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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017/11/d/solution

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Past exam of the physics 2 paper of the University of Oxford / 2017 / 11 / d / Solution by

Codex  0 2026-10-25

Opening the switch removes the source and

R_{1}

but leaves a closed path through the inductor and

R_{2}

. Set the new time origin at switch-off. Inductor current continuity gives

i_{L} (0) = V_{0} / R_{1} = 10 A

. Since the two parallel branches are now isolated, Kirchhoff's current law gives

I_{2} = - i_{L}

, and

L \dot{i}_{L} = - R_{2} i_{L}

. Thus

i_{L} (t) = \frac{V _{0}}{R _{1}} e^{- R_{2} t / L}, τ_{off} = \frac{L}{R _{2}} = 7.90 μ s .

All the initially stored magnetic energy is ultimately converted to Joule heating in

R_{2}

:

Q_{2, off} = \frac{1}{2} L (\frac{V _{0}}{R _{1}})^{2} = 3.95 \times 1 0^{- 2} J .

Direct integration of

R_{2} i_{L}^{2} (t)

gives the same result. The initial shared voltage is

v (0^{+}) = - R_{2} i_{L} (0) = - 1000 V

; its reversal drives the return current. The ideal circuit permits this transient voltage; real switching would also be limited by stray capacitance, breakdown and other parasitic effects.

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