I am having doubt understanding this:

If two voltage sources V1 and V2 are connected in parallel with different voltages like 2v and 3v is it possible or the scenario is wrong.

An example is this circuit

enter image description here

Tried solution:

enter image description here

  • $\begingroup$ Will create a short circuit if no resistor is included--in theory infinite current. $\endgroup$ – user45664 Jun 24 '18 at 3:23
  • $\begingroup$ Exact duplicate of What happens when non-equal voltages are put in parallel? $\endgroup$ – Alfred Centauri Jun 24 '18 at 3:28
  • $\begingroup$ Thanks for your responses ,i have added abd example problem as well to make my question clearer . Here if you see the first connection its showing 4v in parallel with 12 a $\endgroup$ – Laxmikanta Nayak Jun 24 '18 at 3:30
  • $\begingroup$ There is only one voltage source in the image, i.e., there are no parallel connected voltage sources in the schematic you have provided as an example of parallel connected voltage sources. I do, however, see a voltage source in parallel with a current source. $\endgroup$ – Alfred Centauri Jun 24 '18 at 3:30
  • $\begingroup$ If you convert the current aource to voltage source it will be 24v in parallel with 4v or i am wrong ? $\endgroup$ – Laxmikanta Nayak Jun 24 '18 at 3:32

First, let me address what I think is an important conceptual error in your work.

If there were only the 4V voltage source, 2 ohm resistor and 12A current source (in other words, discard all the circuit elements to the right of the 12A current source), then you could make the Norton to Thevenin conversion that you did and then solve for the current through the 4V voltage source.

In other words, with just those three circuit elements, the voltage source 'can't tell the difference' between the 12A current source in parallel with a 2 ohm resistor and a 24V voltage source in series with a 2 ohm resistor.

However, since there are other circuit elements in addition to those three above, the conversion that you did is conceptually wrong. The voltage source, as well as the circuit to the right of the 12A current source, 'can tell the difference' between the original circuit and the one you've drawn.

If what you're trying to do is simplify the circuit in order to solve for $I_S$, then the first step is to completely remove the 2 ohm resistor and 4A current source. Why? Because they are in parallel with a voltage source.

Think carefully about this (I won't work this out here since that would probably be considered doing your homework for you). When you do, you'll find that the only circuit variable the 2 ohm resistor and 12A current source can affect is the current through the 4V voltage source - removing them will not change the value of $I_S$. So remove them and solve for $I_S$.

Put another way, the Thevenin equivalent circuit, looking leftwards towards the 12A current source, is just a 4V voltage source.

I would like for you to work out this Thevenin equivalent for yourself and then post your work as an answer to your own question.

  • $\begingroup$ It really helped but i have not learnt network theorems and its from my basic circuit problem section, With only using kcl or kvl if i want to solve how can i do that? I dont want to get a solution just want some guidance $\endgroup$ – Laxmikanta Nayak Jun 24 '18 at 14:12

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