• Engineering

## Homework Statement

This isn't really a homework question, but I was fiddling around on the virtual electronics lab at 123D circuits to familiarize myself with the various electrical components used in making circuits and just to familiarize myself with building electronics in general since I'm in more of a beginner state right now. Here is a link to the circuit: https://123d.circuits.io/circuits/1414295-simple-led-adjuster#breadboard
Parts used:
9 V Battery
500 Ohm resistor
10 kOhm potentiometer
1 RGB LED - not sure about its voltage rating or its current rating, but I just went off the basic fact that most LEDs can only take 20 milliamps at the most.

My question is - mainly because I cannot remember exactly why it is and have been unlucky in searching for this answer - this: Why would I still need the 500 Ohm resistor to power the LED with the potentiometer since the potentiometer acts as a resistor in itself, so shouldn't I be able to just use a potentiometer to function as this variable resistor without needing an actual resistor?

## Homework Equations

For this one, not really sure but if someone can enlighten me as to how exactly I should go about using calculations to test problems like this, it would be great since I haven't really applied my general knowledge of the following equations to building these circuits (besides Ohm's law of course):
Ohm's Law: U (voltage) = P (current) * R (resistance)
I think Kirchoff's Law's also apply here but haven't really practiced those concepts as much so I'm still unsure about how they apply to this case.

## The Attempt at a Solution

I tried to think of why it wouldn't work, but wasn't sure how to approach this issue. Sorry if this is not satisfactory.

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R is for safety. If potentiometer goes near to zero, your led will burned.

gneill
Mentor
I tried to think of why it wouldn't work, but wasn't sure how to approach this issue. Sorry if this is not satisfactory.
Can you sketch your intended circuit (schematic) and post it here? I don't think your breadboard layout is realizing the circuit you had in mind.

R is for safety. If potentiometer goes near to zero, your led will burned.
OH! That's what I was forgetting. So the main resistor that applies to the basic Ohm's Law equation with U = 9V and I = .020 amps (which gives R = 450 ohms) is basically like the initial resistance/fail safe for the potentiometer to work from? That makes some sense to me in my mind but I don't completely get how that works: does that mean that when the potentiometer is at its resting position, it's functioning as the 10 kOhm resistor and when you turn it counterclockwise, it reduces the resistance of the potentiometer? Thanks for the quick check!

Can you sketch your intended circuit (schematic) and post it here? I don't think your breadboard layout is realizing the circuit you had in mind.
This is probably why I'm not understanding this as easily as I should: I was doing this circuit just based off how I did a simple LED circuit with a resistor in series and went on from there to experimentally(not really - was kinda confused on how to use the virtual multimeter since i only got voltage readings after the current went through the cathode of the LED (or in my mind, before the current flowed into the cathode since its the flow of electrons after all)). I avoided doing a circuit schematic mainly because I'm not sure how to properly analyze such a circuit with something like a potentiometer, or at least how to treat it in such a circuit when it has such a high resistance.
But now I tried, and it's most probably different from my actual circuit but it should still have the same wire connections with the terminals of the potentiometer(used 3 instead of 2). If anything, I'd just like guidance as to how to treat the potentiometer in analyzing this circuit since I'm not sure how to apply it with the equations mentioned earlier due to there also being the main R resistor. I'm self-teaching myself most of the electronic basics in preparation for a class I'm taking next semester since I'm not really familiar with them yet.

*apologies for the poor lighting - took this picture while in a car.

CWatters
Homework Helper
Gold Member
I would expect the potentiometer to be configured as a variable resistor in series with the fixed resistor. It doesn't appear to be like that on the breadboard.

CWatters
Homework Helper
Gold Member
PS I can't see a picture but perhaps that's because I am using a tablet to view the forum at the moment.

PS I can't see a picture but perhaps that's because I am using a tablet to view the forum at the moment.
My connection on the road via mobile wasn't letting me properly upload the image, so here it is again. I did write the schematic so that the potentiometer is in series, but is it actually in parallel in the breadboard? Sorry for such novice questions, fairly new to applying the bare basics.

#### Attachments

• 38.3 KB Views: 254
CWatters
Homework Helper
Gold Member
There are several issue with the breadboard and the schematic but currently they don't match each other.

One complication s that you are using an RGB LED that shares a common pin with the other colours in the same package. It looks like you have a "Common Cathode" RGB LED but can you confirm that? If the LED lights up on the breadboard as it is then I believe it must be a Common Cathode type.

If it is a common cathode type then normally the common pin would be connected to the battery -ve. You have done this on the breadboard but not on the schematic. The breadboard has other problems as well so it's not a simple change.

Personally I would change the schematic so it looks like this and rebuild the breadboard circuit to match..

There are several issue with the breadboard and the schematic but currently they don't match each other.

One complication s that you are using an RGB LED that shares a common pin with the other colours in the same package. It looks like you have a "Common Cathode" RGB LED but can you confirm that? If the LED lights up on the breadboard as it is then I believe it must be a Common Cathode type.

If it is a common cathode type then normally the common pin would be connected to the battery -ve. You have done this on the breadboard but not on the schematic. The breadboard has other problems as well so it's not a simple change.

Personally I would change the schematic so it looks like this and rebuild the breadboard circuit to match..

View attachment 93832
Alright, thank you for providing your recommended schematic - I'll have to just practice properly drawing out the schematic before building a prototype on the breadboard. Thank you all for your answers.

CWatters