Alternating Current

We need to find:For the power dissipated to become half of its maximum value, what is the current amplitude?Step 1: Relation between Power and CurrentThe average power dissipated in an AC circuit is given by:$P = I_{\text{rms}}^2 R$At resonance, the impedance is purely resistive ( $Z = R$ ), and the current is maximum ( $I = I_{\text{ max }}$ ).So maximum power is:$P_{\text{max}} = I_{\text{rms, max}}^2 \;\; R = \left( \frac{I_{\text{max}}}{\sqrt{2}} \right)^2 R = \frac{I_{\text{max}}^2 \;\; R}{2}$But we only need relative power, so we can use $P \propto I^2$.Step 2: Condition for half powerWe are told $P = \frac{1}{2} P_{\text{ max }}$So:$I^2 R = \frac{1}{2} I_{\text{max}}^2 R$Simplify:$I = \frac{I_{\text{max}}}{\sqrt{2}}$So the current amplitude becomes $\frac{1}{\sqrt{2}}$ times its maximum value when the power is half of the maximum.