Hi sorry as I had said I didn't understand what you meant, then I went onto the next part of the question. So I hope what I've done here is correct? Thanks for all your help and patience.

 

Just looking at your bottom diagram reveals some problems -- though they might be minor.

Let's take R2. When you say that the voltage across R2 is -6.24V, you are saying that the voltage of the + end of the resistor is -6.24V higher than the - end of the resistor. Since your voltage is negative, what you are really saying is that the + end of the resistor is 6.24V lower than the - end of the resistor. To keep this straight, just remember that the voltage across two terminals is defined to be the voltage at the positive terminal minus the voltage at the negative terminal, hence you are saying:

(V+) - (V-) = -6.24V

Using the bottom node, (node y) as the 0V reference, we have

(V+) - (Vy) = 12V // Due to the battery
(V+) - (Vx) = -6.24V // Due to stated voltage across R2

From these we get

Vy = (V+) - 12V
Vx = (V+) + 6.24V

What we want is Vxy

Vxy = (Vx) - (Vy) = [(V+) + 6.24V] - [(V+) - 12V]
Vxy = 18.24V

It's the same accounting game as when working with loads on an airframe or pressures in a hydraulic system. To say that there is a 200psi pressure between two points in system, you have to define which point is symbolically at the higher pressure.

 

Are my answers wrong then?

 

Are my answers wrong then?

As indicated by your final diagram they are wrong. If we take Vy to be 0V, then we get three different voltages for Vx depending on which path we take. If we go up the path with R1 we get 6.24V, if we go up the path with R2 we get 18.24V, and if we go up the path with R3 we get 5.76V. Since we know that there is only a single value for the voltage at Vx, we know that we have a problem.

 

I did say you need to understand superposition before you try to use it.

 

You've put in enough effort that I think you are in a good position to understand the points I'm trying to make by comparing your approach to mine -- and keep in mind that there are several correct ways to work the problem.

The first thing to always do is to annotate your diagram so that your terms are unambiguously defined. In this case I first chose a node to act as the reference node and assigned node labels to those nodes that didn't already have them. Then I guessed which direction I thought the currents would probably end up going in the three resistors. Finally, I assigned the voltages across the three resistors in conformance with the passive sign convention. You may or may not need all of these and, with experience, you will be able to decide which ones are and aren't needed. Also, if it turns out you need something that you didn't define, then you can just go back and define it. The thing NOT to do is to just start using terms (voltages, currents, or nodes) that you haven't defined on the diagram.

So here's the first annotated diagram:



Now you can start setting up your solutions. That may involve making new diagrams (and, for superposition, usually does). You can use the terms defined in the original annotated diagram and should, when practical. But sometimes it isn't practical and that's fine -- but if you need new terms, then always name them something different. Never have, for instance, I1 which means something different in two different diagrams -- that's just asking for trouble.

So for our first superposition analysis we turn off the 6V source and end up with the following diagram:


Notice that I kept the original definitions for I2 and I3 but that it seemed more natural to have the current in R1 flow downward, which is opposite the definition of I1. So I simply called this current I1A and note that

I1 = -I1A

I would solve this and get the following:

I2 = 762 mA
I1A = 476 mA
I3 = 286 mA

We would then do the same thing for when we turn off the 12V source and, in doing so, would probably find that it would be natural to keep the original definitions for I1 and I3 but that it would make sense to use I2A such that

I2 = -I2A

(I'll leave drawing that diagram up to you).

and, upon solving this, would get

I1 = 524 mA
I2A = 238 mA
I3 = 286 mA

In combining these, we can either rewrite the currents in terms of I1, I2, and I3 first and then sum them, or we can take care of it all in one step:

SOURCE|I1 (mA)|I1A (mA)|I2 (mA)|I2A (mA)|I3 (mA)
12V|-476|476|762||286
6V|524||-238|238|286
TOTAL|48||524||572

We can now find the voltage that is asked for, Vxy, by noting from the original diagram that it is

Vxy = I3·R3 = (572 mA)(10 Ω) = 5.72 V

We can check our work by finding this same voltage through the other branches:

Vxy = 6V - I1·R1 = 6 V - (48 mA)(6 Ω) = 6 V - 0.288 V = 5.71 V
Vxy = 12V - I2·R2 = 12 V - (524 mA)(12 Ω) = 12 V - 6.288 V = 5.71 V

Always check your work. You WILL make mistakes. In doing this solution I got a value of 286 mA for I3 in the first part but wrote down 256 mA and didn't catch it. So I ended up with Vxy = 5.42 V which is close enough that it didn't raise any alarms. But when I did the first check I got a different result and so I went back and tracked down the error.