Past exam of the physics 2 paper of the University of Oxford / 2013 resit / 10

Use sinusoidal phasors, $v(t)=\mathrm{Re}(\tilde{V}{e}^{i\omega t})$. Differentiation becomes multiplication by $i\omega$, so the time-dependent element laws reduce to algebraic complex impedances:Then $\tilde{V}=Z\tilde{I}$ encodes both amplitude ratio $|Z|$ and voltage-current phase difference $\arg Z$. Series impedances add, while parallel admittances add. This applies to linear sinusoidal steady state; switching transients are separate homogeneous solutions.

The source figure has a series capacitor followed by $R$ in parallel with $L$. Its complex impedance isThe imaginary part vanishes at a finite positive frequency only if $R^{2}C>L$, in which caseAt that frequency, substituting $R^2+{\omega }_{\ast }^2{L}^2={\omega }_{\ast }^{2}L{R}^{2}C$ into the real part givesIf $R^{2}C\le L$, no finite positive-frequency cancellation exists; the impedance merely tends toward its high-frequency limit.

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.