Engineering AC circuit analysis -- mesh and nodal

[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."

 

[Gremlin]

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."

Is there any further you can go without starting to put the figures in?

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

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

 

[The Electrician]

Is there any further you can go without starting to put the figures in?

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

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

You can't do this:
$- \frac{V_{20}}{Z_1} - \frac{V_{20}}{Z_4} - \frac{V_{20}}{Z_5}- \frac{V_{20}}{Z_3}$ becomes $- \frac{V_{20}}{Z_1 + Z_3 + Z_4 + Z_5}$

Try this with numbers to see if these two expressions are the same.
Is $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}$
the same as
$\frac{1}{3+4+5+6}$ ?

Work it out on your calculator. Then review how to combine fractions in algebra.

 

[The Electrician]

Gremlin, it might be worth your while to review this:
http://www.mathsisfun.com/least-common-denominator.html

 

[Gremlin]

Thanks, i shall have a look.

 

[Gremlin]

Is there any further you can go without starting to put the figures in?

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

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

I've looked at that.

To be honest if i was dividing with real numbers as in the above link i wouldn't have a problem (i should have stopped and thought about adding the denominators in post #51), collapsing the 4 fractions with complex numbers however...

The last hint of an idea that i have is collapsing the 4 equations into 2 equations.

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

$\frac{V_{20}}{2} x2 = \frac{2V_{20}}{4}$

$-\frac{2V_{20}}{4} - \frac{V_{20}}{4} = - \frac{3V_{20}}{4}$

$\frac{V_{20}}{-j5} x4 = \frac{4V_{20}}{-j20}$

$\frac{V_{20}}{j4} x5 = \frac{5V_{20}}{j20}$

$- \frac{4V_{20}}{-j20} - \frac{5V_{20}}{j20} = Math Error(!)$

 

[donpacino]

Is there any further you can go without starting to put the figures in?

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

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

yes. example

$\frac{a}{b}+\frac{c}{d}=0$
$\frac{ad+bc}{bd}=0$

In many cases you want to get it in the form of 1 numerator over 1 denominator.

 

[Gremlin]

Looked hard at this tonight and back at my notes and I've nearly cracked it. Hopefully tomorrow i can put it to bed.

Thanks for all of your help.

 

[MrBondx]

1. 
   Hi guys this is my work for part a. What am I doing wrong. Please help thanks
   
   -120 – 2i1 – (i1 – i2 ( -j5) = 0-120 – 2i1 –(-i1j5 + i2j5 = 0-120 – 2i1 + i1j5 – i2j5 = 0i1(-2+j5) – i2(j5) = 120
   (14.142 + j14.142) – (i2 – i3)(j4) + (i2 – i1)(-j5) = 0(14.142 + j14.142) – i2j4 + i3j4 – i2j5 + i1j5 = 0(14.142 + j14.142) = i2j4 - i3j4 + i2j5 - i1j5 (14.142 + j14.142) = i2j4 - i3j4 + i2j5 - i1j5(14.142 + j14.142) = i2(j4+j5) - i3j4 - i1j5
   -4i3 + j120 +(i3 –i2)(j4) = 0-4i3 + j120 +(i3j4 –i2j4) = 0-4i3 + j120 + i3j4 – i2j4 = 0j120 = 4i3 –i3 – i3j4 + i2j4j120 =i3(4-j4) + i2(j4)
   

 

[MrBondx]

I have brought it down to this but my signs still seem different from other posts on this thread. I don't understand why because I have tried to follow loops round properly.-120 - i1(2 - j5) + i2(-j5) = 0-i1(2 - j5) + i2(-j5) = 120 ……………………………………………… 1i1(-j5) – i2 (j4 – j5) + (14.142 + j14.142) + i3(j4)i1(-j5) – i2 (-j) + i3(j4) = - 14.142 - j14.142 …………….. 2i2(j4) – i3(4 + j4) + j120 i2(j4) – i3(4 + j4) = -j120 ……………………………………………. 3

 

[The Electrician]

I have brought it down to this but my signs still seem different from other posts on this thread. I don't understand why because I have tried to follow loops round properly.

Read post #5