Nodal Analysis using Complex Numbers

[willow16v]

Hi,
I would be grateful of any advice on how to solve the problem below.

My aim is to find the voltages at nodes 2, 3, and 4, by means nodal analysis and thus creating simultaneous equations and solving them using matricies or matrix in order to prove that these theories work.

I understand kirchhoffs law etc however its the complex numbers that are throwing me when working out the equations at each node.
, e.g multiplying and dividing by j etc.

I have tried for weeks now and it seems i am beaten yet again on my failure of rearranging algebra.

Any help would be great.

A=60 B=15 C=153


See diagram Below.

Cheers Chris

 

[gnurf]

As an example, consider the isolated voltage divider consisting of E1=100, Z2=10, Z3=j20.

The voltage at point 2 would be

$$V_{2} = E_{1} \times \frac{Z_{3}}{(Z_{2}+Z_{3})}$$

Adding Z2 and Z3 is straight forward: Z2+Z3 = 10 + 20j. This is an ordinary complex number, with an amplitude and phase angle as usual. The fraction E1*Z3/(Z2+Z3) can be reduced by multiplying with the complex conjugate to a more familiar form:

$$V_{2} = 80 + 40j$$

Since Z3 is a complex (i.e reactive) component, the phase angle,$$\phi$$, at V2 is altered:

$$\phi = {tan}^{-1}(\frac{40}{80}) = 26.6 \:deg$$

There's nothing mystical going on here. It's just a series-parallel circuit, but -- due to the reactive components -- you need to carry with you the phase information, which means you need another dimension (the complex plane).

 

[berkeman]

As an example, consider the isolated voltage divider consisting of E1=100, Z2=10, Z3=j20.

The voltage at point 2 would be

$$V_{2} = E_{1} \times \frac{Z_{3}}{(Z_{2}+Z_{3})}$$

Adding Z2 and Z3 is straight forward: Z2+Z3 = 10 + 20j. This is an ordinary complex number, with an amplitude and phase angle as usual. The fraction E1*Z3/(Z2+Z3) can be reduced by multiplying with the complex conjugate to a more familiar form:

$$V_{2} = 80 + 40j$$

Since Z3 is a complex (i.e reactive) component, the phase angle,$$\phi$$, at V2 is altered:

$$\phi = {tan}^{-1}(\frac{40}{80}) = 26.6 \:deg$$

There's nothing mystical going on here. It's just a series-parallel circuit, but -- due to the reactive components -- you need to carry with you the phase information, which means you need another dimension (the complex plane).

Be careful not to do too much of the poster's work for them. This is homework (originally misplaced in EE where you saw it and replied), so the original poster (OP) needs to do the bulk of the work.

However, you left the rest of the problem for him, so I'm going to leave your description for him. Let's see if that's enough for him to follow your lead and finish the solution.

 

[gnurf]

Be careful not to do too much of the poster's work for them. This is homework (originally misplaced in EE where you saw it and replied), so the original poster (OP) needs to do the bulk of the work.

Right, I'll keep that in mind, thanks.

 

[berkeman]

Right, I'll keep that in mind, thanks.

Thanks. Good post BTW. Thanks for the help!

 

[willow16v]

thankyou for your reply, i will try again using your ideas.

The circuit i understand, but its the parralel reistor Z1 that confuses me when working out the voltage at 3 using nodal analysis.

Many thanks
Chris