Confusion about basic connections : Excessive Current?

For the LED blink test, it is advised to use a 220Ohm Resister.

I got the logic behind that, which is to keep the current drawn from the board to .02A. (I(0.2)= 5V/220).

So thats fine.

But for the potentiometer test stated here : http://arduino.cc/en/Tutorial/AnalogReadSerial, I see there is no other resistors than Potentiometer. And it tells about bringing the potentiometer from 0,1 and so. So, if thats 1 Ohm resistance, I = V/1 => 5A.

Am I missing something here? Please correct me :)

For the LED, its really (5V - Vf)/220 = current, with Vf the voltage across the LED.

For the potentiometer, one outer leg is +5, the other outer leg is Gnd, the middle leg will vary in voltage as the input to an analog input pin. The 5V source will only see the full value of the pot all the time.

Vout = 5V*R2/(R1+R2) if you had 2 resistors: 5V R1 -- Vout R2 Gnd.

R1/R2 vary in value as the knob of the pot changes.

CrossRoads: For the LED, its really (5V - Vf)/220 = current, with Vf the voltage across the LED.

For the potentiometer, one outer leg is +5, the other outer leg is Gnd, the middle leg will vary in voltage as the input to an analog input pin. The 5V source will only see the full value of the pot all the time.

Vout = 5V*R2/(R1+R2) if you had 2 resistors: 5V R1 -- Vout R2 Gnd.

R1/R2 vary in value as the knob of the pot changes.

A bit offtopic, but let me ask, If I need to connect a LDR to read the voltage drop. Why do I need an extra 1k Resistor connected to it?

There is a voltage drop across the LDR since it has some resistance. Why cant I measure the drop there?

If the LDR resistance gets too low, the current thru it will burn it up. V/R = I, so as R drops I goes up. The 1K resistor puts a lower limit on the total resistance. V/(Rldr + 1K) = I thus I can never exceed 5mA even if Rldr = 0.

CrossRoads: If the LDR resistance gets too low, the current thru it will burn it up. V/R = I, so as R drops I goes up. The 1K resistor puts a lower limit on the total resistance. V/(Rldr + 1K) = I thus I can never exceed 5mA even if Rldr = 0.

Seems like you are the one answering all my questions. When one answer looks a bit complex, the next reply solves it all.

Cheers!

I do what I can 8)