RL Circuits — Question 4

Question 4

Reversing the source does not reverse inductor current immediately

A series RL circuit has reached steady current V/R under source +V. At t=0 the source is switched to -V with R and L unchanged. Define the old current direction as positive.

1. Find current for t>0.

2. Find when current first crosses zero and its slope immediately after reversal.

Question 4 - Solution

Reversing the source does not reverse inductor current immediately

(a) Method and calculation

$$i_f=-V/R,\quad i(0^+)=V/R,\quad\tau=L/R.$$$$\boxed{i(t)=-V/R+(2V/R)e^{-t/\tau}}.$$

(b) Interpretation and further calculation

$$i=0\Rightarrow e^{-t/\tau}=1/2\Rightarrow\boxed{t=\tau\ln2}.$$$$\boxed{\dot i(0^+)=(-V-R(V/R))/L=-2V/L}.$$

Physical check

During the initial interval after reversal, current still flows in the old direction while the source absorbs electrical power.