Hello I am trying to find the Voltage (U in the photo) using superposition method.

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And the first step is to compute the voltage given by the contribution of the left source of Voltage.enter image description here Is this correct? If yes then I am thinking to use Current divider theorem to find the current near U with the relation: $$I_2'=6\frac{2}{2+2}$$ Also for the step two, when I see the contribution of the right source, are there all resistance in paralel?

  • \$\begingroup\$ Your schematic has 2 power sources but your photo only one. \$\endgroup\$ – Goswin von Brederlow Jun 11 '18 at 11:54
  • \$\begingroup\$ yes sir. That is superposition method. \$\endgroup\$ – Zacky Jun 11 '18 at 11:56
  • \$\begingroup\$ Ahh, got you, it's only the first half of the superposition. The transformation to Rp1 + Rp2 looks right and Rp1 == Rp2. But Rp1 = 1 / (1/2 + 1/2) = 1 Ohm. Not sure where you got that 2 * 2 / (2 + 2) from. With Rp1 = Rp2 = 1 Ohm that 6A result looks right. \$\endgroup\$ – Goswin von Brederlow Jun 11 '18 at 12:01
  • \$\begingroup\$ well $$\frac{1}{R_{p1}}=\frac{1}{2}+\frac{1}{2}=\frac{2+2}{2\cdot 2}\rightarrow R_{p1}=\frac{2\cdot 2}{2+2}=1$$ \$\endgroup\$ – Zacky Jun 11 '18 at 12:04
  • \$\begingroup\$ For the second part I get the same simplification, so another 6A. Then U = 12 A * 2 Ohm = 24V. \$\endgroup\$ – Goswin von Brederlow Jun 11 '18 at 12:12

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