Finding the voltage across the current source terminals.

Hi
I was doing my homework and have some problems i do not understand.

I need to find the voltage across the current source terminals.

I used mesh analysis with the super mesh as there is a common current source, which i got I1=5/2 and I2=-1/2 and i do not get the idea finding the voltage using those elements.

And how do i find the power dissipated in two resistors? like using P = (I^2)(R)??

This is the diagram.

**broken link removed**

This one can be solved by several methods. You may have come across Thevenin and Norton equivalent sources.

The general method is by the Superposition Theorem.
This says that what happens in any branch of a circuit containing multiple sources is the sum of what would have happened from the contribution of each individual sources.
You consider a source, replacing all other voltages sources with short-circuits, and all current sources with a open circuit. You figure the currents.
You do that for each source in turn, then add up the currents in the branches. This is what network solvers do with matrix math.

There are 3 branch currents that happen in your circuit.
One is in the left square, another in the right square, and a third that goes right around the whole outer loop.

BUT - for this one, you do not have to get into solving the 3 simultaneous equations...unless you want to do it the hard way.
You can see that the 3A down the middle must be equal to the sum of the current coming through the 20 Ohms resistor from the 20V source, call it I Amps, plus the current coming from the 25V source through the 10 Ohms resistor, which has to be 3-I Amps. (partly solving the equations)
The voltage drops around any network loop must add up to zero (Kirchoff's law)

Have a look at these links..
https://www.electrical4u.com/superposition-theorem/
https://www.electrical4u.com/norton-theorem-norton-equivalent-current-and-resistance/
https://www.allaboutcircuits.com/vol_1/chpt_10/7.html

Now spot that the other condition that is forced on this circuit is that the voltage at the join point of the resistors must be the same.
So set the voltage drops around the left loop running I Amps equal to the voltage drops in the right loop running 3-I Amps.
You now have the way to find I, and you end up knowing everything about the circuit. Do this bit for yourself.

It was very easy change the two voltage source as current source then 1A, 20ohm parallel and 2.5A and 10ohm parallel so the total current ix 2.5+3-1 = 4.5A.

Correct, using P = (I^2)(R) to find the power dissipated in two resistors.