Quick conceptual question about inductance

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SUMMARY

The discussion centers on the behavior of inductors in electrical circuits, particularly after a switch is opened following a period of closure. It is established that when the switch is closed, the inductor stores energy in its magnetic field, and the current through it cannot change instantaneously. Upon opening the switch, the energy does not remain stored in the inductor indefinitely; rather, it dissipates through resistive elements in the circuit, such as the series resistor. The relevant equations governing this behavior include V(t) = L (dI(t)/dt) and I(t) = (1/L) ∫ V(t) dt.

PREREQUISITES
  • Understanding of inductance and magnetic fields
  • Familiarity with basic circuit components: resistors, inductors, and batteries
  • Knowledge of the equations governing inductors, specifically V(t) = L (dI(t)/dt)
  • Concept of energy dissipation in electrical circuits
NEXT STEPS
  • Study the behavior of inductors in RL circuits
  • Learn about energy storage and dissipation in inductors
  • Explore the implications of the time constant in RL circuits
  • Investigate the effects of different resistor values on current flow in inductive circuits
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Electrical engineering students, circuit designers, and anyone interested in understanding the dynamics of inductors in electrical circuits.

JDiorio
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Ok so i understand that after a switch in a circuit is closed for awhile the inductor acts like a battery, however, what happens after the switch is opened after being closed for awhile.
A review question asks whether the energy:

- is converted to sound and spark
- remains stored in the inductor
- goes into driving the current across the resistor

I want to say it is stored in the inductor, but i know that after the switch is left open for awhile, the emf is zero. So i don't know whether the energy is considered to be stored in the inductor.
 
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JDiorio said:
Ok so i understand that after a switch in a circuit is closed for awhile the inductor acts like a battery, however, what happens after the switch is opened after being closed for awhile.
A review question asks whether the energy:

- is converted to sound and spark
- remains stored in the inductor
- goes into driving the current across the resistor

I want to say it is stored in the inductor, but i know that after the switch is left open for awhile, the emf is zero. So i don't know whether the energy is considered to be stored in the inductor.

Welcome to the PF. Why do you say that the inductor acts like a battery? I hadn't heard that analogy before, and it sounds wrong unless I'm missing some subtlety.

What is the equation that relates the current through an inductor and the voltage across it? And what about a real inductor limits the build-up of current when a constant voltage is put across it?

And what resistor are you alluding to in the problem statement above?
 
Thanks..

Maybe I am wrong. I had the impression that while the switch is closed an inductor takes on a charge and after the switch is closed, the inductor can keep the current flowing until the charge dissipates.

the circuit consists of a battery in series with a resistor in series with an inductor.. and then another resistor in parallel with the inductor.
 
JDiorio said:
Thanks..

Maybe I am wrong. I had the impression that while the switch is closed an inductor takes on a charge and after the switch is closed, the inductor can keep the current flowing until the charge dissipates.

the circuit consists of a battery in series with a resistor in series with an inductor.. and then another resistor in parallel with the inductor.

Where is the switch in the circuit?

Current through an inductor stores energy in the magnetic field of the inductor. The current cannot change instantaneously (just as the voltage across a capacitor cannot change instantaneously). The equation that I was hoping you would write out for the inductor is:

V(t) = L \frac{dI(t)}{dt}

That means that the voltage across the inductor is equal to the inductance L multiplied by the time rate of change of the current through the inductor. The other way to write this equation is in the integral form:

I(t) = \frac{1}{L} \int {V(t) dt}

So if you have a constant voltage across an ideal inductor, the current climbs without limit. In the circuit you describe, the series resistor will limit the current...
 

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