Past exam of the physics 2 paper of the University of Oxford / 2014 resit / 9

Use the phasor convention $v(t)=\mathrm{Re}(\tilde{V}{e}^{i\omega t})$. The complex impedance is $Z=\tilde{V}/\tilde{I}$ in sinusoidal steady state. From $v_R=Ri$, $i_C=Cd{v}_C/dt$, and $v_L=Ld{i}_L/dt$,For this convention an ideal inductor has voltage leading current by $\pi /2$, while an ideal capacitor has current leading voltage by $\pi /2$.

The source diagram has $2R$ in series with two parallel branches: $R+L$ above and $R+C$ below. With $s=\sqrt{15}$, their complex impedances are $Z_{+}=R(1+is)$ and $Z_{-}=R(1-is)$. Their parallel combination isThusThe input electric current is in phase with $V_{AB}$ even though the two branch currents separately have nonzero, opposite phase shifts.

Put ${\tilde{V}}_B=0$ and ${\tilde{V}}_A=V_0$, with ${\tilde{V}}_{PQ}={\tilde{V}}_P-{\tilde{V}}_Q$. The source current is $V_0/(10R)$ and the parallel-network voltage is ${\tilde{V}}_X=4V_0/5$. HenceThe voltage from X to each branch midpoint is its resistor drop, soSince $|1\pm i\sqrt{15}|=4$,Equal amplitude does not mean equal phase.

Relative to the real positive source phasor ${\tilde{V}}_{AB}=V_0$,Thus $V_{XZ}(t)=(V_0/5)\cos (\omega t-75.52^{\circ })$ lags $V_{AB}$. Reversing the voltage ordering to $V_{ZX}$ would add $180^{\circ }$.