Inductors' voltage direction and Lenz Law

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SUMMARY

The discussion centers on the behavior of inductors in a de-energizing RL circuit, specifically addressing Lenz's Law and the direction of induced voltage. When the left switch is opened and the right switch is closed, the inductor generates an opposing voltage to maintain current flow, as indicated by the equation v = Ldi/dt. The voltage induced in the inductor transitions from positive to negative as the current decreases, illustrating that while inductors resist changes in current, they allow for changes in voltage. This behavior is crucial for understanding the dynamics of inductors in electrical circuits.

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  • Understanding of RL circuits and their components
  • Familiarity with Lenz's Law and its implications
  • Knowledge of the relationship between voltage, current, and inductance
  • Basic grasp of electromagnetic principles and magnetic fields
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  • Investigate the role of inductors in AC circuits and their impedance characteristics
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Nikitin
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In the attached figure, a de-energizing RL circuit is drawn (left switch is opened while right switch is closed). In the drawing the voltage induced in the inductor is put to be negative so that the math works out.

However, that would be against Lenz's law, wouldn't it? The inductor should be trying its best to increase the current, not work against it. According to my intuition, the inductor should be "lifting the current" to a higher potential to compensate for the frictional loss in the resistance element.

Hmm, but then again, the electrical field is pointing downwards in the inductor which does accelerate the current.. I'm kinda confused here.
 

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Maybe it would help to first think of how you got to t=0; in other words, a lot happened before the switches changed.

Inductors react to changing current and don't quite like rapid current changes, so...during the time before t=0, the current in the circuit worked its way from zero at t=-N sec to some positive value at t=0 sec and all along the inductor developed an opposing voltage to the source... i and v as shown are both positive since v = Ldi/dt.

Think of this changing current charging the inductor with magnetic field.

At t=0, without the voltage source, the current no longer has a pushing force behind it and can no longer hold the inductor magnetic field...so, now the current without a voltage source behind it wants to start slowing down, but the inductor does not like this and starts using its own magnetic field to try its best to keep the current flowing...it won't succed for long, but the current flows for a little longer starting at t=0 from the same value it had before and working its way to zero.

During this t>0, because the current is now decreasing instead of increasing, the sign of the derivative changes sign and so, the voltage as defined goes instantaneously from positive to negative...and this is fine, because inductors mind currents changing, but don't mind voltages changing (capacitors, on the flip side, don't let voltage change suddenly, but the current can change just fine).

Hope this helps.
 

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