Question: Q. 12. A voltage $V=V_{0} \sin \omega t$ is applied to a series $L C R$ circuit. Derive the expression for the average power dissipate over a cycle.
Under what conditions is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit ?
U[O.D. I, II, III 2014]
Show Answer
Solution:
Ans.
Applied voltage $=V_{0} \sin \omega t$
$1 / 2$
Current in the circuit $=I_{0} \sin (\omega t-\phi)$
where, $\phi$ is the phase lag of the current with respect to the voltage applied.
Hence instantaneous power dissipation
$$ \begin{aligned} & =V_{0} \sin \omega t \times I_{0} \sin (\omega t-\phi) \ & =\frac{V_{0} I_{0}}{2}[2 \sin \omega t \cdot \sin (\omega t-\phi)] \ & =\frac{V_{0} I_{0}}{2}[\cos \phi-\cos (2 \omega t-\phi)]^{1 / 2} \end{aligned} $$
Therefore, average power for one completes cycle $=$ average of $\left[\frac{V_{0} I_{0}}{2}{\cos \phi-\cos (2 \omega t-\phi)}\right.$
The average of the second term over a complete cycle is zero.
Hence, average power dissipated over one complete cycle $=\frac{V_{0} I_{0}}{2} \cos \phi \times \quad 1 / 2$ Conditions :
(i) No power is dissipated when $R=0\left(\text { or } \phi=90^{\circ}\right)^{1 / 2}$ [Note : Also accept if the student writes ‘This condition cannot be satisfied for a series LCR circuit’.]
(ii) Maximum power is dissipated when $X_{L}=X_{C} \cdot 1 / 2$ Or
$$ \omega L=\frac{1}{\omega C}(\text { or } \phi=0) $$
[CBSE Marking Scheme 2014]
