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The current passing through a choke coil of 5 henry is decreasing at the rate of 2 ampere per second. The e.m.f. developed across the coil is
- 10 V
- – 10 V
- 2.5 V
- – 2.5 V
Correct answer: 10 V
Solution
For an inductor, the induced emf is $e=-L\,di/dt$. Here $L=5$ H and the current is decreasing at 2 A/s, so $di/dt=-2$ A/s. Therefore $e=-5(-2)=10$ V.
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