Mutual inductance as a psuedo-capacitor?

In summary, the conversation discusses the concept of mutual inductance as a pseudo-capacitor and the possibility of using capacitor-like equations to analyze the situation. The post also suggests visiting various resources for more information and clarification on the differences between inductance and capacitance.
  • #1
DMcG
6
0
Mutual inductance as a psuedo-capacitor?

1. I am testing how reduced the mutual inductance btwn two parallel ring-shaped inductors is when a material is placed btwn them. The primary coil is connected to an AC generator in series with an ammeter (on the order of mA), and the secondary in series with an ammeter.
[/b]
What I'm trying to do is calculate either the permittivity of the material, it's dielectric constant, or some ratio of them, by measuring the reduction in current thru the secondary. [/b]


2. I may have a solution to determining the ratio of epsilon/kappa by the equation epsilonnaught = epsilon / kappa:



3. The primary inductor L1 directs a B-field through the coils of L2, the strength of the field along the axis of the inductor is (munaught / 4 pi) [(2pi*R^2*I)/(x^2 + R^2)^3/2].
But I see elements of a parallell-plate capacitor btwn these coils:[/b]
[/b]
The primary current is alternating, but at a given time t the current dq1/dt will possesses q1 charges involved in inducting the current dq2/dt, and the secondary will possesses q2 charge, which must be less than q1. There should be a potential difference btwn this charge difference, and by the electromagnetic eqn E = -grad(V) there is an electric field pointing along the same axis as the gradient of this potential difference. [/b]
[/b]
Now a parallel-plate capacitor has a charge difference and so it also possesses a potential difference and has a (ideally) uniform electric field through it's separation distance. So if I CAN treat my mutual inductors as a cap, THEN I can use cap-like equations. By the duality relationship btwn inductance and capacitance, the work stored in the electric field of a cap is equal to the work stored in the magnetic field of an inductor: [/b]
(1/2)(L I^2) = (1/2)(C V^2) [/b]
Now V = E*d, where d is sepeartion btwn inductors. [/b]
Also, E = Q /(A*epsilonnaught) [/b]
My prof doesn't really trust my logic on this one. Any help?
 
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  • #2


1) This post doesn't belong in Advanced Physics; Electrical Engineering might be more appropriate.
2) To DMcG: I think you are quite confused about capacitance, inductance, and the differences between electric and magnetic materials. You generally can't treat mutual inductance like capacitance. Visit the following sites for some introductory material on these topics:

Elementary:
http://www.tpub.com/neets/book2/2.htm"
http://www.tpub.com/neets/book2/3.htm"

Wikipedia:
http://en.wikipedia.org/wiki/Inductance"
and http://en.wikipedia.org/wiki/Capacitance"

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html"
Click on *inductance on the index to the right, then the Inductance Concepts link. Clicking on the resulting balloons will take you through the various topics. Repeat for *capacitor.

Come back with questions and we'll do our best to answer them.
 
Last edited by a moderator:

1. What is mutual inductance as a pseudo-capacitor?

Mutual inductance as a pseudo-capacitor refers to the phenomenon in which two or more inductors are placed in close proximity and interact with each other, resulting in a capacitance-like effect. This is due to the changing magnetic fields created by one inductor inducing a voltage in the other, similar to how a capacitor stores charge.

2. How does mutual inductance as a pseudo-capacitor differ from a traditional capacitor?

While both mutual inductance and traditional capacitors store energy, they do so through different mechanisms. A traditional capacitor stores energy in an electric field between two conductive plates, while mutual inductance stores energy in a magnetic field between two inductors. Additionally, mutual inductance as a pseudo-capacitor typically has a higher capacitance value than a traditional capacitor of the same physical size.

3. What factors affect the mutual inductance as a pseudo-capacitor?

The mutual inductance as a pseudo-capacitor is affected by the number of turns in each inductor, the distance between them, the material used for the inductor cores, and the frequency of the AC signal passing through the circuit. Increasing the number of turns and decreasing the distance between the inductors will result in a higher mutual inductance value.

4. How is mutual inductance as a pseudo-capacitor used in practical applications?

Mutual inductance as a pseudo-capacitor has various practical applications, such as in power electronics and radio frequency circuits. It can be used to filter out unwanted signals or to couple signals between different circuits. It is also commonly used in transformers, where the changing magnetic fields between the primary and secondary coils result in energy transfer.

5. What are the limitations of mutual inductance as a pseudo-capacitor?

One limitation of mutual inductance as a pseudo-capacitor is its frequency dependence. As the frequency of the AC signal passing through the circuit increases, the mutual inductance value decreases. Additionally, mutual inductance can only store energy for a short period of time, unlike traditional capacitors which can hold a charge for longer periods. This makes it less effective for certain applications, such as energy storage.

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