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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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`E=-L*(dI)/(dt) or |E|=L*(dI)/(dt)`
Calculation of self induced e.f.m. `E=|E|=0.25V`
Detailed Answer:
Inductance of the coil `L=5xx10^(-3)H`
Initial current in the coil `=2.5A`
Final current in the coil `=2.0A`
Change in current `dI=2.0-2.5=-0.5A`
Time taken for the change, `dt=0.01`
`E=-L(dI)/(dt)=(-5xx10^(-3)(-0.5))/(0.01)=0.25V`
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