Problems with superpositsion.

[sevenfold4]

Hey i have this problem with superpositsion. I understand it and everything but when you short circuit a certain voltage source it confused me I am pretty new to electronics so i do not really know how the circuit will behave. I hope i can find some quick help here.
Picture of the circuit is here https://imgur.com/vaCLSVs (the voltage source on the left is 40v)

The biggest problem for me is that now that the voltage source on the left has been shorted i have no idea on how to calculate the resistor values since i dont know if i should just calculate them in series and ignore that line or what.
Sorry for the silly question. I am sure the solution is really easy

 

[chuckey]

On the extreme right you have a 560+1.5k = 2.060K. then going to the left you add 8.2K, so the resistors in this bit come to 8.2 +2.060 = 10.26K. On the extreme left you have 1K+ 1k = 2K. So you current generator sends most of its current down the 2K path and some down the 10.26K path.
The common thing is the voltage across the resistors are equal, so V/2K + V/10.26K = I = .5A, so multiplying both sides by 2K X 10.26K gives 10.26k X V + 2k X V = .5 X 2K X 10.26k, so 12.26 k X V = 1k X 10.26 k. So V = 1k X 10.26k/12.26k.
Does this get you going?
Frank

 

[sevenfold4]

On the extreme right you have a 560+1.5k = 2.060K. then going to the left you add 8.2K, so the resistors in this bit come to 8.2 +2.060 = 10.26K. On the extreme left you have 1K+ 1k = 2K. So you current generator sends most of its current down the 2K path and some down the 10.26K path.
The common thing is the voltage across the resistors are equal, so V/2K + V/10.26K = I = .5A, so multiplying both sides by 2K X 10.26K gives 10.26k X V + 2k X V = .5 X 2K X 10.26k, so 12.26 k X V = 1k X 10.26 k. So V = 1k X 10.26k/12.26k.
Does this get you going?
Frank

something becomes clearer but what are the formulas you use(looks like ohms). The answer would be 836 and that seems too high. That really doesnt matter since i have to find the current using the superpositsion

 

[albbg]

When you short the two voltage generators as you drawn, on the left side of the current generator you have the two resistor, each of 1k, in series for a total of 2k. Looking on the right side, instead the two resistor 560 and 1.5 are short circuited that means a current of 0A will flow through them and a voltage of 0V is across them. So at the right side you have only the 8.2k resistor in parallel with the current generator.
In practice you have two meshes: Ig-2k and Ig-8.2k. This means a current I1=Ig*8.2k/(2k+8.2k) will flows through the two 1k resistors while a current of I2=Ig*2k/(2k+8.2k) will flow through the 8.2k resistor. Now you can easily calculate the voltage across each resistor.
Now you have to do the analysis of the other 2 circuits and sum the results.
Be careful in applying in all three circuits the same current direction and voltage polarity.

 

[sevenfold4]

When you short the two voltage generators as you drawn, on the left side of the current generator you have the two resistor, each of 1k, in series for a total of 2k. Looking on the right side, instead the two resistor 560 and 1.5 are short circuited that means a current of 0A will flow through them and a voltage of 0V is across them. So at the right side you have only the 8.2k resistor in parallel with the current generator.
In practice you have two meshes: Ig-2k and Ig-8.2k. This means a current I1=Ig*8.2k/(2k+8.2k) will flows through the two 1k resistors while a current of I2=Ig*2k/(2k+8.2k) will flow through the 8.2k resistor. Now you can easily calculate the voltage across each resistor.
Now you have to do the analysis of the other 2 circuits and sum the results.
Be careful in applying in all three circuits the same current direction and voltage polarity.

Oh beautiful ok i understand now. I thought that because of the short there was no current through those resistors but i was not sure and this confirms it. Thank you