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

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

Codex  0 2026-10-25

R_{2}

has instantaneous power

P_{2} = R_{2} [V_{0} / (R_{1} + R_{2})]^{2} e^{- 2 t / τ}

. Integrating the finite transient gives

Q_{2, on} = \int_{0}^{\infty} P_{2} d t = \frac{R _{2} V _{0}^{2}}{( R _{1} + R _{2} ) ^{2}} \frac{τ}{2} = \frac{L V _{0}^{2}}{2 R _{1} ( R _{1} + R _{2} )} = 3.59 \times 1 0^{- 3} J .

(1)

Although the source continues supplying power to

R_{1}

indefinitely,

R_{2}

is eventually shorted and its heat integral converges. It would be incorrect to identify this heat with the final stored magnetic energy, because the source and

R_{1}

also participate during switch-on.

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