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A 12 V battery connected to a coil of resistance 6 Ω through a switch, drives a constant current in the circuit. The switch is opened in 1 ms. The induced emf across the coil is 20 V. The inductance of the coil is
- 10 mH
- 8 mH
- 5 mH
- 12 mH
Correct answer: 10 mH
Solution
The induced emf in an inductor is given by the formula emf = -L (di)/(dt). Given the induced emf of 20 V and the change in current over the time interval of 1 ms, we can rearrange the formula to find the inductance L. Substituting the values leads to an inductance of 10 mH.
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