Question

The power dissipated in an LCR series circuit connected to an ac source of emf $ \varepsilon $ is
(A) $ \dfrac{{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega - \dfrac{1}{{C\omega }}} \right)}^2}} }}{R} $
(B) $ \dfrac{{{\varepsilon ^2}\left[ {{R^2} + {{\left( {L\omega - \dfrac{1}{{C\omega }}} \right)}^2}} \right]}}{R} $
(C) $ \dfrac{{{\varepsilon ^2}R}}{{\sqrt {{R^2} + {{\left( {L\omega - \dfrac{1}{{C\omega }}} \right)}^2}} }} $
(D) $ \dfrac{{{\varepsilon ^2}R}}{{\left[ {{R^2} + {{\left( {L\omega - \dfrac{1}{{C\omega }}} \right)}^2}} \right]}} $

Answer 1

Hint:
 In a series LCR circuit, we can calculate the impedance of the circuit. Then using that impedance we can calculate current in the circuit. The power dissipated is the product of the square of the current and the resistance.
Formula Used: In this solution we will be using the following formula,
 $ P = {I^2}R $
where $ P $ is the power dissipated, $ I $ is the current in the circuit and the resistance in the circuit is $ R $ .

Complete step by step answer:
In a series LCR circuit, the impedance is given by the impedance of the inductor and the capacitor and the value of the resistance.
Therefore, we have
 $ Z = R + j\left( {{X_L} - {X_C}} \right) $
Here $ {X_L} $ is the impedance of the inductor and the impedance across the capacitor is $ {X_C} $ .
Therefore we have the square of the impedance as,
 $ {Z^2} = {R^2} + {\left( {{X_L} - {X_C}} \right)^2} $
Now the value of the impedance across the inductor is given as, $ {X_L} = \omega L $ . Here $ L $ is the inductance.
And the value of the impedance across the capacitor is given as, $ {X_C} = \dfrac{1}{{\omega C}} $ where $ C $ is the capacitance of the capacitor.
Now substituting these values in the formula for the square of inductance we get,
 $ {Z^2} = {R^2} + {\left( {\omega L - \dfrac{1}{{\omega C}}} \right)^2} $
On taking root on both the sides we have,
 $ Z = \sqrt {{R^2} + {{\left( {\omega L - \dfrac{1}{{\omega C}}} \right)}^2}} $
Now in the circuit, the current will be given by the emf of the circuit divided by the impedance.
Therefore we have,
 $ I = \dfrac{\varepsilon }{Z} $
Hence on substituting the values we have,
 $ I = \dfrac{\varepsilon }{{\sqrt {{R^2} + {{\left( {\omega L - \dfrac{1}{{\omega C}}} \right)}^2}} }} $
The formula for the power dissipated in an LCR circuit is given by the formula, $ P = {I^2}R $ .
Now substituting the value of the current in the formula for the power we get,
 $ P = {\left( {\dfrac{\varepsilon }{{\sqrt {{R^2} + {{\left( {\omega L - \dfrac{1}{{\omega C}}} \right)}^2}} }}} \right)^2}R $
Therefore we get,
 $ P = \dfrac{{{\varepsilon ^2}R}}{{\left[ {{R^2} + {{\left( {\omega L - \dfrac{1}{{\omega C}}} \right)}^2}} \right]}}R $
Therefore the correct answer is option D.

Note:
 A series LCR circuit consists of a resistor, capacitor and an inductor in series. It is also known as a resonant circuit or a tuned circuit. In the case of the inductor, the voltage leads the current by a phase angle of $ 90^\circ $ and for the capacitor, the voltage lags behind the current by $ 90^\circ $ .