9
\$\begingroup\$

I have read that current is always the same within a circuit, but as far as I understood voltage is not. Every electronic part I use lowers the voltage more or less, even simple wires do this. So far, so good.

Now I wonder why it does not matter if a resistor comes before or behind an LED with respect to voltage drops caused by the parts of a circuit.

Supposed I have a very simple circuit:

9V Battery -> Resistor -> LED -> 9V Battery

Also supposed that the LED has a maximum voltage of 3V and 20 mA. So I need to calculate the desired resistor:

9V - 3V = 6V

So I need a resistor that takes 6V out, and since I want 20 mA and current is the same in the entire circuit, it's according to Ohm's law:

U = R * I
6V = R * 0,02A
R = 6V / 0,02A
R = 300 Ohm

Again, so far, so good.

Now, with a resistor taking out 6V this makes sure that only 3V are left for the LED: The battery provides 9V, the resistor uses 6V, the LED gets the remaining 3V. Everything's fine.

What I do not get is why it also works the same way if I have the resistor behind the LED. Wouldn't that mean that we have a battery providing 9V, the LED getting all the 9V, using 3V, and then 6V for the resistor remaining?

Why does this work? Shouldn't 9V be way too much for the LED? Why doesn't it matter if the resistor is set up before or behind the LED?

| improve this question | |
\$\endgroup\$
  • 4
    \$\begingroup\$ See electronics.stackexchange.com/questions/19759/… and electronics.stackexchange.com/questions/13746/… - do those make it clearer? \$\endgroup\$ – pjc50 Mar 5 '15 at 9:27
  • \$\begingroup\$ Hm, well, I'm not too sure (but I'm really a beginner, so to be honest, most of the explanations weren't clear enough (yet) for me). What I think I got: Does that effectively mean that the resistor does not lower the voltage behind it, but on both sides? \$\endgroup\$ – Golo Roden Mar 5 '15 at 9:52
  • \$\begingroup\$ I think this answer helped me :-) \$\endgroup\$ – Golo Roden Mar 5 '15 at 9:57
  • 1
    \$\begingroup\$ One way the LED has 9 and 6 volts at its terminals, the other way 3 and 0. Still a 3V voltage drop either way. \$\endgroup\$ – Marquis of Lorne Mar 5 '15 at 11:30
26
\$\begingroup\$

One thing you need to remember is that voltage is relative. Voltage is a potential difference and it makes no sense to discuss voltage without a 'zero' reference.

In the case of your LED circuit there will be a voltage across the battery, a voltage across the LED, and a voltage across the resistor. If you add up all the voltages as you go around the loop, you get zero – up 9 at the battery, down 6 at the resistor, down 3 at the LED, the total is zero and you're back at the same point in the circuit. The LED only sees the difference in voltage between its two leads, as does the resistor. Since only the difference is important it makes no difference what order the parts are connected in.

As for current, it is only the same along a continuous path. Electrons are not created or destroyed (what goes in must come out). Since there is only one possible path for the electrons to take in your circuit, the current will be the same through all of the components. In a parallel circuit you see the opposite: all of the components have the same voltage, but the currents will be different.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Perfect answer, I completely understood id, thanks for making it that clear and easy to grasp :-)) \$\endgroup\$ – Golo Roden Mar 5 '15 at 12:00
8
\$\begingroup\$

If you want 3V across a LED it doesn't matter if you apply 13V to the anode and 10V to the cathode. There is still 3V across the LED.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks as well for your answer, it also explains it very nicely :-) \$\endgroup\$ – Golo Roden Mar 5 '15 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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