0
$\begingroup$

enter image description here

Take the above circuit. I'm trying to understand how current flows through this potential divider circuit. I used the following simulation to help me visualize this.

enter image description here

In the simulation the electrons flow up the middle wire ( Towards the bulbs ) between the two bulbs. Yet I can't understand why it does this. As the electrons move past the first bulb could they not travel down the middle wire? Why doesn't this happen. Electrons leaving the first resistor have two paths to choose from while Electrons leaving the first bulb should have two paths to choose from as well. Yet somehow the Electrons do not travel down that middle wire. Why?

Furthermore when I decrease the resistance of resistor 2 the brightness of the bulb decreases. Is this because a decrease in resistance means a decrease in voltage, so the bulb will be dimmer?

$\endgroup$
4
  • 1
    $\begingroup$ You are trying to explain how the currents flow without using the concept of potential difference or voltage. This is depriving yourself of an almost essential tool. $\endgroup$ Aug 27, 2022 at 21:39
  • $\begingroup$ If we think of current as water, current leaving the first bulb will fork and take one of the two paths. Either to the other bulb or towards the resistor. Current passing the 1st resistor will also fork and can either move up towards the lightbulbs and also towards the other resistor. What I don't understand is, according to this logic we have current both flowing down that middle wire and up that middle wire simultaneously which is not possible, so what is actually happening? $\endgroup$ Aug 29, 2022 at 22:48
  • $\begingroup$ Could you please use the notation in your top diagram, (XK and KY for the resistors; L1 and L2 for the lamps). Call the junction between the lamps "L", so you have a wire KL. That way I'll be able to follow your question. $\endgroup$ Aug 29, 2022 at 22:56
  • $\begingroup$ As I tried to explain in my answer, if K is nearer X than Y, then current will go through the wire from K to L. So the current coming down through L1 will not split when it reaches L. Instead it will be joined by current reaching L from K, and the sum of these currents will go through L2. $\endgroup$ Aug 30, 2022 at 14:25

2 Answers 2

0
$\begingroup$

Imagine you have a water pipe that forks out into two pipes, one completely clean, the other partially clogged. When water runs through the main pipe, what do you think it is going to do at the fork? Move as one through the unclogged pipe or split between the two pipes, the clogged one receiving less water than the clean one?

$\endgroup$
0
$\begingroup$

That lower diagram scares me! Looking at the top diagram, the battery places a potential difference (voltage) across the variable potential divider, XY, and the same potential difference across the two lamps in series.

If I'm not mistaken, K is shown nearer X than Y (that is the resistance of XK is less than that of KY). If there were no wire between K and the junction – call it 'L' – between the lamps (assumed identical), K would be nearer X than Y in terms of potential whereas L would be midway in potential between X and Y. Therefore there would be a potential difference, $\Delta V$, between K and L, with K at the higher potential. That's why there's a current from K to L when the wire is re-instated. [Electrons, being negative, flow from L to K.]

You might object that when the wire is re-instated, the potential difference between K and L disappears (because $\Delta V = IR$ and $R$ is zero for the wire.) But $R$ will not quite be zero and $\Delta V$ won't quite be zero. $I=\frac{\Delta V}R$ will have a finite value.

Turning to your last paragraph, if we move the slider, K, up to X, L2 is connected straight across the battery, so will be bright, but L1 will have no voltage across it and will not give out any light.

• Think about what will happen if K is moved down to Y,

• If K is moved to midway between X and Y, the bulbs will have the same 'medium' brightness, just as they would be if the wire between K and L were removed. Can you see, in terms of potential difference, why this should be so?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.