Question

A device is turned on and 3.00 A flows through it 0.100 ms later. What is the self-inductance of the device if an induced 150 V emf opposes this?

Final Answer

$5.00\textrm{ mH}$

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 23, Problem 58 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A device starts off with no current flowing through it and then 0.100 milliseconds later, 3.00 amps are flowing through it so this change in current occurs over a time of 0.100 times 10 to the minus 3 seconds and the change in current is the final current because it starts off at zero so the change is 3.00 amps and the induced

*emf*in the device is 150 volts and we are asked to figure out what is its self-inductance? So I have an absolute value symbol around here just because the formula is*emf*induced is negative times self-inductance times the rate of change of current but the negative sign is from Lenz's law to indicate the direction of the*emf*and that direction is not important for our purposes and*L*is just always a positive number and so we want*L*to work out to be positive so we put absolute value around this*emf*. So the magnitude of the*emf*then is the self-inductance times the change in current divided by change in time and we can solve this for*L*by multiplying both sides by*Δt*over*ΔI*and then we get that the self-inductance is 150 volts times 0.100 times 10 to the minus 3 seconds divided by 3.00 amps, which is 5.00 millihenries.