0
\$\begingroup\$

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

enter image description here

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?

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.