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A current of I=10 A is passed through th...

A current of I=10 A is passed through the part of a circuit shown in the figure . What will be the potential difference between A and B when I is decreased at constant rate of `10^(2)` amp/s at the beginning ?

Text Solution

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Applying the law of potential between the points A and B we obtain,
`V_(B)-V_(A) = -IR +E-L""(di)/(dt)`
`implies V_(a) -V_(A)=-10xx2+12 -5xx10^(-3)xx(-10^(2))=-20+12-0.5=-7.5` volt.
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