Question

When the 20.0 A current through an inductor is turned off in 1.50 ms, an 800 V emf is induced, opposing the change. What is the value of the self-inductance?

Final Answer

$60.0 \textrm{ mH}$

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 23, Problem 65 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A 20 amp current in an inductor is turned off which means the change in current will be the final current of 0 minus the initial current of 20 making a change in current of negative 20 amps but I am just writing positive 20 here coz we don’t really care about the sign of the induced EMF. So, change in current is 20 amps. This current just turned off in a time of one and a half milliseconds written as times 10 to the minus 3 seconds and the induced EMF we are told is 800 volts and so the question is, what is the self-inductance? So, our formula for induced EMF is inductance times rate of change of current and we will solve for

*L*by multiplying both sides by delta*T*over delta*I*and so switch the sides around. So, inductance is EMF times change in time over change in current. So, that’s 800 volts times 1 and a half times 10 to the minus 3 seconds divided by 20 amps giving an inductance of 60.0 millihenrys.