 Solution

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

The source diagram has

2 R

in series with two parallel branches: one containing

R

then

L

, and the other containing

2 R

. Let their common voltage be

u (t)

, measured from the upper node to the lower return. Kirchhoff's current law and the branch laws give

I_{0} = I_{1} + I_{2}, u = 2 R I_{2} = R I_{1} + L \frac{d I _{1}}{d t}, V = 2 R I_{0} + u .

Using

I_{2} = u / (2 R)

in the source equation yields

u = V /2 - R I_{1}

. Substitution into the coil branch gives

L \frac{d I _{1}}{d t} + 2 R I_{1} = \frac{V}{2} .

The switch was open for a long time, so

I_{1} (0^{-}) = 0

I_{1} (0^{+}) = 0

for a finite applied voltage.

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