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

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 $180^{\circ }$.