# Why changing the potentiometer affects the whole circuit?

simulate this circuit – Schematic created using CircuitLab

I have the following simple circuit where the potentiometer is used as a rheostat.

9V battery (-) --- 0.470K Resistor --- LED --- 1K Potentiometer --- LED --- (+)

When I turn the shaft of the potentiometer both LEDs change their brightness.

My problem arise from my own understanding that electric charge flows from (-) to (+) so whatever happens after the first LED should not affect it (should not matter).

Where is the problem in my mental picture?

• The schematic editor is available when you edit your question. You should add a diagram to go with your question. It may help someone else in future. Dec 12 '15 at 17:16
• Let's suppose that "whatever is connected after the first LED should not matter". In that case, it wouldn't matter if your circuit was even complete (returned to the battery's other pole) at all. Does that sound possible? What kind of trouble would happen if a complete circuit wasn't necessary, and just touching the negative terminal of the battery would get power flowing? Could you touch the battery terminal without getting shocked? Do your observations of real life support your conceptual understanding? Dec 12 '15 at 19:46

The brightness of a LED is controlled by the current flowing through it. Current flows through the whole circuit and each element in your circuit gets an equal amount of current, meaning that identical LEDs in series will have an identical brightness, disregarding minor manufacturing process variations.

Now, the current through your complete circuit is determined (in this case) by the total resistance and the voltage of the battery, so when you change your rheostat, the total series resistance changes, and the total current changes.

Electric current flows in a circuit a loop if you like. It moves because one or more elements of the circuit, such as batteries or generators create an 'electro motive force' (EMF) - measured in Volts that drives a current round the loop. The same current will pass through each element in the circuit, and the amount of current will depend on the total resistance of the elements in the loop.

It is true that electrons in the circuit conductors drift from the negative terminal towards the positive terminal of the driving EMF but this happens quite slowly, on the other hand the effect of the EMF creates a field that travels through the conductors at close to the speed of light. There is no real reason why, but conventionally we say that current is flowing from the positive terminal to the negative terminal.

I think this circuit might exhibit a response more like you expected.

simulate this circuit – Schematic created using CircuitLab

As others have commented D1 has to pass all the current in the circuit. In the modified circuit the voltage across D2 + R1 varies with the potentiometer and so the current through D2 does as well. As the pot is turned down D2 will dim to the point where it turns off.

The circuit still won't be exactly what you expected because D1's current still depends on the pot setting as the pot wiper is moved up D2 will glow brighter but so will D1 because all the circuit current passes through it.

Try it and see. It will be instructive to connect your voltmeter between the wiper (+) and the bottom of the pot (-). If your meter has a mA range then try connecting that in series above D1, below D1 (should be the same), at the bottom of the pot and in the D2 - R1 branch.

You can also learn some of the characteristics of an LED if you connect the meter across D2. You should see the voltage rise a good bit before you see any light. You will notice that the voltage maxes out at some point (about 1.8 V for a red LED). All this should teach you that the LED's voltage and current relationship is like that of a diode and not like a resistor.

Have fun!