An inductor is connected to a battery through $e_{1}$ switch induced emf is a when the switch is pressed and $e_{2}$ when the switch is opened. Then:

e_{1} = e_{2}

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e_{1} > e_{2}

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e_{1} < e_{2}

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e_{1} > / < e_{2}

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Solution

The correct option is C $e_{1} < e_{2}$
When the switch is closed, the current grows in the circuit but due to self-inductance of the switch inductor the growth is slow. On disclosing the switch the current suddenly decrease to zero on account of the infinite resistance of the circuit. Thus, rate of growth of the current is lower as compared to the rate of decays.

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An inductor is connected to a battery through e1 switch induced emf is a when the switch is pressed and e2 when the switch is opened. Then:

An inductor is connected to a battery through $e_{1}$ switch induced emf is a when the switch is pressed and $e_{2}$ when the switch is opened. Then: