1.

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?

Answer»

Solution :`E = L(dI)/dt `
`THEREFORE L = e(dt)/dI`
`therefore L = 8xx5/100xx1/2` = 0.2H.


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