# resistors on either side of LED

Right when I think I am understanding ohms law...

When I build a 9V battery + LED circuit, why is it that I can put my resistor on either side of my LED and have it still work? Is it because the LED is a non-ohmic device?

Or is it that the water-analogy of current falls down here? If current was like water, you would need to narrow the pipe to the proper range for the light BEFORE it reached, not after.

Since electrical current always flows in a closed circuit, it does not matter which order you place the parts.

In a parallel circuit the voltage is the same across each component, and the current is shared proportionally between each one.

In a series circuit the current is the same through each component, and the voltage is shared proportionally across each one.

You have a series circuit, so you can put the component's in any order you like and the result will be the same.

Imagine you have a circuit made of a 9V battery, a 2 ohm, 3 ohm and 4ohm resistor. Connect them all in series, any order you like. Now your 9V battery has 9V across it. According to Ohm's Law the current is I=V/R = 9/(2+3+4) = 1 Amp. Work that backwards and you have V=IR = 1A * 2 ohm = 2V across that resistor. V=IR = 1A * 3 ohm = 3V across that resistor. And V=I*R = 1A * 4 ohm = 4V across that resistor. Add them all together and you get the originla 9V.

In electronics you need to think about the "voltage drop" across a component. Also you're right, you could call LEDs non ohmic. In very simplified terms a red LED wants 20mA to flow through it. This usually means 2.2V but it does vary with temperature, manufacturing tolerances, LED colour and a bunch of other things. With the resistor you are assuming the voltage will be pretty close to 2.2V and the reisstor will limit the current to 20mA, or close to it.

``````--------------
|            |
|            [] resistor
batt+        |
batt+       LED
|            |
--------------
``````

With the resistor you are assuming the voltage will be pretty close to 2.2V and the reisstor will limit the current to 20mA, or close to it.

No, the voltage can be anything you desire (within reason), the object is to develop 20mA of current. Look up the LED voltage drop Vf, subtract that from your battery. R=E/I If your supply is 5V you will need (5-Vf)/.02 or 180 ohms if Vf = 1.4V.

You might find it useful to look at the various series of photos I posted in this thread, especially replies 7, 8 and 9.

My meter shows the same current no matter where the circuit is broken, and also that the voltage across the led is the same regardless of which side of the resistor it's fitted.

The water analogy does not break down here. With a continuous loop of pipe with a constant pressure pump driving it and the water returning to the pump, you can put a restriction in the pipe anywhere and it limits water flow (current) at all other points in the pipe.

jremington:
Since electrical current always flows in a closed circuit, it does not matter which order you place the parts.

Unless one of them is a transistor. ]

rmetzner49:
No, the voltage can be anything you desire (within reason), the object is to develop 20mA of current. Look up the LED voltage drop Vf, subtract that from your battery. R=E/I If your supply is 5V you will need (5-Vf)/.02 or 180 ohms if Vf = 1.4V.

That's what I was getting at. Bear in mind that the LED voltage does vary a little between LEDs and with environmental factors. This means the voltage across the resistor and the current will vary a little too.

fungus:
Unless one of them is a transistor. ]

Leakage current still flows