 Solution

The parameters satisfy $R^{2}C=100\mu \mathrm{H}>L=10\mu \mathrm{H}$, andThus the source current is $10$ A RMS. The ideal reactive elements dissipate no average power, so all mean input power is lost in $R$:The source's instantaneous input power also has maximum $200$ W because its total current is in phase with its voltage, although its power maximum need not occur at the resistor's power maximum: the reactive elements exchange stored energy.

With only $R$ across the source,The network transforms the real input load from $10\Omega$ to $1\Omega$. At ${\omega }_{\ast }$ the parallel-branch impedance is $1+3i\Omega$, giving resistor RMS voltage $10\sqrt{10}$ V and again $(10\sqrt{10}{)}^2/10=100$ W. The extra power is supplied by the voltage source; it is not produced by the ideal capacitor or inductor.