 Solution

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

Put

V_{B} = 0

and

V_{A} = V_{0}

, with

V_{PQ} = V_{P} - V_{Q}

. The source current is

V_{0} / (10 R)

and the parallel-network voltage is

V_{X} = 4 V_{0} /5

. Hence

I_{+} = \frac{V _{0}}{20 R} (1 - i 15), I_{-} = \frac{V _{0}}{20 R} (1 + i 15) .

The voltage from X to each branch midpoint is its resistor drop, so

V_{XZ} = \frac{V _{0}}{20} (1 - i 15), V_{X Y} = \frac{V _{0}}{20} (1 + i 15), V_{X A} = - \frac{V _{0}}{5} .

Since

∣1 \pm i 15 ∣ = 4

∣ V_{XZ} ∣ = ∣ V_{X Y} ∣ = ∣ V_{X A} ∣ = \frac{V _{0}}{5} .

Equal amplitude does not mean equal phase.

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