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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?

  • 2
    \$\begingroup\$ 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. \$\endgroup\$ – 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.

  • \$\begingroup\$ 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? \$\endgroup\$ – c0rv Dec 11 '17 at 0:23
  • \$\begingroup\$ @c0rv There are two currents. What are you getting as the current through \$R_0\$ when shorting out \$E\$? \$\endgroup\$ – jonk Dec 11 '17 at 0:33
  • \$\begingroup\$ To compute U1 i need Resistance and I, and value of I Is J? Or i am wrong? \$\endgroup\$ – c0rv Dec 11 '17 at 1:23
  • \$\begingroup\$ @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? \$\endgroup\$ – jonk Dec 11 '17 at 1:38

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