This may be a stupid question but it seems that I don't have a clear image about how this works exactly.

So my question is:

In the attached schematic, why will there always be a higher current in the A loop than in the B loop. What I mean is: why the current from the 10V source won't take the path marked with yellow in the image?

I understand that a higher voltage (and the same resistance) will give a higher current but I am confused about the path.



The current from the 10 V source won't follow the yellow arrow because that doesn't lead back to the power source. Current only will flow from a voltage or current source through a circuit if it can go back to that source, so that a closed loop is formed.

enter image description here

This schematic may help you to understand this. It's exactly the same schematic as in your question. So why would the current coming from R3 go to the left part of the circuit? It's a dead end, there's no return path.

Let's call Kirchhoff in. Kirchhoff's Current Law (KCL) says that the total current in a node must be zero; the sum of the incoming currents = the sum of the outgoing currents.

enter image description here

Applied to node A this gives us I1 + I3 = I2. But when we follow I1 through V2 and R3 we see that it's the same as I2, since there are no branches on the way. So if I1 = I2 then I3 must be 0. Even more simple when you know that KCL doesn't only apply to nodes, but to any closed boundary:

enter image description here

The sum of all currents going out of the blue rectangle = the sum of all currents coming in. There's only one possible path, so I3 must be zero to satisfy the criterion.

  • \$\begingroup\$ I am not sure I understand this correctly. Could you please elaborate a bit more on the "dead end"? One more question: basically the amount of charge that goes out of a voltage/current source must also go back in? No more or no less charge? \$\endgroup\$ – Buzai Andras Aug 23 '12 at 13:06
  • \$\begingroup\$ Dead end: if current would flow from the right half to the left half, how would it return? It has to, current can only flow in a closed loop. That's your second question about. Yes, it has to go back in, otherwise there wouldn't be a charge balance; V2 would remain positively charged, and V1 negatively. Charge won't do this all by itself. It will always go for electrical neutral. \$\endgroup\$ – stevenvh Aug 23 '12 at 13:16
  • \$\begingroup\$ @BuzaiAndras: It might also help to think of it this way: A current exists (i.e., electrons flow in a loop) when and because there is a difference in potential (voltage) between one point and another -- This difference (described as 10V and 5V for the two separate sources in your circuit) is what motivates the electron flow in the first place, and also is what keeps it going. BTW, it isn't the SAME electrons that travel all the way from the positive to the negative terminal of the right-side source, but it can be thought of in essence as one separate loop in your comparison of the currents. \$\endgroup\$ – boardbite Aug 23 '12 at 14:09
  • \$\begingroup\$ @stevenvh Just to make sure that I understand this correctly: basically the current won't go on the path marked with yellow because, in its attempt to achieve a neutral state, the V2 (10V) voltage source will try to attract the same amount of charge that got out from it? And the same thing applies to V1 (5V) source? Is this correct or I am way out of track? \$\endgroup\$ – Buzai Andras Aug 25 '12 at 13:33
  • \$\begingroup\$ @Buzai - That's right. Replace the word "charge" by "current" and you have KCL. I added to my answer. \$\endgroup\$ – stevenvh Aug 25 '12 at 14:38

basically the amount of charge that goes out of a voltage/current source must also go back in? No more or no less charge?

Yes. This is because of the law of charge conservation.

Charge conservation is a physical law that states that the change in the amount of electric charge in any volume of space is exactly equal to the amount of charge flowing into the volume minus the amount of charge flowing out of the volume.

  • \$\begingroup\$ The charge conservation law only says that total charge doesn't change, but not where that charge should be. So if there would flow charge from the +10 V to the +5 V it wouldn't violate the law: there would have charge left the 10 V source, and the difference between the starting charge and the remaining charge would be equal to the charge that left. \$\endgroup\$ – stevenvh Aug 23 '12 at 18:04
  • \$\begingroup\$ @stevenvh I had to read your comment a few times but I think I understand what you're saying. I was reading the law as "the amount of charge in any volume is equal to the amount flowing into AND the amount flowing out of that same volume and the amount doesn't change". But what you're saying is that charge can flow into OR out of a volume and if it's out of balance then the value of the charge (positive or negative) of the volume would then change accordingly. Am I understanding you correctly? \$\endgroup\$ – embedded.kyle Aug 23 '12 at 18:20
  • \$\begingroup\$ Yes, that's what I meant. The quotation says "the change in the amount of electric charge", and as I understand it your interpretation doesn't allow change in charge. Sorry if it wasn't very clear. \$\endgroup\$ – stevenvh Aug 23 '12 at 18:58
  • \$\begingroup\$ @stevenvh Thank you for clarifying. This is a concept I'm trying to understand more about myself and I guess it's application to this question was a bit off. \$\endgroup\$ – embedded.kyle Aug 23 '12 at 19:11

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