Self-Induction — Question 4

Question 4

A finite voltage pulse changes current by its signed volt-second area

An ideal .0500-H inductor initially carries .400 A. Its passive voltage is +3.00 V for .0100 s and then -1.00 V for .0200 s.

1. Find current after each interval.

2. Find the duration of the second voltage that would instead restore the initial current.

Question 4 - Solution

A finite voltage pulse changes current by its signed volt-second area

(a) Method and calculation

$$\Delta i_1=\frac1L\int vdt=3.00(.0100)/.0500=.600\,A\Rightarrow i_1=1.00\,A.$$$$\Delta i_2=-1.00(.0200)/.0500=-.400\,A\Rightarrow\boxed{i_2=.600\,A}.$$

(b) Interpretation and further calculation

$$\Delta i_{\mathrm{cycle}}=0\Rightarrow3.00(.0100)-1.00T_2=0.$$$$\boxed{T_2=.0300\,s}.$$

Physical check

Finite voltage changes current continuously; equal signed voltage-time areas cancel the net current change.