potential at a point due to two voltages

hi everyone,
we were given a simple circuit as shown in the attachment, and we are asked to find out voltage across 5ohm resistance and then we can find out current.


my question is if by superposition theorem i find out effect of the voltage sources individually then:
1)potential difference(p.d.) of 2 ohm resistance due to 10v battery alone is 4.736885v so the potential at 5ohm resistance is 10-4.736885=5.26315v.
2)potential at 5 ohm sue to 5v acting alone is p.d. of 4ohm resistance is 3.6842v so the potential at 5ohm resistance is 5-3.6842=1.31578v.
i can find all this up-to here i have no doubt. then, it's mentioned that we have to find out total voltage across 5ohm so here i subtract the two potentials at that point i.e 5.26315v-1.31578v=3947337v. here my answer is wrong. because the correct answer is :5.26315v+1.31578v=6.57893v.
i am not understanding this how can two voltages here be additive? because u can see that 10v>5v hence,it's as if 10v is charging 5v & 5v is refusing to charge by imposing it's own effect,hence the voltages should be subtractive in nature.

There is no "charging" and "refusing" of voltage sources. Just use Kirchoff's laws to solve the problem. Don't be distracted if some voltage is negative (like V2 in Your case).

Zooming in on the V2 source, it appears to be labelled -5 V.

Yet its icon is oriented the same direction as the +10V source. Maybe it is intended to throw us off?

sorry that is actually +5v i have actually forgotten to rectify it.
so assuming that it is +5v i have made all calculations above..
and please read my question once again and consider that as +5v

Solving Kirchoff law equations You get: I3=(R2*V1+R1*V2)/(R1*R2+R1*R3+R2*R3)

Here is another way of looking at it.

The node of interest is influenced by 3 different voltages seen through 3 different resistances.

To calculate the net influence, we begin by calculating the equivalent resistance seen through all legs. We do this by taking the reciprocal of the sum of the reciprocals (as though all legs are in a parallel circuit).

1 / (1/2 + 1/5 + 1/4) = 1.05. That is the effective resistance which influences the node.

Now we calculate the weighting factor for each leg attached to the node.
Secondly, multiply the voltage seen through that leg times its weighting factor.
Lastly we will sum all the weighted voltages.

The leftmost leg has 2 ohms. Its weighting factor is 1.05/2. Thus it contributes 52.6 percent of whatever voltage it sees. It sees 10V. So multiply that by .526, equals 5.26V.

The leg containing 5 ohms goes to ground. It has 21% weighting ( 1.05/5 ). It sends 0V. Its weighted contribution is 0V.

The rightmost leg is 4 ohms. It has 26.3% weighting ( 1.05/4 ). It sends 5V. So it contributes 5 * .263, equals 1.32 V.

Now we sum all values. 5.26+ 1.32 = 6.58 V.

yes i got what u said but i still have some doubt (your answer is actually right).
according to me it should be 5.26-1.32v. I will justify my answer: as u have already found out the voltage due to 10v& it's series resistance acting alone has created a """POTENTIAL""" of ""+5.26v"" at the start of the 4ohm(let me name this node as ""A""),then due to 5v and it's series resistance acting alone there will be a potential of ""+1.32v"" being created at node ""A"". so as per u these voltages are additive & is 5.26v+1.32v=6.58v right it's correct, but according to my imagination these two voltages are opposing each other at that node so, their effect will be subtractive in nature & is 5.26v-1.32v i know i am wrong but please explain me without using any laws...

Here is another way of looking at it.

Suppose we greatly increase the value of the center resistor, so that we effectively remove it...

Then yes, the 5V will subtract from the 10V going around the resulting loop. (The one source could be said to oppose the other, in a manner of speaking.)

Then it is simple to calculate the volt level at the node.

Now we start to reduce the ohm value of the center resistor. It begins to pull down the volt level.

The less its value, the closer the voltage is pulled to zero ground.