edgemoron:
I'd bet you put the meter in PARALLEL with the LED, ( a SHORT CIRCUIT), reading DC current requires the meter to be in SERIES with the load being tested.
Oh my god, That is EXACTLY what I was doing. My bad.. Thanks so much for your help! I'm getting accurate readings now.
olf2012:
How does your ceiling light know to only draw 100W from the power plant providing several Gigawatts? Read about Ohm's law.
Yes I think I'm getting a much better grasp on this now. If I understand this correctly, the Arduino regulates the voltage to 5V, so without factoring in how much resistance is on the Arduino board itself, the amperage drawn would be equal to whatever resistance is in the circuit. So with a simple circuit containing only a 1000 Ohm resistor the amperage would be somewhere around 5mA, as per the following calculation:
V = IR
I = V/R
I = 5V/1000 Ohms
I = 0.005A/5mA.
This does beg the question. How much resistance is on the Arduino board itself/How many amps would be drawn if I removed the resistor from this circuit? I tried to test this out, and measured around 280mA. So using the following calculation I get a resistance of 17.86 Ohms (rounded to 2 decimal places):
V = IR
R = V/I
R = 5V/0.28A
R = 17.86 Ohms.
Is this correct?
I should note that when I carried out this test the onboard Arduino lights faded off as the amp measurement lowered, so they were completely off by the time I hit 280mA. Is this because the multimeter became the path of least resistance? And therefore 17.86 Ohms is actually the resistance on the multimeter rather than the Arduino?
Sorry for the n00b questions.. Appreciate you guys helping me.