Help me understand - Page 1
Forum section: electronics / beginners

author: lazyusername
author_rank: Newbie
timestamp: September 19, 2015, 08:34:09 pm
content: Dave reinvigorated an interest in electronics in me, and so I was mucking around with some page 1 difficulty circuits in my old RadioShack learning lab (that's my level).Something weird happened to me because I didn't bother to remove a previous circuit before starting a new one, and I eventually figured out the essence of what was going on. In the image I took, you can see that the switch in the top left is set to 'Off', and I have connected the ground spring of the DC milliamp meter to +3 Volts rather than ground. I have then connected the other spring to +6 volts with an 8.2k resistor in series.Current is flowing! It reads .4, but if I calibrate it with a potentiometer, it will give me the difference between whatever two voltages I hook it up to, divided by ten.Why would it do this? I can't replicate it with batteries and wires, because if you only connect each spring to one or more cathodes, no current should flow because no circuit is created, right? It seems like a flaw of whatever needlessly complicated wiring they used to give these multiple of 1.5 volt increments.Thoughts?

author: IanB
author_rank: Super Contributor
timestamp: September 19, 2015, 08:56:39 pm
content: Quote from: lazyusername on September 19, 2015, 08:34:09 pmThere's your answer. Voltage is a measure of difference between two points.6 - 3 = 33 - 0 = 3Same difference!

author: Joule Thief
author_rank: Regular Contributor
timestamp: September 19, 2015, 09:02:56 pm
content: and 3 volts with 8200 ohms in series with the ammeter will flow .36 mA = so there is your current reading

author: Nerull
author_rank: Frequent Contributor
timestamp: September 19, 2015, 09:18:31 pm
content: Quote from: lazyusername on September 19, 2015, 08:34:09 pmBut this setup isn't like two isolated batteries. You're dealing with a common ground. If you connected the negative terminals of a 9v battery and 12v battery together, and then measured between the positive terminals, you would see 3V. And if you connected those terminals together, current would flow. Depending on the ESR of the batteries you chose, potentially quite a lot of it...I would not recommend doing this, as the batteries might protest violently depending on the type, but its the same thing. Voltage is a measure of the difference in electrical potential between two points - any two points, and if you complete a circuit between them, current will flow.

author: IanB
author_rank: Super Contributor
timestamp: September 19, 2015, 09:39:24 pm
content: Quote from: lazyusername on September 19, 2015, 08:34:09 pmWell, you can, actually.Get four 1.5 V cells and line them up in series, positive to negative, to make a 6 V stack of cells (a 6 V battery).Now connect a wire to the middle of the stack, between cells 2 and 3, and connect another wire to the top of the battery at the positive end. Between these two wires is a 3 V battery. Connect a 3 V bulb and it will light up.This is basically the same thing you have done with the RadioShack lab.The bulb lights up because there is a circuit, and by doing this experiment with batteries and wires you can see where the circuit is.

author: Nerull
author_rank: Frequent Contributor
timestamp: September 19, 2015, 09:51:55 pm
content: Here's an animated diagram showing the circuit you made and the current flowing around it - as you can see, there is a complete circuit. If your breadboards voltage outputs were isolated and did not have a common ground, no current would flow, but that is not the case.

author: IanB
author_rank: Super Contributor
timestamp: September 19, 2015, 10:11:05 pm
content: Quote from: Nerull on September 19, 2015, 09:51:55 pmNo, I don't think that is the circuit. I think the breadboard hasfoursix 1.5 V cells in series with voltage taps at 0 V, 1.5 V, 3 V, 4.5 V, 6 V, 7.5 V and 9 V. I believe the OP connected a circuit between the 3 V tap and the 6 V tap. The OP was confused because the circuit did not include the 0 V tap, or "ground". This caused the OP to think there was no return path for the current, and therefore no circuit.

author: lazyusername
author_rank: Newbie
timestamp: September 19, 2015, 10:12:13 pm
content: Quote from: Nerull on September 19, 2015, 09:51:55 pmI'm confused. The different voltage rows are shown as isolated from each other, with a broken black line printed over each one. They have a common ground at the bottom, but I made no connection to it.

author: IanB
author_rank: Super Contributor
timestamp: September 19, 2015, 10:17:29 pm
content: Quote from: lazyusername on September 19, 2015, 10:12:13 pmSee my post #4 above. Take the batteries out of the lab and try the experiment with just batteries and wires so you can see what is going on.

author: lazyusername
author_rank: Newbie
timestamp: September 19, 2015, 10:22:42 pm
content: Quote from: IanB on September 19, 2015, 10:17:29 pmI was imagining the batteries were wired in parallel with each other, because I know nothing. But your explanation makes some sense.

author: Nerull
author_rank: Frequent Contributor
timestamp: September 19, 2015, 10:27:59 pm
content: Quote from: IanB on September 19, 2015, 10:11:05 pmI had assumed they were using voltage regulation, but I found some pictures of the lab itself and it looks like that's exactly how it works. The lab is battery powered, and the voltage outputs are directly wired to the battery holder - there is no regulation circuitry of any kind in this thing. The batteries are wired in series, with the voltage tapped between each cell to provide the different voltages on the breadboard.In that case, the circuit looks like this:

author: IanB
author_rank: Super Contributor
timestamp: September 19, 2015, 10:38:36 pm
content: Here's a picture of how to replicate the experiment: