1
\$\begingroup\$

I just used Ohm's law and I was puzzled by this. I am quite new to resistors, and I am only going to use it for an Ionocraft project to reduce the flow of current to ground.

So if I have a 9v battery, and had an impossible LED that could get 1 amp. Which means it would be 9v divided by 1 amp which would ask for 9 ohms of resistance. The problem is, a nine volt battery doesn't have 1 amp of current.

How would this work??? This question has really been distracting me and it would be nice if someone could explain it to me.

\$\endgroup\$
5
\$\begingroup\$

All batteries have an internal resistance. You can imagine it as an ideal voltage source with a resistor in series with it. So the more current you try and draw, the more voltage is dropped across this resistance. In turn this means the voltage you see across your circuit will reduce. At some point you will reach an equilibrium.

Imagine the following:

schematic

simulate this circuit – Schematic created using CircuitLab

If you have a 9V battery as you say, then Vs=9V. In the schematic, Rs is the internal resistance of the battery, lets say it is 50Ohms (About realistic for a PP3 battery).

Now when you have no load attached (Rload = Infinity), then there will be no current flowing and so no voltage will be dropped across Rs (Ohms law). So Vload would be 9V. If you measure the battery with a multimeter, you would see 9V as the meter has a very high resistance.

Now lets say you attached your 9Ohm resistor, Rload. This would mean you now have Vs driving a circuit containing a 50Ohm resistor Rs and 9Ohm resistor Rload. Using ohms load, you can find that I = V/R = 9/(50+9) ~= 150mA. What gives? Surely there should be 1A flowing through your load as you calculated.

Lets look a bit further. Using ohms law again, you can find that Vload = I*Rload = 0.15*9 = 1.4V. Ah, there we go. Essentially what has happened in the voltage at the terminal of your battery (Vload) has dropped significantly, so now you essentially have a 1.4V voltage source driving a 9Ohm resistor.

If you lower the resistance, the current will go down and the load voltage will go up.

If you were to instead use a 9V power supply which had an internal resistance of say, 0.1 Ohms, you would indeed get almost 1A flowing through your resistor. Notice I said almost. Why? Same reason as the 9V battery example, the load voltage will drop slightly due to the internal resistance.

\$\endgroup\$
1
\$\begingroup\$

LEDs do have a voltage drop that depends on colour, and is fairly independent of current. Red LEDS are about 1.8 volts, Yellow 2, and green 2.2 - I call them all 2 volts, for convenience. Blue and white LEDs are aobut 3 volts.

So, to determine the current limiting resistor for an LED, you have to subtrack the LED voltage from the supply voltage to get the actual voltage across the resistor. If you have a 1 Amp Yellow LED, you allow 2 volts for the LED, leaving 7 volts for the resistor.

However, all batteries have an internal resistance that depends on cell size, chemistry, and usage (the internal resistance of a new battery will be much less than that of a battery near end-of-life).

When you connect your 1 Amp LED and 7 ohm resistor to a common 9 volt battery, it will attempt to draw 1 Amp, but the internal resistance of the battery will be in series with your resistor, and will limit the current. Due to the small size of the cells in the 9 volt battery, the internal resistance will rapidly rise, further reducing the current through the LED.

\$\endgroup\$
0
\$\begingroup\$

Ohm's law applies to specific conditions. If you say there is 9 volts, then there is, and they don't care how you get it. When you get into the real world, you find that you may get different results. And in your example, you will find that the battery voltage drops to a lower voltage when you attempt to draw that amp out of it. If you measure the resulting voltage and current, they would agree with the 9 ohm load provided by the LED (But note that in real life an LED does not look anything like a resistor).

Bonus info: If you look at the difference between the battery voltage before you connect it, and the voltage you get when it's connected, you can say that the "missing" voltage appeared across another resistor inside the battery. And you can use ohm's law calculate its value, too, using the measured current and that "missing" voltage.

We call that the battery's "internal resistance".

\$\endgroup\$

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.