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Past exam of the physics 2 paper of the University of Oxford

2014 resit

Put $\widetilde V_B=0$ and $\widetilde V_A=V_0$, with $\widetilde V_{PQ}=\widetilde V_P-\widetilde V_Q$. The source current is $V_0/(10R)$ and the parallel-network voltage is $\widetilde V_X=4V_0/5$. Hence
$$
\widetilde I_+=\frac{V_0}{20R}(1-i\sqrt{15}),\qquad
\widetilde I_-=\frac{V_0}{20R}(1+i\sqrt{15}).
$$
The voltage from X to each branch midpoint is its resistor drop, so
$$
\widetilde V_{XZ}=\frac{V_0}{20}(1-i\sqrt{15}),\quad
\widetilde V_{XY}=\frac{V_0}{20}(1+i\sqrt{15}),\quad
\widetilde V_{XA}=-\frac{V_0}{5}.
$$
Since $|1\pm i\sqrt{15}|=4$,
$$
\boxed{|\widetilde V_{XZ}|=|\widetilde V_{XY}|=|\widetilde V_{XA}|=\frac{V_0}{5}.}
$$
Equal amplitude does not mean equal phase.


Solution

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

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

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

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