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I'm trying to understand the behavior of the circuit below containing both voltage and current sources.

I don't really see how can we get 30 V at the node in the middle when we have only 20 V voltage sources. I assume this would be the effect of the current source...

Does anyone know how to analyze the circuit analytically to get the 30 V? I tried it without success.

enter image description here

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    \$\begingroup\$ An ideal current source can produce an infinite voltage. \$\endgroup\$
    – Finbarr
    Commented May 17, 2023 at 12:13
  • \$\begingroup\$ "Without success" doesn't help us to help you. Show the steps you tried and we can see where you went wrong. \$\endgroup\$
    – Finbarr
    Commented May 17, 2023 at 12:14
  • \$\begingroup\$ @Finbarr, So would you say that this situation (having 30 V at middle node) would not be realistic if I had a non-ideal current source? \$\endgroup\$
    – Likely
    Commented May 17, 2023 at 12:21
  • \$\begingroup\$ @Finbarr, Right, I should've added more details. Basically, I did a simple analysis where I said current will only be coming from the current source and divided equally between left and right since we have no voltage difference between left and right. This leads to V_R1 = V_R2 = 10*1 V and because current comes from current source only, voltage at middle node should be higher than left or right nodes. Hence, 30 V!! \$\endgroup\$
    – Likely
    Commented May 17, 2023 at 12:24
  • \$\begingroup\$ "Voltage source" should really be called a "fixed voltage" because in circuits like this the power is actually flowing into the so-called source. If it helps, you can imagine a battery getting recharged. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented May 17, 2023 at 12:26

1 Answer 1

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Well, using KCL we can see that:

$$0=\text{I}_1+\text{I}_{\text{R}_1}+\text{I}_{\text{R}_2}\tag1$$

The voltage across the resistors is given by Ohm's law:

  • $$\text{I}_{\text{R}_1}=\frac{\text{V}_1-\text{V}_{\text{e}_3}}{\text{R}_2}\tag2$$
  • $$\text{I}_{\text{R}_2}=\frac{\text{V}_2-\text{V}_{\text{e}_3}}{\text{R}_2}\tag3$$

So, we get:

$$0=2+\frac{20-\text{V}_{\text{e}_3}}{10}+\frac{20-\text{V}_{\text{e}_3}}{10}\space\Longleftrightarrow\space\text{V}_{\text{e}_3}=30\space\text{V}\tag4$$

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