It'll be my last question.

A massive thanks to all that have contributed

Don't worry about asking questions.

As long as you show some work and a schematic, there are many here that are more than happy to help somebody learn!

Other people will learn from seeing the questions you asked and the problems you ran into, so it is a benefit to all. It also keeps us geezers on our toes.

 

You can solve it using mesh equations:

\(\left[ \begin{array}{3}10&-10 &0\\-10&10+20+30&-30\\0&-30&30+60\end{array}\right]*\left[ \begin{array}{4}Itot\\I2\\I4\end{array}\right]=\left[ \begin{array}{4}25\\ 50\\-50\end{array}\right]\)

which will give you the clockwise currents Itot, I2 and I4 in the three obvious meshes, left to right.

Or you can use the nodal method:

\(\left[ \begin{array}{2}1 & 0 \\\frac {-1}{20} & \frac {1}{20} + \frac {1}{30}+ \frac {1}{60} \end{array}\right]*\left[ \begin{array}{1}N1\\N2\\\end{array}\right]=\left[ \begin{array}{1}25\\ \frac {-50}{30}\\\end{array}\right]\)

which will give you the voltages at the nodes N1 and N2.

I understand the mesh/loop analysis (I actually did this, before trying to use ohms law to justify I2). What I mean is this;

V1/R1=2.5A = I1.

I was unsure on how to get I2 yet I new from mesh/loop analysis it was 3.958A, so I minused 2.5 from this value to give me I2, then I2 x R2 = Vr2, then V1 - VR2 = N2, as moving away from V1, the V2 has an impact on R2 as it's Vdrop is greater then V1, so to deal just with V2, is why I did this.

Then N2 - V2 = Vr3, where N2 is Vdrop over R4.

But for whatever reason I couldn't find I2, but I see now, that the only way is to use nodal, or directly to use mesh/loop; my assumption was to break it down to it's lowest form, but i see there's various ways, just not as I expected.

Also, and this is were I initially came undone, I don't get the 1st part of your nodal analysis, I get the 2nd part that equates -50/30, but, I don't see why you ended up wioth [1, 0 = 25]

as if I work it out, I get;

V1-25 + V1-V2/20 = 0
1-25 + 1/20 + (-1/20) = 0
(1 + 1/20)V1 + (-1/20)V2 = 25

1.05V1 + (-0.05)V2 = 25 (1)

I don't see or understand how you got;

[1V1 + 0V2 = 25]

BUT, when I worked out your values, I got N1=25V, N2=29.16V,

Yet N2 is -4.16V.



I'm abit confused

 

Also, and this is were I initially came undone, I don't get the 1st part of your nodal analysis, I get the 2nd part that equates -50/30, but, I don't see why you ended up wioth [1, 0 = 25]

When using the nodal method, a voltage source connected directly to a node prevents you from using KCL to write an equation summing currents at that node. Instead you write a "constraint" equation. Since the 25 volt source is connected to N1, we already know the voltage there so we write an equation V1 = 25. It's that simple.

Our first equation is just:

1*V1 + 0*V2 = 25

The second equation is the result of applying KCL at node N2:

(V2-V1)/20 + (V2+50)/30 + (V2-0)/60 = 0

Rearranging, we have:

(-1/20)*V1 + (1/20+1/30+1/60)*V2 = -50/30

If you substitute 25 for V1, you'll get:

(-1/20)*25 + (1/20+1/30+1/60)*V2 = -50/30

Do the algebra and you'll get V2 = -25/6 = -4.1666666667

as if I work it out, I get;

V1-25 + V1-V2/20 = 0
1-25 + 1/20 + (-1/20) = 0
(1 + 1/20)V1 + (-1/20)V2 = 25

1.05V1 + (-0.05)V2 = 25 (1)

I don't see or understand how you got;

[1V1 + 0V2 = 25]

BUT, when I worked out your values, I got N1=25V, N2=29.16V,

Yet N2 is -4.16V.



I'm abit confused

I don't follow what you're doing in this last bit, but it's wrong so I won't try to figure it out. Just do what I describe above.

 

I fully understand now, thanks a lot.

To highlight what I never knew in the constraint equation, has clarified this completely.

As to say the constraint of N1=25V will make anything to do with V2 at N1 obsolete, as you said V1 = 25 to no need to calculate anything more out using v2 at N1.

I will remember this.

Thanks to all that have contributed.