# Magnetically Coupled Circuit

1. Aug 10, 2011

### berryberry

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 [Broken]

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: May 5, 2017
2. Aug 10, 2011

### Staff: Mentor

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. Aug 11, 2011

### berryberry

Would that be to convert it into a T-Circuit?

4. Aug 11, 2011

### Staff: Mentor

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. Aug 11, 2011

### berryberry

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 [Broken]

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: May 5, 2017
6. Aug 11, 2011

### Staff: Mentor

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: May 5, 2017
7. Aug 11, 2011

### berryberry

I'm not sure what you mean by this.

I assume replacing di/dt for just I?

8. Aug 11, 2011

### Staff: Mentor

Yup.

9. Aug 11, 2011

### mheslep

Impedance form: V=I*Z where Z=j*omega*M