# Complex numbers, matricies and Kirchoff's laws

1. Jan 10, 2007

### Tone

Hello everyone..

I have quite a problem regarding A.C. circuit analysis using complex numbers and 2x2 matricies. :yuck:

* The aim is to find the current in each of the two loops and apply Kirchoff's laws. I believe the overall aim is just to prove that the laws are actually in place..

(SEE ATTATCHMENT)

_________________________________________________________________

Kirchoff's Law's:

1) The potential difference (voltage) in each loop must add up to zero

2) At any node (i.e. any circuit junction) the current flowing into the junction must equal the current flowing away from it.
_________________________________________________________________

I've spent a long time looking at this but am finding it very confusing...

At first I thought it was a simple case of Ohm's law (for AC) to find the current ie. I = E / Z (aka I = V / R )

So I began looking for the current in the first loop (I1):

I1 = E1 / (Z1 + Z3) in the complex number form (using conjutives etc)

However this resulted in very obscure results and the j-operators (imaginary numbers) would not cancel out. I was hoping for a simple whole number.

SO now I'm back to square one. I'm sure I have to create a complex number formula, and then put these into a 2x2 matrix. But I don't know the necessary steps to take, or how exactly I would find the current.

I'm still not really sure on the purpose of Matricies either, could they be applied onto each circuit junction/node to evaluate currents going in and out and hence prove Kirchoff's laws? Or am I way off the mark.. is it more to do with phasors and sineforms?

Can anybody help? Thank you

#### Attached Files:

• ###### kirchoff.jpg
File size:
19.1 KB
Views:
77
Last edited: Jan 10, 2007
2. Jan 10, 2007

### interested_learner

The approach of taking the power sources independently is correct. However, the voltage source must be shorted. Remember in superposition, open current sources, short voltage sources. This means that I1 = E1/(Z1 + Z3 || Z2), not I1 = E1/(Z1+Z3). Do that for each power source and things should work out better.

Matricies are just an easy way to solve simultaneous equations. They may look more difficult now, but believe me they will save you much time in the future.

Last edited: Jan 10, 2007
3. Jan 11, 2007

### BobG

The arrows are showing direction of current for your loops? Or are they supposed to be showing which is the positive/negative terminal of the voltage sources? (you do need to know that since reversing the terminals will change your answer).

In any event, node voltage analysis would work better on this particular circuit. You need a ground, which you can place at one node, leaving you only one node to solve.

If you have to use mesh current (which you do have to know), you will have two unknowns and two equations which can be solved a couple of different ways. One way is to create a 2 x 2 matrix and solve it using Cramer's Law, which should be in your book somewhere (in the appendix?) or can be found on-line.

There's no particular reason the imaginary part of the answer should cancel out.

4. Jan 13, 2007

### mjsd

Yes, I agree that node-voltage analysis would be much better. In fact, you can solve all problems without knowing mesh current anaylsis anyway (mesh current doesn't work on non-planar networks). by the way, this circuit is probably too simple to invoke the use of superposition principle in my opinion. but as BobG mentioned really need to know the reference direction of the currents.

Assuming that both $$I_1, I_T-I_1$$ flows away from the node, ie. $$I_T$$ flows up the middle branch containing Z3, and assuming that the arrows indicate polarity of the phasor sources with arrow-head meaning "+" (higher potential), then these are the equations: (hope the LaTeX codes work)

$$\displaystyle{\frac{V_1 - E_1}{Z_1}+\frac{V_1}{Z_3}+\frac{V_1 + E_2}{Z_2}=0}$$

can now solve for $$V_1$$ which in general is a complex number.

NB: $$V_1$$ is node voltage at top node, while the bottom node is set to GND. also the first term is just $$I_1$$, second is $$-I_T$$, third is $$I_T-I_1$$ (all with the flow-out-of-node reference direction.)

hope this helps.