## A doubt about inductors and capacitors

[/itex]Hi I am new here

I know (correct me if wrong please ) that inductors store current in its magnetic field, and capacitors store voltage in its electric field

But, why when disconnecting a inductor charged from a DC circuit, a spark is produced at the disconnected point and it loses its energy while the capacitor not and the electricity is stored like a battery? I understand that $I = \frac{V}{R}$ and the spark in inductors are produced because Voltage rises to compensate the resistance when opening the circuit but, why does a capacitor voltage stay the same despite no current and infinite resistance and a inductor discharges immediately at the same circumstances?
Many thanks.
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Recognitions:
Homework Help
 inductors store current in its magnetic field, and capacitors store voltage in its electric field
That would be a very confused way of putting it yes.

Capacitors store energy in their electric fields.
Inductors store energy in their magnetic fields.

Inductors cannot become "charged" in a DC circuit ... in a steady-state DC circuit they look like a short circuit and act as an electromagnet. If you disconnect one, the collapsing magnetic field induces further current in the coil.

Where inductors are just a wire (short circuit) in DC, capacitors are a gap (open circuit). The power supply just moves charge from one side of the gap to the other. Disconnect the supply and there is no path for the charges to get back ... so the charges get stored.

In the fluid-analogy for electric circuits, a capacitor is a water-tank. I don't think there is an equivalent to an inductor.

Recognitions:
 Quote by Simon Bridge I don't think there is an equivalent to an inductor.
If a flow or water corresponds to the current, a sort of analogy would be a propellor in the water pipe that all the water has to flow through, turning a heavy wheel. When you apply some water pressure, the flow/current can only increase slowly because it takes time to get the wheel spinning. If the flow is steady, the wheel doesn't affect it (ignoring friction, which corresponds to the resistance of the wire in a real inductor).

If you try to stop the water flow suddenly, you get a huge pressure (voltage) because the kinetic energy in the wheel is still trying to pump the water through the propellor, but the water has nowhere to go - except possiblly to burst the pipe (send a spark across the electrical switch).

OK, that's not so good as some of Maxwell's mechanical analogies for EM fields built from infinte arrays of gyroscopes coupled together with linkages - but it's a start.

It could be fun making a real device like this plus a water tank to simulate an LC circuit - a project for a science fair, maybe? Demonstrate resonance by rasiing and lowering the tank a bit, and sloshing water everywhere if you do it at the right frequency....

Recognitions:
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## A doubt about inductors and capacitors

That's not bad: water-wheels are usually used to represent light-bulbs. But you are right - when the flow stops the wheel wants to keep going.
 A collapsing magnetic field induces an emf. There can only be an induced current if there is a complete circuit.

Recognitions:
Homework Help
 A collapsing magnetic field induces an emf. There can only be an induced current if there is a complete circuit.
Well OK - though you can have a current without a complete circuit, the cited example did involve a spark so presumably the emf was strong enough to push charges across the gap.
 Mentor The analogy of a disconnected inductor (static case: no current flow) is a short-circuit for a capacitor (static case: no voltage). And a disconnected capacitor (infinite resistance) corresponds to a inductor with a short-circuit (no resistance).
 Faradays laws state how to calculate induced emf not induced current.

 Quote by Simon Bridge In the fluid-analogy for electric circuits, a capacitor is a water-tank.
not the best fluid analogy for a capacitor. a better analogy is a cylinder with a spring-loaded piston inside.

 I don't think there is an equivalent to an inductor.
sure there is. a water turbine with the turbine shaft connected to a flywheel.

 Quote by mfb The analogy of a disconnected inductor (static case: no current flow) is a short-circuit for a capacitor (static case: no voltage). And a disconnected capacitor (infinite resistance) corresponds to a inductor with a short-circuit (no resistance).
Yes dualogs and analogs - I think you mean dualog. Analogs rather refers to system elements of the same type.

Wikipedia must have a good long discussion for any reader wishing to expand their knowledge.

 Quote by derek10 [/itex]Hi I am new here I know (correct me if wrong please ) that inductors store current in its magnetic field, and capacitors store voltage in its electric field But, why when disconnecting a inductor charged from a DC circuit, a spark is produced at the disconnected point and it loses its energy while the capacitor not and the electricity is stored like a battery? I understand that $I = \frac{V}{R}$ and the spark in inductors are produced because Voltage rises to compensate the resistance when opening the circuit but, why does a capacitor voltage stay the same despite no current and infinite resistance and a inductor discharges immediately at the same circumstances? Many thanks.
We usually think of a DC circuit being connected to a voltage source. A capacitor will charge up only to the value of the voltage of the battery and then no more current will flow. For an inductor connected to a battery the rate of change in current is what gives the 'resistance' to the battery, so if you have a strong enough battery and your inductor is constructed well enough, then very high currents can be achieved.

But, if you consider a current source, then since the current is not changing then no voltage will be observed across the inductor. On the other hand, the current source is continiously pumping current into a capacitor which can then charge up to very high potential. You can then witness sparks across a gap when the voltage becomes high enough. I am not absolutely sure but storm clouds would acquire the high potentials this way and you would call the discharge lightning and thunder.

 Quote by truesearch Faradays laws state how to calculate induced emf not induced current.
I don't wish to be rude, but that is only partially correct. Faraday's Law, herein "FL", relates magnetic flux to voltage. Specifically, FL relates net flux to net voltage. When a time-varying magnetic flux is present in the vicinity of a conductor, a Lorentz force acts on the conductor's free electrons, resulting in charge motion, which is current. If the path is open, current does indeed flow, but it is called a displacement current, which can be small, or at very high frequencies, quite large.

If the loop is open, charges accumulate at each end, and there is an emf across the gap between the 2 ends of the conductor. As the mag flux varies with time, these charges move and displacement current continues. Even with an open path there is current as well as voltage.

If the loop is closed with a resistance R, obviously we have conduction current as well as displacement current. Assuming it is low enough frequency where conduction is much greater than displacement. The current in the loop has its own magnetic flux encircling the conductor. Per law of Lenz, LL herein, its polarity is opposite to that of the external field. So the actual flux enclosed by the loop is the sum total of the external plus the internal flux associated with the current in the wire.

We have mutual & self induction going on simultaneously. FL states that the voltage one time around the loop equals the negative of (LL) the time derivative of magnetic flux. But we must remember that the voltage, or emf if you prefer, is a net quantity, as is the flux.

A good example is to let the loop be superconducting. The voltage once around the loop becomes zero. This is due to the current in the loop having a magnetic flux around the wire that cancels the incoming external flux. Hence we have zero emf.

If the loop resistance R, is not zero, but very small, we have about the same current, and a very small voltage. Again, per LL, cancellation takes place. But if the loop R is made larger than a critical value so that the loop current is too weak to produce a magnetic flux that cancels the external flux, then the flux "phi" in FL, is essentially the external flux. Increasing R results in a loop emf that hardly changes. When R is large, the internal flux generated by the loop current is very small compared to the external incoming flux, and the net flux is very nearly that of the external flux. Under these conditions, we can state that the induced voltage, i.e. work per unit charge once around the loop, is the negative of the time derivative of the external flux.

But for low values of R, we cannot make that assumption. I don't wish to nitpick, but that is a point that should be made. I will elaborate if needed. Best regards.

Claude
 I have a limited range of about 12 A level text books and none of the explanations therein relate to anything posted here. I know that an emf involves charges separated and that means there must have been a 'current' for however long. I will confine my interest to the text book explanation of faradays law in terms of induced emf and rate of change of flux linkage. I am vaguely interested in how you would calculate the currents you allude to (or the voltages) but it looks like it will be beyond me. The original post is from someone new here and I hope that he is able to pick his way through everything posted and come out wiser.

 Quote by truesearch I have a limited range of about 12 A level text books and none of the explanations therein relate to anything posted here. I know that an emf involves charges separated and that means there must have been a 'current' for however long. I will confine my interest to the text book explanation of faradays law in terms of induced emf and rate of change of flux linkage. I am vaguely interested in how you would calculate the currents you allude to (or the voltages) but it looks like it will be beyond me. The original post is from someone new here and I hope that he is able to pick his way through everything posted and come out wiser.
http://www.physicsforums.com/showthr...46#post2163546

The above link is to a computation sheet I posted a few years ago. A loop which has a resistance R, inductance L, frequency omega, & external flux phie, can be computed for current, voltage, & internal flux phii. I will elaborate if needed.

Claude
 I read your computation sheet which is interesting. Is it published in any recognised text books? I cannot find a recognised text book that states anything other than Induced emf = rate of change of flux linkage. none state any reference to induced current as a version of faradays law. There must be a good reason for that ! It is misleading to confuse the 'displacement' current, ie the movement of electrons giving rise to the induced emf with the induced current in a complete circuit resulting from said induced emf. anyone wanting to get to grips with faradays law should stick to the text book explanations.