Explanation

(b) The average power output of the emf source is
$P = \frac{1}{2}\left| V_{0} \right|\left| I_{0} \right|\cos\theta$
Since, $V_{0} = I_{0}R$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P = \frac{1}{2}R\left| I_{0} \right|^{2}$
It is clear that only the resistor dissipates energy in the circuit. The inductor and capacitor both store energy but they eventually return it to the circuit without dissipation.