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The current in coil of self inductance 5...

The current in coil of self inductance 5 mH changes from 2.5 A to 2.0 A is 0.01 second. Calculate the value of self induced e.m.f.

Text Solution

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Given : L-5mH, di- 0.5A dt -0.01s.
Self induced emf
`=epsi=-L(dl)/(dt)=5xx10^(-3)xx(0.5)/(0.01)`
`epsi=250xx10^(-3)V=0.25V`
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