I am learning about circuit analysis now, but there is something that I can't wrap my head around.

Imagine this simple circuit:

A -------------------------
|                          |
|                          |
V                          R
|                          | 
|                          |

Imagine that V is a voltage source that goes up from B to A. That means the potential difference Va - Vb is positive and that a current will flow through that branch. Because this is a closed circuit, this current will flow back from B to A through the right side (via resistor R). But how is this possible, since that path has the same potential difference (Va - Vb is positive). How can the current flow that way?


Let us crudely imagine the voltage source as a pump pumping water up to the top of a water slide, and the resistor as the slide itself with water flowing down through it.

The height difference between the top of the slide and the bottom of the slide is the same as the height difference between the top of the pump and the bottom of the pump.

The voltage source is adding energy to the system the same way the pump is adding energy to the water, that energy is used up as it flows through the resistor or down the slide.

The water must travel in a closed loop up the pump and down the slide and back to the start up the pump, if it doesn't you will eventually run out of water at the bottom of the pump. Similarly the current must flow "up" the voltage source and "down" the resistor so it forms a closed loop.


Now lets apply Kirchhoff's circuit laws to the slide analogy.

Kirchhoff's Current Law states that all of the "water" flowing into one point must flow out again:

So the amount of water entering the pump at the bottom is equal to the amount of water leaving it at the top.

And the amount of water that enters the slide at the top is the amount that leaves at the bottom.

Kirchhoff's Voltage Law states that if you follow a loop in a mesh the total voltage drop will sum up to zero.

There is only one loop that the water can take, up the pump and down the slide and back to the start, and what this law states is that if you add the height the water travels up the pump and subtract the height the water drops down the slide and get back to the start then this will add to zero.

Which is completely obvious in the slide case since if you travel in a loop back to where you started from the obviously if you are back at the start then at the end you are zero meters above or below where you started from.

  • $\begingroup$ I see, but doesn't that interfere with Kirchoff's law? The potential difference on the both branches is the same, right? Why does the current move 'up' via the left branch and down via the right branch? In one branch it goes via the potential difference, and in the other branch it goes the other side. That means that if the voltage source adds 1V to one branch, and the current flows back via the other branch, the total potential difference of the path would be 2V, but Kirchoff's law states that that should be 0V $\endgroup$ – Alexander Cogneau Oct 6 '14 at 9:48
  • $\begingroup$ In the loop B -> V -> A -> R -> B. V has a positive potential lets say one volt, which means that A has (+1 V) volts above B, so R must have a NEGATIVE potential drop of minus one volt (-1 V) to get back down to B, so +1 -1 =0 $\endgroup$ – user288447 Oct 6 '14 at 9:54
  • $\begingroup$ How can that be? Kirchhoff's law says that the potential difference across a closed loop is always 0 $\endgroup$ – Alexander Cogneau Oct 6 '14 at 9:56
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    $\begingroup$ @AlexanderCogneau, the loop must have a potential difference of 0, but segments of that loop may have differing relative potentials. $\endgroup$ – Mr. Mascaro Oct 6 '14 at 13:32
  • $\begingroup$ @AlexanderCogneau The flow in the battery goes 'up' because the battery acts like a pump; a source of energy to drive the circulation. You can't apply simply Ohm's law to it to work out which way the current flows in it, because it is not a passive component like a resistor. Kirchoff's law still holds, though. $\endgroup$ – richardb Oct 6 '14 at 17:01

Consider the current as the flow of positive charges. Positive chrges flow from higher potential to lower potential naturally if they are free to do so. But you can force them to move in the opposite direction too with the help of an extenal agency. Actually this is the work of the battery, it will move the charges from lower potential to higher potential at the cost of its own energy just like a pump as described by @ user288447. A kind of battery force acts on the charges which has non electostatic origin and is in the opposite direction of the electic field due positive and negative poles. This force is responsible for taking the charges from lower ptential to higher poential and work done by this force is the energy supplied by battery.


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