Past exam of the physics 2 paper of the University of Oxford / 2014 / 9 / iii

Table of contents
- Solution 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}) .

At

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.

Figure 1.
Switch-on currents in the resistive network containing one inductor
. Currents are normalized by V/R. Values at switching and at equilibrium are marked for all three source-diagram current directions. Details.

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