# Find the value of U by the means of superposition

I want to do that by superposition rule.

Firstly i need to short the E source and compute U0 in case of current divider formula?

So i will get u1 which is voltage loss when the voltage source is short $$U1=6mA\cdot \frac{2R}{2R+R0}$$ Then in second case i will use only E source. Now i have to find total current in that circuit which is equal $$It=\frac{E}{2R+R0}$$ And then i just compute U2 which is voltage loss when we have our current source open $$U2=1mA\cdot 2kOhm$$ And my result is $$U0=U2+U1=5V$$ Am i right?

• Start by correcting U1. There's 2R in parallel with Ro. Multiply the resulted resistance by J to get the J-part of Uo. Ask then again. – user287001 Dec 10 '17 at 23:41

With $E$ shorted out, you have $R$ in parallel with $2 R$. The $J$ current will divide between these. Select only the portion that goes through $R_0$ here and write it down.

Now return $E$ and remove (open) $J$. Now you have $E$ in a series loop containing two $R$ resistors and one $R_0$ resistor. Again, compute the current through $R_0$ here and write it down.

Now sum the two values. What do you get?

P.S. It may help a little that I also suggest you look at the direction of the current arrow for $R_0$. This may indicate a sign reversal to be thinking about.

• Oh, okay i thought that second resistor is not needed, ten value od my u1 will be 3V. Current when J is open is 4V/4kOhm=1mA and now im calculating voltage loss on R0 which is R0*1mA=2V. Suming all up i get 5V. Am I correct now? – c0rv Dec 11 '17 at 0:23
• @c0rv There are two currents. What are you getting as the current through $R_0$ when shorting out $E$? – jonk Dec 11 '17 at 0:33
• To compute U1 i need Resistance and I, and value of I Is J? Or i am wrong? – c0rv Dec 11 '17 at 1:23
• @c0rv You have enough information to compute the current through $R_0$ with the voltage source shorted out. Do you see that fact? Or not? – jonk Dec 11 '17 at 1:38