Question:
An inductor is connected to a battery through a switch. Induced emf is \(e_1\) when the switch is pressed and \(e_2\) when the switch is opened. Then
An inductor is connected to a battery through a switch. Induced emf is \(e_1\) when the switch is pressed and \(e_2\) when the switch is opened. Then
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Induced emf always opposes the change in current (Lenz's law).
Updated On: Apr 23, 2026
- \(e_1 = e_2\)
- \(e_1>e_2\)
- \(e_1<e_2\)
- \(e_1>e_2\) or \(e_1<e_2\) depending on battery voltage
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Induced emf \(e = -L\frac{di}{dt}\). It depends on the rate of change of current.
Step 2: Detailed Explanation:
When the switch is pressed, current increases and induces emf opposing the rise.
When the switch is opened, current decreases and induces emf opposing the fall.
In both cases, the induced emf acts to oppose the change in current, and is considered equal in magnitude under ideal conditions.
Step 3: Final Answer:
Thus, \(e_1 = e_2\).
Induced emf \(e = -L\frac{di}{dt}\). It depends on the rate of change of current.
Step 2: Detailed Explanation:
When the switch is pressed, current increases and induces emf opposing the rise.
When the switch is opened, current decreases and induces emf opposing the fall.
In both cases, the induced emf acts to oppose the change in current, and is considered equal in magnitude under ideal conditions.
Step 3: Final Answer:
Thus, \(e_1 = e_2\).
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