Lesson · Problem 1
Light the LED
One series loop sets the LED current. The anode faces the source.
Try it
The arrow is conventional current, from the positive terminal back to ground. A is the anode, where current enters. K is the cathode. The bar marks K.
I = (5.0 V − 2.0 V) / 330 Ω = 9.1 mA, inside the 8–12 mA window. The resistor can sit on either side of the LED. The current stays 9.1 mA.
Why it works
The LED holds about 2.0 V while it is on. The resistor gets the rest of the 5.0 V, and that voltage divided by resistance is the current in the whole loop. Putting the resistor on the other side of the LED does not change that current. Reversing the LED points the cathode at the source, so this training LED stays dark.
Key takeaway
One current flows through the source, the resistor, and the LED. I = (5.0 V − 2.0 V) / R. The anode faces the source.
Question 1 of 2