Question 487: AC fundamentals

Question

Power dissipated in an LCR series circuit connected to an A.C. source of emf $$\varepsilon $$ is

Accepted Answer

Power dissipated, P = E
rms
I
rms
cos$$\phi $$
cos$$\phi $$ = $${R \over Z}$$
But I
rms
= $${{{E_{rms}}} \over Z}$$
$$ 	herefore $$ P = $$E_{rms}^2.{R \over {{Z^2}}}$$
Also we know, Z = $$\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} ight)}^2}} $$
So, P = $$E_{rms}^2.{R \over {{R^2} + {{\left( {{X_L} - {X_C}} ight)}^2}}}$$
= $${{{\varepsilon ^2}R} \over {\left[ {{R^2} + {{\left( {L\omega  - {1 \over {C\omega }}} ight)}^2}} ight]}}$$

Answer

$${{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega  - {1 \over {\omega }}} \right)}^2}} } \over R}$$

Answer

$${{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega  - {1 \over {C\omega }}} \right)}^2}} } \over R}$$

Answer

$${{{\varepsilon ^2}R} \over {\sqrt {{R^2} + {{\left( {L  - {1 \over {C\omega }}} \right)}^2}} }}$$

Answer

$${{{\varepsilon ^2}R} \over {\left[ {{R^2} + {{\left( {L\omega  - {1 \over {C\omega }}} \right)}^2}} \right]}}$$

AC fundamentals

Power dissipated in an LCR series circuit connected to an A.C. source of emf $\varepsilon $ is

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