At tbe risk of offending everyone who has posted, i think the essential fundamental concept has been overlooked.
Ohm's Law
Resistors follow Ohm's Law. Voltage, resistance and current are related by
V = I×R where
V = volts
R = resistance in ohms
I = current in amps
Lets say we have a 100 ohm resistor with a current of 100mA. What will the voltage across the resistor (i.e., voltage drop) be?
V = 0.1 × 100 = 10 volts - remember, current must be in terms of amps,
We can rearrange the equatuon to calculate the desired quantity. Considering tour circuit of battery, LED and resistor. How can with determine the current through the circuit? Starting with
V = I × R,
we can rearrwnge it:
I = V / R
Lets say we have a 10 ohm resistor and we measure 2 volts across it while the circuit is hooked up and the LED is on.
I = V / R =2 / 10 = 0.2 amps = 200mA
Ohm's Law applies to resistors. The LED is an active device, so Ohm's Law doesn't apply to it.
Power Dissipation
The power dissipated by an element in a circuit is
P = I ×V where
P = power in watts
I = current in amps
V = volts
For the circuit of battery, LED and resistot, the current is the same everywhere in the circuit. In the above example, we calculated the current as 0.2 amps. If we measurethe voltage across the LED and find it to be 3.0 volts, then
P = I × V = 0.2 × 3.0 = 0.6 watts.
The power dissipated by the resistor may be found the same way. Combining the Ohm's Law equation and the power dissipation equation, we can get a couple shortcuts to calculate the power dissipated by the resistor.
P = V^2 / R
P = i^2 × R