I'm still trying to wrap my head around current, voltage and resistance, and slowly I have the feeling that I start to get some things. Hence, I would like to describe my understanding so far and ask whether it's correct so far.
Current, voltage and resistance
First of all, inside of a circuit there's a stream of electrons. While technically they go from the negative to the positive end, we usually consider it to be the other way round.
Now since it's a stream, we can count how many electrons pass an arbitrary point in the circuit in a specific amount of time. This is what is called current, and it's measured in Ampere.
So: 1A is so and so many electrons per time frame.
Something has to make the electrons move. The force that makes them move is the voltage, and it's measured in Volt. Finally, there is resistance, and it's measured in Ohm. While voltage "drives" electrons, resistance slows them down.
We have: 1V is the force needed to move 1A over a resistance of 1 Ohm.
If we increase the voltage, we get a higher current; and if we increase the resistance, we get a lower current. If we multiply current and voltage, we get the actual power, which is measured in Watt.
Is this correct so far?
Now, if we have a 9V battery, and we connect both of its poles directly using a wire, there is almost no resistance. Using Ohm's law, we get a very high current due to the extremely low resistance. This results in a huge amount of energy moved in a very short time, hence the wire and the battery heat up, and this is what we call a short circuit.
Adding an LED
Now let's make an LED brighten up. Since an LED is a diode, electrons can only flow through it in one direction. So let's connect the LED's anode and cathode to the battery's poles, and see what is going to happen.
The LED causes a voltage drop of, let's say 2.5V, and is able to handle a maximum of 20 mA. Now if we connect it directly to the battery, yes, well, what is going to happen?
I know that we should add a resistor, and I even know how to calculate it, but I can not explain why this is, or to put it another way round: What would happen if we did not add a resistor.
Let me explain this:
I know that if we have a 9V battery, and the LED drops 2.5V, there still need 6.5V to be dropped. Since the LED handles 20 mA, and current is everywhere the same in a circuit, we need to calculate:
6.5 / 0.02 = 325
Hence, we need a resistance with 325 Ohm. Since the resistor now drops 6.5V voltage, overall we drop
6.5V + 2.5V = 9V
which is exactly the potential of the battery. Also, it doesn't matter whether we put the resistor before or behind the LED into the circuit, as all that counts is the drop overall. So everything is fine.
But what happens if we do not add the resistor? In this case I have some questions that I am not able to answer:
- The LED drops 2.5V, but the battery's potential is 9V. What about the "missing" 6.5V? Where are they? What happens to them? What happens to the battery, and what to the LED?
- If we do not add a resistor, I guess the current will be way too high for the LED, so I guess we will have a short circuit, but now with an additional blowing LED, right?
- What is the actual current in the circuit?