 Solution

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

Put

e_{t} = e^{- t / τ}

. From

u = V /2 - R I_{1}

and

I_{0} = I_{1} + I_{2}

I_{1} = \frac{V}{4 R} (1 - e_{t}), I_{2} = \frac{V}{8 R} (1 + e_{t}), I_{0} = \frac{V}{8 R} (3 - e_{t}) .

t = 0^{+}

the three values are

I_{1} = 0

and

I_{2} = I_{0} = V / (4 R)

. At late times they are

I_{1} = V / (4 R)

I_{2} = V / (8 R)

, and

I_{0} = 3 V / (8 R)

. Thus

I_{1}

and

I_{0}

increase monotonically while

I_{2}

decreases; every transient has the same time constant. By

10 τ

, the exponential correction is only

e^{- 10} ≃ 4.54 \times 1 0^{- 5}

of its initial amplitude.

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