An inductor, a capacitor and a resistor are connected in series with an ac source $\displaystyle \mathrm{v}=\mathrm{v}_{\mathrm{m}} \sin \omega \mathrm{t}$. Derive an expression for the average power dissipated in the circuit. Also obtain the expression for the resonant frequency of the circuit.

Deriving an expression for the average power dissipated in series LCR circuit Obtaining expression for the resonant frequency \[\mathrm{v}=\mathrm{v}_{\mathrm{m}} \sin \omega \mathrm{t} \] \[\mathrm{i}=\mathrm{i}_{\mathrm{m}} \sin (\omega \mathrm{t}+\varphi) \] \[\text { Power, } \begin{aligned} \mathrm{P}=\mathrm{v} \mathrm{i} & =\left(\mathrm{v}_{\mathrm{m}} \sin \omega \mathrm{t}\right) \times\left[\mathrm{i}_{\mathrm{m}} \sin (\omega \mathrm{t}+\varphi)\right] \\ & =\frac{\mathrm{v}_{\mathrm{m}} \mathrm{i}_{\mathrm{m}}}{2}[\cos \varphi-\cos (2 \omega \mathrm{t}+\varphi)] \end{aligned} \] The average power over a cycle is given by the average of the two terms in RHS of eqn ($\displaystyle 1$). It is only the \(\displaystyle 2^{\text {nd }}\) term which is time dependent. It's average is zero. Therefore, \[\mathrm{P}=\frac{\mathrm{v}_{\mathrm{m}} \mathrm{i}_{\mathrm{m}}}{2} \cos \varphi \]