 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014-resit/9/c/solution

Relative to the real positive source phasor

V_{A B} = V_{0}

ar g \frac{V _{XZ}}{V _{A B}} = - tan^{- 1} 15 ≃ - 75.5 2^{\circ} .

Thus

V_{XZ} (t) = (V_{0} /5) cos (ω t - 75.5 2^{\circ})

lags

V_{A B}

. Reversing the voltage ordering to

V_{ZX}

would add

18 0^{\circ}

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