Let's go back and remember our younger days, when we were just connecting up different resistors to the LEDs willy-nilly. You may recall this breadboard, where we used three red LEDs and a 100 ohm, 1000 ohm and 10,000 ohm resistor to set the brightness.
Since its so important for you to have practice using those laws you just learned, we're going to have another quiz. Please try to solve the problems using the diagrams above. Yes there are calculators online that will do this for you but part of learning electronics is being able to do the calculations even if on a desert island.
In the above breadboard, the voltage generator here is +5V, the forward voltage of the LEDs are 2.2V. What is the voltage across the 10,000 ohm resistor?
This is a KVL question. 5V = 2.2V + Vresistor so Vresistor = 2.8V.
Now that you know the voltage, how much current (I) is going through the 10,000 ohm resistor (R)?
We determined the voltage (V) is 2.8V, we use Ohm's Law I = V/R = 2.8/10000 = 0.00028 Amps = 0.28 milliAmps.
In the same breadboard, the voltage generator is +5V, the forward voltage is 2.2V, what is the voltage across the 1000 ohm resistor?
You didn't get fooled by this one right? The voltage is exactly the same, the resistance does not matter for KVL, it is still 2.8V.
Now that you know the voltage, how much current (I) is going through the 1,000 ohm resistor (R)?
We determined the voltage (V) is 2.8V, we use Ohm's Law I = V/R = 2.8/1000 = 0.0028 Amps = 2.8 milliAmps.
How much current (I) is going through the 100 ohm resistor (R)?
We determined the voltage (V) is 2.8V, we use Ohm's Law I = V/R = 2.8/100 = 0.028 Amps = 28 milliAmps.
Now you can see how changing the resistor effects the brightness and why a smaller resistor means a brighter LED.
Last updated on 2015-11-20 at 05.39.25 PM
Published on 2013-02-11 at 04.20.09 PM