power dissipated in an $$L-C-R$$ series circuit connected to an $$AC$$ source of emf $$\varepsilon $$ is
A.
$$\frac{{{\varepsilon ^2}R}}{{\left[ {{R^2} + {{\left( {L\omega - \frac{1}{{C\omega }}} \right)}^2}} \right]}}$$
B.
$$\frac{{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega - \frac{1}{{C\omega }}} \right)}^2}} }}{R}$$
C.
$$\frac{{{\varepsilon ^2}\left[ {{R^2} + {{\left( {L\omega - \frac{1}{{C\omega }}} \right)}^2}} \right]}}{R}$$
D.
$$\frac{{{\varepsilon ^2}R}}{{\sqrt {{R^2} + {{\left( {L\omega - \frac{1}{{C\omega }}} \right)}^2}} }}$$
Answer :
$$\frac{{{\varepsilon ^2}R}}{{\left[ {{R^2} + {{\left( {L\omega - \frac{1}{{C\omega }}} \right)}^2}} \right]}}$$
Solution :
Power dissipated in series $$LCR.$$
$$\eqalign{
& P = I_{rms}^2R \cr
& = \frac{{\varepsilon _{rms}^2R}}{{{{\left| Z \right|}^2}}} = \frac{{{\varepsilon ^2}R}}{{\left[ {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} \right]}} \cr} $$
As $${I_{rms}} = \frac{{{V_{rms}}}}{Z}$$