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When the current in a coil changes from 0 to 2A to 4A in 0.05sec, then e.m.f. developed in the coil is 8V. What is the coefficient of self-induction of the coil? |
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Answer» Solution :`E = L(dI)/dt ` `THEREFORE L = e(dt)/dI` `therefore L = 8xx5/100xx1/2` = 0.2H. |
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