 Solution

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

The original diagram places

R_{2}

in parallel with the inductor;

R_{1}

connects their common node to the source only when the switch is in position 2. Let

v (t)

be their shared voltage and

i_{L} (t)

i_{L} (0^{-}) = 0

and a finite voltage cannot change it instantaneously, the inductor initially behaves as an open branch. Thus

I_{2} (0^{+}) = \frac{V _{0}}{R _{1} + R _{2}} = \frac{100}{110} A = 0.909 A .

At long times, an ideal inductor carrying constant electric current has

v = L \dot{i}_{L} = 0

, shorting

R_{2}

. Therefore

I_{2} (\infty) = 0

and

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

For the full transient, Kirchhoff's current law gives

(V_{0} - v) / R_{1} = i_{L} + v / R_{2}

with

v = L \dot{i}_{L}

. Equivalently, the source seen by the inductor has the Thévenin equivalent

V_{th} = \frac{V _{0} R _{2}}{R _{1} + R _{2}}, R_{th} = R_{1} ∥ R_{2} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} .

Solving

L \dot{i}_{L} + R_{th} i_{L} = V_{th}

gives

i_{L} (t) = \frac{V _{0}}{R _{1}} (1 - e^{- t / τ}), I_{2} (t) = \frac{V _{0}}{R _{1} + R _{2}} e^{- t / τ}, τ = \frac{L}{R _{th}} = 86.9 μ s .

The relevant resistance is the parallel combination with the ideal voltage source suppressed, not

R_{1} + R_{2}

and not

R_{1}

alone.

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