Help with mesh current analysis

[Odysseas]

Hello everyone,

I have a nasty circuit here, and I have been trying to perform a mesh current analysis (as described in the wikipedia article) on it, with little success... I have drawn the curcuit and have put up the equations.

**broken link removed**

The provided values are: R1 = 1kΩ, R2 = 2kΩ, R3 = 3kΩ, Iq1 = 1mA, Iq4 = 4mA, Uq2 = 2V and Uq3 = 3V, I have also noted them in the drawing.

I have some trouble putting up the equations for IM1 and IM2 (currents within Mesh1 and Mesh2). Also, can you please tell me if those few equations I managed to get out of the schematic are correct?

Thanks a lot!

 

[Ratchit]

Odysseas,

I have a nasty circuit here, and I have been trying to perform a mesh current analysis (as described in the wikipedia article) on it, with little success... I have drawn the curcuit and have put up the equations.

What is nasty about it? You already have the answer just by looking at it. You have a 1 ma current source in the west loop, and a 4 ma current source is the east loop. That gives 4 ma -1 ma = 3 ma in the branch dividing the east and and west loops. What more do you need?

Ratch

 

[Odysseas]

So I can safely assume that IM1 = 1mA, and IM2 = 4mA? Are you sure it is that simple? Sounds almost too easy...

 

[cowana]

Almost.

IM1 = -1mA
IM3 = 4mA

Then you can calculate IM2.

 

[Odysseas]

What is IM3? There are only two meshes...

 

[Odysseas]

Another question: The assigment asks for the internal resistance of this circuit between A and B. Shouldn't that be zero, due to the constant current source across A and B???

 

[crutschow]

Another question: The assigment asks for the internal resistance of this circuit between A and B. Shouldn't that be zero, due to the constant current source across A and B???

You are thinking of a voltage source. An ideal constant-current source has infinite impedance.

 

[Odysseas]

Ah, there we go Now I remember. To calculate the internal resistance, I have to break the connection at every constant current source, and place a short circuit across any voltage source. Is that correct?

 

[crutschow]

Ah, there we go Now I remember. To calculate the internal resistance, I have to break the connection at every constant current source, and place a short circuit across any voltage source. Is that correct?

That is correct.