Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section C / C11 / Solution

In series the same current passes through both resistors, while their voltage drops add: $V=I(R_1+R_2)$. In parallel the same voltage drives currents $V/R_1$ and $V/R_2$, which add. Thus $R_s=R_1+R_2$ and $R_p^{-1}=R_1^{-1}+R_2^{-1}$.

Use phasors proportional to $e^{i\omega t}$. Then $Z_L=i\omega L$ and $Z_C=1/(i\omega C)$. The printed diagram places C in parallel with the series RL branch, so the input admittance isIts nonzero zero-phase frequency exists for $R^2<L/C$:At zero frequency, $|Y|=1/R$ and $I_0=V_0/R$. At ${\omega }_1$, $R^2+{\omega }_1^2{L}^2=L/C$, givingFor $0<\omega <{\omega }_1$, $\mathrm{Im}Y<0$, so current lags voltage. At very high frequency, $Y\sim i\omega C$, so current leads by $\pi /2$. Zero phase need not coincide exactly with the minimum current magnitude in this lossy network.

The three measured limits determine the components in order:For the changed resistance, the quoted $0.6\mathrm{kHz}$ is an ordinary frequency, so ${\omega }^{\prime }=2\pi (600{\mathrm{s}}^{-1})$. Keeping L and C fixed gives