Past exam of the physics 2 paper of the University of Oxford / 2022 / 1 / ii

Solution

Put

s = e^{- R Δ t / L}

and

J = V / R

, with

0 < s < 1

I_{1} = s I_{0}

. Position B reconnects the battery, so the driven RL circuit obeys

L \dot{I} + R I = V

e^{Rt / L}

, or subtraction of the constant solution

J

, gives

I (t) = J + (s I_{0} - J) e^{- R (t - Δ t) / L}, Δ t < t < 2Δ t .

Consequently

I_{2} = s^{2} I_{0} + J (1 - s) .

The same two intervals give the recurrence relation

I_{2 n + 2} = s^{2} I_{2 n} + J (1 - s)

I_{*} = J / (1 + s)

. Subtracting this fixed point makes the recurrence relation homogeneous:

I_{2 n} - I_{*} = s^{2 n} (I_{0} - I_{*}), n \to \infty lim I_{2 n} = \frac{V / R}{1 + e ^{- R Δ t / L}} .

This is the upper endpoint of the eventual periodically switched RL circuit response, immediately before switching from B to A. The lower endpoint is

s J / (1 + s)

. Their sum is

J

, consistent with equal charging and discharging intervals.

There are no discussions about this article yet.

There are currently no matching articles.