Engineering AC circuit analysis -- mesh and nodal

[his_tonyness]

Homework Statement

Determine, using the values given in TABLE A, the current I in the circuit of FIGURE 2 by:
(a) mesh analysis
(b) nodal analysis.

Any help verify for part (a)? And to help me get started with (b) as V3 is driving me nuts.

Homework Equations

The Attempt at a Solution

For (a) my equations are
120=2I1+(I1-I2)(-j5) ---Loop I1
(-j5)(I2-I1)+(-j5)(I2-I4)+(j4)(I2-I3)=0 ---Loop I2
(j4)(I3-I2)+(4)(I3)=j120 ---- Loop I3
-14.14+j14.14=(-j5)(I4-I2) ---Loop I4

Answers
I1=23.542+j17.5695 A
I2=16.5142+j2.98628 A
I3=21.764+j24.7503 A
I4=13.6862+j5.81428 A

 

[gneill]

Note that V₃ creates a fixed potential difference across Z₂. This means for loop 2 you can forget about Z₂ and just use that known potential difference in its loop equation. Since you're looking for current $I$, you won't even need to include loop 4 in your analysis (or if you like, exchange the positions of V₃ and Z₂ on the diagram. Since their node connections remain the same as before, the circuit operation does not change at all and the current through Z₂ becomes obvious...).

 

[his_tonyness]

Note that V₃ creates a fixed potential difference across Z₂. This means for loop 2 you can forget about Z₂ and just use that known potential difference in its loop equation. Since you're looking for current $I$, you won't even need to include loop 4 in your analysis (or if you like, exchange the positions of V₃ and Z₂ on the diagram. Since their node connections remain the same as before, the circuit operation does not change at all and the current through Z₂ becomes obvious...).

Thankyou for your kind assistance, I have little confidence with circuits. I did have a suspicion that the current through Z2 would be I=V3/Z2. I will attempt another solution.

 

[his_tonyness]

Hello, I swapped Z2 with V3 and came up with the following newer equations.;
(a)
Loop 1 --- 120-(2)I1-(I1-I2)(-j5)=0
Loop2 --- (-j5)(I2-I1)+(j4)(I2-I3)-14.14+j14.14=0
Loop3 --- (j4)(I3-I2)+(4)I3-j120=0

Answers;
I1=19.1624+j12.5642
I2= 14.1367-j3.77088
I3= 23.9538+j20.1829

Not 100% percent sure how what to do with those loop currents to get current through Z4.

Part (b)(Updated PDF file detailing labelled nodes.)
V20-V30=14.14+j14.14 (supernode)
and ((120-V20)/2)+(0-V20)/-j5)+(0-V30)/j4)+(j120-V30)/4=0
So..., V20= 86.3761+j45.7584 and V30=72.2361+j31.6184

Current I is V20/Z4= -9.15168+j17.2752 A

Am I alright with my approach so far?

 

[gneill]

You need to be a bit careful with the voltage supplies when you walk the loop. If you're summing potential drops (as positive numbers) then passing through a voltage supply from + to - counts as a positive drop. If you're summing potential rises, then each potential change through a resistance in the direction of the current is a negative value, and passing through a voltage source from + to - counts as a negative change, too.

So take another look at your Loop 3 equation. It looks like you're summing drops as positive values as you "walk" clockwise around the loop, but then you make the drop across V2 negative...

For Loop 2, I don't understand how your V3 components ended up with different signs.

Your nodal analysis looks fine to me. You should be able to compare results between mesh and nodal if you determine $I$ by both methods.

 

[his_tonyness]

Ok, for loop analysis I have, Current I is I1 - I2 = (16.8119-22.8792i)-(25.9636-40.1544i)=-9.1517+j17.2752

Mesh analysis= Current I is V20/Z4= -9.15168+j17.2752 A

Sounds good to me don't you think?

 

[gneill]

Ok, for loop analysis I have, Current I is I1 - I2 = (16.8119-22.8792i)-(25.9636-40.1544i)=-9.1517+j17.2752

Mesh analysis= Current I is V20/Z4= -9.15168+j17.2752 A

Sounds good to me don't you think?

it not only sounds good, but it looks good too :)

 

[his_tonyness]

Thanks for your help!:)

 

[justadaftspark]

Hi Folks, Could anyone help please? I'm trying to find the same loop currents as His_Tonyness and this is what I have so far

LOOP 1 V1=Z1(i1)+Z4(i1-i2) = 120+J0 = i1(2-J5)-i2(0-J5)
LOOP 2 -V3=Z5(i2-i3)+Z4(i2-i1) = -14.14+J14.14 = -i1(0-J5)+i2(0-J1)-i3(0+j4)
LOOP 3 -V2=Z5(i3-i2)+Z3(i3) = - 0+J120 = -i2(0+J4)+i3(4+J4)

When I plug these into a spreadsheet they don't work out when putting them back into the equation...Thanks

 

[KatieMariie]

Hi Guys, I've got the same as above^^; but am not sure on how to proceed with my mesh analysis - how do i solve the system of equations?

 

[gneill]

Hi Guys, I've got the same as above^^; but am not sure on how to proceed with my mesh analysis - how do i solve the system of equations?

Exactly the same? Because I've spotted a sign error in justadaftspark's second loop equation's numerics (the symbolic version is fine).

Once that's sorted out, use whatever technique you're familiar with to solve the three equations in three unknowns.

 

[KatieMariie]

No not exactly, I've factored the equation differently. but no worry! in the mean time i'd had a chat with my work colleague who's on the same course and between us we've managed to figure it out. :) thanks!

 

[KatieMariie]

Now we're working on part b and are stumped. not sure where to begin with our reference node?

 

[gneill]

Now we're working on part b and are stumped. not sure where to begin with our reference node?

You'll have to show more effort there. Have you identified the essential nodes at least?

 

[KatieMariie]

Sorry, i thought this forum was for guidance? or am i mistaken? I'm not looking the the whole answer obviously, just a bit of guidance would be appreciated.

The two nodes connecting Z4, Z2 and Z3 are both essential, along with the lowermost. but they all appear to have four branches. normally we'd have gone for the node with the most branches? Or do we not include the branch to V3?

 

[gneill]

Sorry, i thought this forum was for guidance? or am i mistaken? I'm not looking the the whole answer obviously, just a bit of guidance would be appreciated.

The forum is indeed for guidance. That does not include handing out answers, even bit by bit. The idea is to guide you towards solving the problem by your own efforts. By the forum rules, "I'm stumped" or "I don't know where to begin" are not counted as showing effort. Hints and suggestions are given once you've demonstrated your effort and ideas.

The two nodes connecting Z4, Z2 and Z3 are both essential, along with the lowermost. but they all appear to have four branches. normally we'd have gone for the node with the most branches? Or do we not include the branch to V3?

Choosing a reference node is sometimes a bit of an art if you have a particular potential or current to solve for and a particular choice of reference node happens to make that simpler. But for the most part you choose a node with the most connections as you say. When number of connections isn't a deciding factor then you need to go by instinct which develops with practice. Often the person who designed the circuit makes the choice obvious by drawing the circuit with a practical reference node at the bottom.

Now for a hint. The V3 source connecting the two upper nodes forces your hand in how you approach this problem using nodal analysis. The two nodes have a fixed potential difference, so they cannot be independent nodes. Check your course notes or text for the Supernode concept.

 

[Gremlin]

(or if you like, exchange the positions of V₃ and Z₂ on the diagram. Since their node connections remain the same as before, the circuit operation does not change at all and the current through Z₂ becomes obvious...).

If you do this, which way is the voltage going? In the original circuit the voltage is going in an anti-clockwise direction (as indicated by the arrow above it) - if we swap V3 and Z2 do we kept the voltage going in an anti-clockwise direction - ie arrowhead pointing towards Z3 - or do we effectively pull the symbol straight down and keep it in the same direction - ie arrowhead pointing towards Z1 and the voltage flowing into the same node?

Thanks.

 

[Gremlin]

My incorrect working out so far looks like this:

Loop 1:

V₁ - Z₁I₁ - Z₄ (I₁ - ₂) = 0

V₁ - (Z₁ + Z₄) I₁ + Z₄I₂ = 0

V₁ = (Z₁ + Z₄) I₁ - Z₄I₂

120 = (2 + (-j5)) I₁ - (-j5)I₂

120 = (2 - j5) I₁ - (-j5) I₂

Loop 2:

-V₃ - Z₅ (I₂ - I₃) - Z₄ (I₂ - I₁)

-V₃ + Z₄ I₁ - (Z₅ + Z₄) I₂ + Z₅ I₃

(-14.142 + j14.142) + (-j5) I₁ - (j4 + (-j5)) I₂ + (j4) I₃

-14.142 + j14.142 = - (-j5) I₁ + (-j1) I₂ - (j4) I₃

Loop 3:

-V₂ - Z₅ (I₃ - I₂) - Z₃ I₃

-V₂ - Z₅ I₂ - (Z₅ + Z₃) I₃

-j120 + (j4) I₂ - (j4 + 4) I₃

-j120 = -(j4) I₂ + (4 + j4) I₃

Now i know that's incorrect as when i plug it into my spreadsheet it won't have it - can anyone tell me where it goes astray? I imagine that it's an error with signs somewhere.

Thanks.

 

[gneill]

If you do this, which way is the voltage going? In the original circuit the voltage is going in an anti-clockwise direction (as indicated by the arrow above it) - if we swap V3 and Z2 do we kept the voltage going in an anti-clockwise direction - ie arrowhead pointing towards Z3 - or do we effectively pull the symbol straight down and keep it in the same direction - ie arrowhead pointing towards Z1 and the voltage flowing into the same node?

Thanks.

All connections must remain the same in order to preserve the circuit topology. So the voltage source connections must remain as before as well. That is, the polarity arrow must point to the same node as it did before.

 

[Gremlin]

That's what i suspected, thanks.

I see on loop 3 that V₂ should have been -V₂, but that still isn't correct by the looks of it.

My incorrect working out so far looks like this:

Loop 1:

V₁ - Z₁I₁ - Z₄ (I₁ - ₂) = 0

V₁ - (Z₁ + Z₄) I₁ + Z₄I₂ = 0

V₁ = (Z₁ + Z₄) I₁ - Z₄I₂

120 = (2 + (-j5)) I₁ - (-j5)I₂

120 = (2 - j5) I₁ - (-j5) I₂

Loop 2:

-V₃ - Z₅ (I₂ - I₃) - Z₄ (I₂ - I₁)

-V₃ + Z₄ I₁ - (Z₅ + Z₄) I₂ + Z₅ I₃

(-14.142 + j14.142) + (-j5) I₁ - (j4 + (-j5)) I₂ + (j4) I₃

-14.142 + j14.142 = - (-j5) I₁ + (-j1) I₂ - (j4) I₃

Loop 3:

V₂ - Z₅ (I₃ - I₂) - Z₃ I₃

V₂ - Z₅ I₂ - (Z₅ + Z₃) I₃

j120 + (j4) I₂ - (j4 + 4) I₃

j120 = -(j4) I₂ + (4 + j4) I₃

Now i know that's incorrect as when i plug it into my spreadsheet it won't have it - can anyone tell me where it goes astray? I imagine that it's an error with signs somewhere.

Thanks.

 

[gneill]

You have a sign problem with the V3 voltage source components in Loop 2.

You are right when you state that V2 should have been -V2 in the Loop 3 equation.

 

[Gremlin]

Loop 2:

-V₃ - Z₅ (I₂ - I₃) - Z₄ (I₂ - I₁)

-V₃ + Z₄ I₁ - (Z₅ + Z₄) I₂ + Z₅ I₃

(-14.142 + j14.142) + (-j5) I₁ - (j4 + (-j5)) I₂ + (j4) I₃

-14.142 + j14.142 = - (-j5) I₁ + (-j1) I₂ - (j4) I₃

I honestly can't see where.

V3 = 20∠45 which is 14.142 + j14.142.

If we start at midnight in the loop you go against V₃ so -14.142 + j14.142, then you walk into Z₅ so -(0+j4) x (I₂ - I₃) and finally you walk into Z₄ so - (0-j5) x (I₂ - I₁).

Do we have to include Z₂ somehow?

 

[gneill]

I honestly can't see where.

V3 = 20∠45 which is 14.142 + j14.142.

Yes.

If we start at midnight in the loop you go against V₃ so -14.142 + j14.142,

No. Why do you negate only the real term of the complex value? It's all one number!

Do we have to include Z₂ somehow?

Nope. It belongs to a separate loop and is not in the path of the current in loop 2.

 

[Gremlin]

No. Why do you negate only the real term of the complex value? It's all one number!

Because i don't fully understand complex numbers i suppose is the answer. I can't wait to see the back of them. Although i look at part b of this question and i can see that I'm going to have to simplify a simultaneous equation full of them... great.

Still, even if i change it to -14.142 - j14.142 in my complex matrix spreadsheet it still tells me that it can not solve the equation.

 

[gneill]

Because i don't fully understand complex numbers i suppose is the answer. I can't wait to see the back of them.

Don't hold your breath! Complex numbers are ubiquitous in science and engineering!

Although i look at part b of this question and i can see that I'm going to have to simplify a simultaneous equation full of them... great.

You can always simplify symbolically first. Plug in the complex values at the end.

Still, even if i change it to -14.142 - j14.142 in my complex matrix spreadsheet it still tells me that it can not solve the equation.

Then you must have other issues entering the values for the solver. The equations that you presented look fine if you've made the corrections that I pointed out. Failing all else, show us the nine entries of the 3 x 3 impedance matrix.[/quote]

 

[Gremlin]

Don't hold your breath! Complex numbers are ubiquitous in science and engineering!

You can always simplify symbolically first. Plug in the complex values at the end.

Then you must have other issues entering the values for the solver. The equations that you presented look fine if you've made the corrections that I pointed out. Failing all else, show us the nine entries of the 3 x 3 impedance matrix.

[/QUOTE]

Here is what's in the spreadsheet.

http://s10.postimg.org/6tixvn7d5/Spreadsheet.png

 

[gneill]

There are multiple incorrect signs in the imaginary components of the impedance matrix.

 

[Gremlin]

There are multiple incorrect signs in the imaginary components of the impedance matrix.

I'm not sure how as I've put them in as they are outlined in my post on the previous page but i'll go back over it tomorrow, it's done my head in enough for one day.

Thanks for your help.

 

[Gremlin]

So to pick up on this again we're happy with the simultaneous equation of loop2 at this point:

(-14.142 - j14.142) + (-j5) I₁ - (j4 + (-j5)) I₂ + (j4) I₃ = 0

Then we go:

(-14.142 - j14.142) = -(j5) I₁ + (j4 + (-j5))I₂ - (j4)I₃

Which is:

(-14.142 - j14.142) = -(j5) I₁ + (-J1)I₂ - (j4)I₃

Is it correct to that point can you tell me or has it already gone astray?

 

[gneill]

So to pick up on this again we're happy with the simultaneous equation of loop2 at this point:

(-14.142 - j14.142) + (-j5) I₁ - (j4 + (-j5)) I₂ + (j4) I₃ = 0

Then we go:

(-14.142 - j14.142) = -(j5) I₁ + (j4 + (-j5))I₂ - (j4)I₃

The sign has become wrong on the I₁ term. You've essentially moved it from the LHS of the equation to the RHS, so its sign should change.

I think that part of your problem is that you're dealing with terms like -(-j5) and the double negatives can be confusing when you go to negate them again. So if I may suggest, before you start to move terms around expand these terms to get rid of the compound signs. For example, the term " + (-j5) I₁" becomes simply " - j5 I₁".

 

[Gremlin]

The sign has become wrong on the I₁ term. You've essentially moved it from the LHS of the equation to the RHS, so its sign should change.

I think that part of your problem is that you're dealing with terms like -(-j5) and the double negatives can be confusing when you go to negate them again. So if I may suggest, before you start to move terms around expand these terms to get rid of the compound signs. For example, the term " + (-j5) I₁" becomes simply " - j5 I₁".

That's a great help. So that's:

(-14.142 - j14.142) + (-j5) I₁ - (j4 + (-j5)) I₂ + (j4) I₃ = 0

(-14.142 - j14.142) - j5 I₁ - (-j1)I₂ + j4I₃ = 0

(-14.142 - j14.142) - j5I₁ + j1I₂ + j4I₃ = 0

(-14.142 - j14.142) = j5I₁ - j1I₂ - j4I₃

 

[gneill]

Yup. Much better

 

[Gremlin]

Yup. Much better

Great, so loop 1 is:

V₁ - Z₁I₁ - Z₄ (I₁ - I₂) = 0

V₁ - (Z₁ + Z₄) I₁ + Z₄I₂ = 0

120 - (2 + (-j5)) I₁ + (-j5) I₂ = 0

120 - (2 -j5) I₁ - j5I₂ = 0

120 = (2 - j5) I₁ + j5I₂

And loop3 is:

-V₂ - Z₅ (I₃ - I₂) - Z₃I₃ = 0

- V₂ + Z₅I₂ - (Z₃ + Z₅) I₃ = 0

- j120 + j4I₂ - (4 + j4) I₃ = 0

-j120 = -j4I₂ + (4 + j4) I₃

Correct?

 

[gneill]

Yes, looks good.

 

[Gremlin]

Yes, looks good.

Yes it is. Thanks for your help on that.

I'll be asking you about simplifying symbolically before you know it.

 

[Gremlin]

Yes it is. Thanks for your help on that.

I'll be asking you about simplifying symbolically before you know it.

For part b). If you number the nodes V₁₀, V₂₀, V₃₀ & V₄₀ left to right we know that:

V₁₀ = 120V
V₄₀ = j120V

They're trivial nodes.

V₂₀ & V₃₀ we can combine to form a supernode. I get the equation:

(120 - V₂₀ / 2) + (0 - V₂₀ / -j5) + (0 - V₃₀ / j4) + (j120 - V₃₀ / 4) = 0

And i know i need to times by something to get rid of the divisors, but because they're a mix of real and imaginary numbers I'm unsure how to go at it - what's the symbolic method that you referred to if you don't mind explaining it?

 

[gneill]

Just leave all the terms as symbols; don't plug in any numbers. Then it's a matter of algebra to isolate the unknown term. Plug in the numbers afterwords.

By the way, you need one more equation to define the supernode. How are V₂₀ and V₃₀ related?

 

[Gremlin]

Just leave all the terms as symbols; don't plug in any numbers. Then it's a matter of algebra to isolate the unknown term. Plug in the numbers afterwords.

By the way, you need one more equation to define the supernode. How are V₂₀ and V₃₀ related?

V₂₀ - V₃₀ = 14.142 + j14.142 V

On the symbols/algebra front I'm afraid i don't follow. Leave the terms as symbols, do you mean Z₁ rather than 2, Z₄ rather than -j5, etc?

 

[gneill]

V₂₀ - V₃₀ = 14.142 + j14.142 V

On the symbols/algebra front I'm afraid i don't follow. Leave the terms as symbols, do you mean Z₁ rather than 2, Z₄ rather than -j5, etc?

Yes. And leave V₁, V₂, and V₃ as symbols, too.

 

[Gremlin]

Yes. And leave V₁, V₂, and V₃ as symbols, too.

Ok, so it becomes:

(V₁₀ - V₂₀ / Z1) + (0 - V₂₀ / Z4) + (0 - V₃₀ / Z5) + (V₄₀ - V₃₀ / Z3) + (V₂₀ - V₃₀ / Z2) = 0

Again, should ' / Z2' be in that last part of the equation?

After that, i still need to get rid of the divisors as far as I'm aware - how do i go about this?

 

[gneill]

Ok, so it becomes:

(V₁₀ - V₂₀ / Z1) + (0 - V₂₀ / Z4) + (0 - V₃₀ / Z5) + (V₄₀ - V₃₀ / Z3) + (V₂₀ - V₃₀ / Z2) = 0

Don't use trivial nodes. V₄₀ is just V₂, a known value, so use V₂. The same goes for V₁₀ which is just V₁. In other words, don't multiply your variables unnecessarily. And replace either V₂₀ or V₃₀ using your second equation that relates them. That will reduce your equation to one unknown. I suggest getting rid of V₃₀ since it's V₂₀ you need in order to find $I$.

Again, should ' / Z2' be in that last part of the equation?

No, Z2 is entirely internal to the supernode so it doesn't present a current that enters or leaves the supernode. Remove that term.

After that, i still need to get rid of the divisors as far as I'm aware - how do i go about this?

You can get rid of the divisors when you plug in the numbers, reducing terms to individual complex values.
Start by expanding the terms into into simple fractions:

$\frac{V_1}{Z_1} - \frac{V_{20}}{Z_1} - \frac{V_{20}}{Z_4} + ...$

 

[Gremlin]

Don't use trivial nodes. V₄₀ is just V₂, a known value, so use V₂. The same goes for V₁₀ which is just V₁. In other words, don't multiply your variables unnecessarily. And replace either V₂₀ or V₃₀ using your second equation that relates them. That will reduce your equation to one unknown. I suggest getting rid of V₃₀ since it's V₂₀ you need in order to find $I$.

No, Z2 is entirely internal to the supernode so it doesn't present a current that enters or leaves the supernode. Remove that term.

You can get rid of the divisors when you plug in the numbers, reducing terms to individual complex values.
Start by expanding the terms into into simple fractions:

$\frac{V_1}{Z_1} - \frac{V_{20}}{Z_1} - \frac{V_{20}}{Z_4} + ...$

Ok, so:

V₃₀ = V₂₀ - (14.142 - j14.142)

$\frac{V_1}{Z_1} - \frac{V20}{Z_1} - \frac{V20}{Z_4} + \frac{V20}{Z_5} - \frac{14.142 - j14.142}{Z_5} + \frac{V20}{Z_3} + \frac{V20}{Z_3} - \frac{14.142 + j14.142}{Z_3} + (14.142 + j14.142) = 0$

$\frac{120}{2} - \frac{V20}{2} - \frac{V20}{-j5} + \frac{V20}{j4} + \frac{14.142 - j14.142}{j4} + \frac{j120}{4} + \frac{V20}{4} - \frac{14.142 - j14.142}{4} + (14.142 + j14.142) = 0$

$60 - \frac{V_20}{2} - \frac{V20}{-j5} + \frac{V20}{j4} - (3.54 + j0.29) + j30 + \frac{V20}{4} - (3.54 - j3.54) + (14.142 + j14.142) = 0$

$(67.1 + j47.4) - \frac{V20}{2} - \frac{V20}{j5} + \frac{V20}{j4} + \frac{V20}{4} = 0$

$(67.1 + j47.4) - \frac{V20}{2-j5} + \frac{V20}{4+j4} = 0$

Does that look good to you so far?

 

[gneill]

Ok, so:

V₃₀ = V₂₀ - (14.142 - j14.142)

Leave $V_3$ as $V_3$ for now! That way you don't have to carry around any complex digits through the initial simplifications. Don't be in a hurry to plug in numbers, especially complex numbers; That just gives more opportunities to make transcription and sign errors, and makes it more difficult to spot mistakes in the workings.

You should then have:

$\frac{V_1 - V_{20}}{Z_1} + \frac{-V_{20}}{Z_4} + \frac{-(V_{20} - V_3)}{Z_5} + \frac{V_2 - (V_{20} - V_3)}{Z_3} = 0$

which becomes, after distributing the signs:

$\frac{V_1 - V_{20}}{Z_1} - \frac{V_{20}}{Z_4} + \frac{V_3 - V_{20}}{Z_5} + \frac{V_2 - V_{20} + V_3}{Z_3} = 0$

Can you carry on from there to isolate $V_{20}$? You should be able to reach a point where you have something like:

$V_{20}[$ some terms $] + [$ more terms $] = 0$

Then you can work on reducing the "some terms" and "more terms" down to individual complex numbers, since then all the variables in them will be known values.

<snip>

Does that look good to you so far?

Some sign errors snuck in. So no, not so good

 

[Gremlin]

$\frac{V_1 - V_{20}}{Z_1} + \frac{-V_{20}}{Z_4} + \frac{-(V_{20} - V_3)}{Z_5} + \frac{V_2 - (V_{20} - V_3)}{Z_3} = 0$

which becomes, after distributing the signs:

$\frac{V_1 - V_{20}}{Z_1} - \frac{V_{20}}{Z_4} + \frac{V_3 - V_{20}}{Z_5} + \frac{V_2 - V_{20} + V_3}{Z_3} = 0$

I'm going to have a go now, but before i start how has + $\frac{V_2 - (V_{20} - V_3)}{Z_3}$ become +$\frac{V_2 - V_{20} + V_3}{Z_3}$

Is that correct?

 

[gneill]

I'm going to have a go now, but before i start how has + $\frac{V_2 - (V_{20} - V_3)}{Z_3}$ become +$\frac{V_2 - V_{20} + V_3}{Z_3}$

Is that correct?

It is correct. The minus sign in front of the parentheses is distributed over the terms within.

 

[The Electrician]

Ok, so it becomes:

(V₁₀ - V₂₀ / Z1) + (0 - V₂₀ / Z4) + (0 - V₃₀ / Z5) + (V₄₀ - V₃₀ / Z3) + (V₂₀ - V₃₀ / Z2) = 0

Again, should ' / Z2' be in that last part of the equation?

After that, i still need to get rid of the divisors as far as I'm aware - how do i go about this?

It should be mentioned that your expressions (when you're not using latex) need some more parentheses. If you input your expressions into a computer solver just as you have them, you won't get the right result because of the usual precedence of operators in computer algebra. You should have it like this:

((V₁₀ - V₂₀) / Z1) + ((0 - V₂₀) / Z4) + ((0 - V₃₀) / Z5) + ((V₄₀ - V₃₀) / Z3) +((V₂₀ - V₃₀) / Z2) = 0

Here's how a computer solver solves both forms; you can see the difference:

 

[KatieMariie]

With regards to simplifying: (V1-V20/Z1)-(V20/Z4)+(V20-V3/Z5)+(V2-V20+V3/Z3)=0 to isolate V20;

At what point do we substitute the known values? before or after simplification?

I've substituted before simplification, and started to simplify, so now have the following:
(60-V20)-(V20/-J5)+(V20-3.353+J3.353)+(J120-V20+3.353+J3.353)=0

I'm not entirely sure if this is the right way to go? if so, how do i proceed from here - the /-j5 is throwing me off?! and if not, what measures would i need to take to be able to begin isolating the V20 in the original symbolic equation?

 

[gneill]

Personally I would expand the expression (symbolically!) into individual terms, then gather all the terms with V20 in the numerator, factoring out the V20 (as I indicated in post #43 above). After that you can plug in numerical values for the terms and reduce the sums of terms to single complex values.

You remove a complex denominator by multiplying the numerator and denominator by the complex conjugate of the denominator. That is, if you have a complex fraction:

$\frac{a + b i}{c + d i}$

Then you can "normalize" it by:

$\frac{a + b i}{c + d i} \frac{(c - d i)}{(c - d i)} = \frac{ac + bd + (bc-ad)i}{c^2 + d^2}$

Note that the denominator is then entirely real. If this is not clear then you will have to review complex arithmetic.

 

[Gremlin]

$\frac{V_1 - V_{20}}{Z_1} - \frac{V_{20}}{Z_4} + \frac{V_3 - V_{20}}{Z_5} + \frac{V_2 - V_{20} + V_3}{Z_3} = 0$

Can you carry on from there to isolate $V_{20}$? You should be able to reach a point where you have something like:

$V_{20}[$ some terms $] + [$ more terms $] = 0$

$\frac{V_1}{Z_1} - \frac{V_{20}}{Z_1} - \frac{V_{20}}{Z_4} + \frac{V_3}{Z_5} - \frac{V_{20}}{Z_5} + \frac{V_2}{Z_3} - \frac- {V_{20}}{Z_3} + \frac{V_3}{Z_3} = 0$

$\frac{120}{2} - \frac{V_{20}}{2} - \frac{V_{20}}{-j5} + \frac{14.142 + j14.142}{j4} - \frac{V_{20}}{j4} + \frac{j120}{4} - \frac{V_{20}}{4} + \frac{14.142 + j14.142}{4} = 0$

$60 - \frac{V_{20}}{2} - \frac{V_{20}}{-j5} + (3.54 - j3.54) - \frac{V_{20}}{j4} + j30 - \frac{V_{20}}{4} + (3.54 + j3.54) = 0$

$(67.08 + j30) - \frac{V_{20}}{2} - \frac{V_{20}}{-j5} - \frac{V_{20}}{j4} -\frac{V_{20}}{4} = 0$

I'd then be tempted to go:

$(67.08 + j30) - \frac{V_{20}}{2 + (-j5) + j4 + 4} = 0$

$(67.08 + j30) - \frac{V_{20}}{6-j1} = 0$

 

[The Electrician]

$\frac{V_1}{Z_1} - \frac{V_{20}}{Z_1} - \frac{V_{20}}{Z_4} + \frac{V_3}{Z_5} - \frac{V_{20}}{Z_5} + \frac{V_2}{Z_3} - \frac- {V_{20}}{Z_3} + \frac{V_3}{Z_3} = 0$

$\frac{120}{2} - \frac{V_{20}}{2} - \frac{V_{20}}{-j5} + \frac{14.142 + j14.142}{j4} - \frac{V_{20}}{j4} + \frac{j120}{4} - \frac{V_{20}}{4} + \frac{14.142 + j14.142}{4} = 0$

$60 - \frac{V_{20}}{2} - \frac{V_{20}}{-j5} + (3.54 - j3.54) - \frac{V_{20}}{j4} + j30 - \frac{V_{20}}{4} + (3.54 + j3.54) = 0$

$(67.08 + j30) - \frac{V_{20}}{2} - \frac{V_{20}}{-j5} - \frac{V_{20}}{j4} -\frac{V_{20}}{4} = 0$

I'd then be tempted to go:

$(67.08 + j30) - \frac{V_{20}}{2 + (-j5) + j4 + 4} = 0$

$(67.08 + j30) - \frac{V_{20}}{6-j1} = 0$

You're ignoring gneill's advice from post #43, where he said:

"Leave V3 as V3 for now! That way you don't have to carry around any complex digits through the initial simplifications. Don't be in a hurry to plug in numbers, especially complex numbers; That just gives more opportunities to make transcription and sign errors, and makes it more difficult to spot mistakes in the workings."