RL Circuits — Question 9

Question 9

A linearly increasing source produces an RL current that lags its resistive limit

A series RL circuit starts with zero current and is driven by $V(t)=\alpha t$ for t nonnegative. Let $\tau=L/R$.

1. Solve for i(t).

2. Find the early-time leading behavior and the long-time difference from the purely resistive current alpha t/R.

Question 9 - Solution

A linearly increasing source produces an RL current that lags its resistive limit

(a) Method and calculation

$$L\dot i+Ri=\alpha t\Rightarrow\boxed{i(t)=\frac\alpha R[t-\tau(1-e^{-t/\tau})]}.$$$$t\ll\tau:\quad e^{-t/\tau}=1-t/\tau+t^2/(2\tau^2)+\cdots.$$

(b) Interpretation and further calculation

$$\boxed{i(t)\simeq\alpha t^2/(2L)}\quad(t\ll\tau).$$$$t\gg\tau:\quad\boxed{\alpha t/R-i(t)\to\alpha L/R^2}.$$

Physical check

The initial quadratic rise follows because both initial current and initial source voltage are zero.