Difficulty: Easy
Correct Answer: A large voltage is induced (a high counter-emf spike appears) across the inductor
Explanation:
Introduction / Context:
When current through an inductor is interrupted, the stored magnetic energy must dissipate. The mechanism by which energy leaves the inductor involves a transient voltage often called flyback or counter-emf. Recognizing this behavior is crucial for protecting switches and semiconductor devices from overvoltage stress.
Given Data / Assumptions:
Concept / Approach:
Inductors oppose changes in current. If current attempts to drop rapidly to zero, the inductor develops whatever voltage is necessary (within circuit limits) to keep current flowing briefly. This produces a high counter-emf spike whose polarity sustains the former current direction, potentially arcing across switches or overstressing transistors unless a clamp (diode, snubber, TVS) is provided.
Step-by-Step Solution:
Stored energy: E = 0.5 * L * I^2 resides in the magnetic field.At turn-off, di/dt is large and negative.Inductor voltage: v_L = L * di/dt; a large magnitude results when di/dt is large.Hence a high-voltage spike (counter-emf) appears across the inductor terminals.
Verification / Alternative check:
Oscilloscope measurements of relay coils or buck converter inductors during switch turn-off show sharp voltage spikes. Adding a flyback diode limits the spike near one diode drop, confirming the underlying behavior.
Why Other Options Are Wrong:
Current doubles instantaneously: inductors prevent instantaneous current change; it cannot double.A new larger field expands: the field collapses; it does not expand.No voltage because source is gone: induced voltage is produced by changing flux, not by the source after turn-off.
Common Pitfalls:
Underestimating the magnitude of the spike and omitting protective components. Always consider clamp paths to control v_L during turn-off.
Final Answer:
A large voltage is induced (a high counter-emf spike appears) across the inductor
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