The current passing through a choke coil of 5 henry is decreasing at the rate of 2 ampere/sec. The e.m.f. developing across the coil is

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10 V

– 10 V

2.5 V

– 2.5 V

answer is A.

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Detailed Solution

Given didt=2A/sec.,L=5H∴e=Ldidt=5×2=10V

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