- #1

- 11

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However , I cannot comprehend how this phenomenon can be explained purely based on induced voltages and currents. An intuitive understanding basically.

Can anyone help me out?

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- Thread starter suhasm
- Start date

- #1

- 11

- 0

However , I cannot comprehend how this phenomenon can be explained purely based on induced voltages and currents. An intuitive understanding basically.

Can anyone help me out?

- #2

- 12,134

- 161

If there is a nonzero current, then by Ohm's Law there is a nonzero voltage (due to the resistor).

Since there is a nonzero voltage, then by Faraday's law of induction the magnetic flux through the inductor must be changing.

The magnetic flux through the inductor is produced by the current. So a changing flux goes hand-in-hand with a changing current.

To figure out whether the current change causes an increase or decrease, use Lenz's Law and one of the right-hand-rules -- specifically, the one that tells you the direction of

Hope that helps.

- #3

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Current is merely the net flow of electrons through a material. If the material has resistance then we know that electrons are scattering off of the electrons which make up the conductor and thus loose speed in the process.

However , I cannot comprehend how this phenomenon can be explained purely based on induced voltages and currents. An intuitive understanding basically.

Can anyone help me out?

- #4

- 113

- 0

So long as there is no resistance in the circuit then the current will flow forever because no energy is discharged. This is a superconductor situation.

Put a low resistance into the circuit and with the given current as a starting value then energy will discharge slowly into the resistance and we will see the current decay as the energy is transferred from the magnetic field of the L to heat in the resistor.

Put a high resistance into the circuit and with the given starting current there will be higher loss rates in the resistor. There is still the same amount of energy in the L to start with so because the higher resistive load consumes it faster the current decays faster.

- #5

BruceW

Homework Helper

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And the typical timescale for the decay is given by the inductance, divided by resistance. So in the limit of very small inductance, we get what we expect - the current goes to zero very rapidly, so that we would hardly see the curve, it would look almost like a step function. (Well, the exponential tail never really disappears, but you know what I mean, it looks more like a step function, for systems with smaller inductance).

Edit: well, the shape of the decay function is always the same. But the timescale depends on the inductance, so if you had a circuit with inductance much smaller than resistance, then you would need to have a better 'time resolution' to be able to see the curve before it gets very close to zero.

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