 Solution

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

The source diagram has

2 R

in series with two parallel branches:

R + L

above and

R + C

below. With

s = 15

Z_{+} = R (1 + i s)

and

Z_{-} = R (1 - i s)

. Their parallel combination is

Z_{p} = \frac{Z _{+} Z _{-}}{Z _{+} + Z _{-}} = \frac{R ^{2} ( 1 + s ^{2} )}{2 R} = 8 R .

Thus

Z_{A B} = 2 R + 8 R = 10 R .

The input electric current is in phase with

V_{A B}

even though the two branch currents separately have nonzero, opposite phase shifts.

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