 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit/9/b/solution

(V - V_{a}) / R = V_{a} / (2 R) + I_{2}

. Since

I_{1} = V_{a} / (2 R)

I_{1} (t) = \frac{V}{3 R} - \frac{1}{3} I_{2} (t) = \frac{V}{3 R} - \frac{2 V}{15 R} e^{- t / τ} .

Thus

I_{1}

rises from

V / (5 R)

V / (3 R)

, whereas

I_{2}

falls from

2 V / (5 R)

to zero. Their derivatives have opposite signs, and both approach their limiting values on the same time scale

τ = 5 RC /3

V / R

and time in units of

τ

 New to topics? Read the docs here!