How do I solve this circuit using the impedance form?

In summary: If you're dealing with sinusoidal sources of the same frequency and you're given the impedances of the reactive components (the inductances), you can dispense with the differential form and use the impedances directly. This will give you two equations in two unknowns (the currents).
  • #1
berryberry
5
0
I am currently revising over some past papers and this question has come up which I am having problems with. My notes don't show how to solve this and I don't have any worked examples.

So far from looking on the internet, I have seen two methods. The first is using an equation for energy and the second is to use KVL for each coil and then substituting the equations.

I've tried to understand using both of these methods but I still can't apply to this question.

http://img714.imageshack.us/img714/3693/elecprob.jpg

Uploaded with ImageShack.us

If anyone could give me pointers on what I need to be doing, or a brief example of how it's done, it would be greatly appreciated as I'm having a hard time getting my head around it.

Thanks.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
How might you adjust the circuit diagram to incorporate the effects of the mutual inductance (the j3Ω indicated above the inductors) into each loop?
 
  • #3
Would that be to convert it into a T-Circuit?
 
  • #4
berryberry said:
Would that be to convert it into a T-Circuit?

Nope. The mutual inductance acts in each loop as though it were a current-controlled dependent voltage source in series with that loops inductor, with the control currents being the currents in the other loop.

For example, if M is the mutual inductance, i1 the current in the first loop, and i2 the current in the second loop, then in the first loop you could insert a voltage source in series with the inductor with a value M*i2. The same can be done for the second loop with the control current being i1.

This manipulation of the circuit diagram should allow you to write the KVL loop equations for each loop in a straightforward fashion.
 
  • #5
If I'm understanding you correctly, that it is to use these equations (with the inclusion of R*I for the resistors)...
http://img189.imageshack.us/img189/750/unled1dg.jpg

Uploaded with ImageShack.us

If so what I have done so far is...

100/_0 = 10I1 + j10 dI1/dt + j3 dI2/dt
50/_45 = 5I2 +j5 dI2/dt + j3 dI1/dt

Although I'm not quite sure on how I would rearrange this, if this is the corrent way forward.
 
Last edited by a moderator:
  • #6
berryberry said:
If I'm understanding you correctly, that it is to use these equations (with the inclusion of R*I for the resistors)...
http://img189.imageshack.us/img189/750/unled1dg.jpg

Uploaded with ImageShack.us

If so what I have done so far is...

100/_0 = 10I1 + j10 dI1/dt + j3 dI2/dt
50/_45 = 5I2 +j5 dI2/dt + j3 dI1/dt

Although I'm not quite sure on how I would rearrange this, if this is the corrent way forward.

Yes, you're on the right track. Since you're dealing with sinusoidal sources of the same frequency and you're given the impedances of the reactive components (the inductances), you can dispense with the differential form and use the impedances directly.

You'll have two equations in two unknowns (the currents).
 
Last edited by a moderator:
  • #7
gneill said:
(the inductances), you can dispense with the differential form and use the impedances directly.

I'm not sure what you mean by this.

I assume replacing di/dt for just I?
 
  • #8
berryberry said:
I'm not sure what you mean by this.

I assume replacing di/dt for just I?

Yup.
 
  • #9
Impedance form: V=I*Z where Z=j*omega*M
 

FAQ: How do I solve this circuit using the impedance form?

What is a magnetically coupled circuit?

A magnetically coupled circuit is a type of electrical circuit where two or more components are connected through a shared magnetic field. This allows for the transfer of energy or signals between the components without the need for direct electrical connections.

How does a magnetically coupled circuit work?

In a magnetically coupled circuit, the magnetic field generated by one component induces a current in the other component, allowing for the transfer of energy or signals. This is known as mutual inductance and is based on Faraday's law of induction.

What are the advantages of a magnetically coupled circuit?

One of the main advantages of a magnetically coupled circuit is its ability to transfer energy or signals without the need for direct electrical connections. This can reduce the complexity and size of the circuit and also eliminate the risk of electrical shock. Additionally, magnetic coupling can provide isolation between components, allowing for different voltages to be used without interference.

What are some applications of magnetically coupled circuits?

Magnetically coupled circuits are commonly used in transformers, which are essential for power distribution and voltage transformation in electrical systems. They are also used in inductive coupling for wireless charging, data transfer in communication systems, and sensing in electronic devices.

How can I ensure efficient coupling in a magnetically coupled circuit?

Efficient coupling in a magnetically coupled circuit can be achieved by having a high mutual inductance between the components, which can be achieved by having a close physical proximity and proper alignment of the components. Additionally, using materials with high magnetic permeability can enhance the coupling efficiency.

Similar threads

Replies
4
Views
1K
Replies
35
Views
3K
Replies
3
Views
1K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
12
Views
1K
Replies
2
Views
1K
Replies
7
Views
1K
Back
Top