# Voltage and Current across a basic DC circuit

Joined Nov 29, 2018
40
Hi folks,

I believe I'm OK with respect to understanding what happens to current along a basic DC circuit and I believe I'm now OK with the voltage drop resulting from resistors but I'm struggling to understand what happens to voltage along the entirety of a basic DC circuit. I thought it best to include 2 example circuits with a reed switch included; one circuit has no resistors while the other has 1 5600 ohm 20% resistor. I've also included 4 different points on each circuit to represent where voltage & current measurement might take place. I understand from what I've read that the method of measurement can distort the circuit but for the sake of helping me understand voltage along the length of the circuit lets leave this distortion out.

Below is the diagram of the 2 circuits and further below i've listed out what I think would be observed at each of the 4 points on the circuits. I suspect that voltage drops proportionally over the length of the circuit since once it reaches the - here V should be 0...again if i've understood what i've read. What I am ultimately trying to accomplish is monitor the circuit and detect when the switch is opened and closed but before I dive back into that I want to clearly understand this basic concept. • Circuit A
• Open
• Position 1: V = 12.00, A = 0.000
• Position 2: V = 12.00, A = 0.000
• Position 3: V = 00.00, A = 0.000
• Position 4: V = 00.00, A = 0.000
• Closed
• Position 1: V = 12.00, A = 0.001
• Position 2: V = 12.00, A = 0.001
• Position 3: V = 11.58, A = 0.001
• Position 4: V = 11.58, A = 0.001
• Circuit B
• Open
• Position 1: V = 12.00, A = 0.001
• Position 2: V = 12.00, A = 0.001
• Position 3: V = 06.38, A = 0.001
• Position 4: V = 06.38, A = 0.001
• Closed
• Position 1: V = 12.00, A = 0.001
• Position 2: V = 12.00, A = 0.001
• Position 3: V = 11.58, A = 0.001
• Position 4: V = 11.58, A = 0.001
many thanks!

Sean

#### MrChips

Joined Oct 2, 2009
19,747
Everything is wrong in all of the two circuits.

When you close the switch, you are introducing zero resistance.
Ohm's Law dictates that I = V / R.
When R = 0, I = infinity.
Your measurements would not be realistic.

You need to rethink your experiment.

#### wayneh

Joined Sep 9, 2010
16,127
Looks basically correct but I don't like your source. It should either be 12V or 1mA, but it cannot be both. And are you saying there is 20Ω introduced by the switch itself, and only when closed?

A "real" power source has an impedance and that means its voltage will sag as the load increases. But to call your 12V supply also a 1mA supply means its voltage sags to zero when 1mA of current flows. I don't think that's what you meant.

#### MrChips

Joined Oct 2, 2009
19,747
Looks basically correct but I don't like your source. It should either be 12V or 1mA, but it cannot be both. And are you saying there is 20Ω introduced by the switch itself, and only when closed?

What is the voltage output of an ideal constant 12VDC supply when the supply is shorted?

Joined Nov 29, 2018
40
thanks guys. I may have a misunderstanding with respect to defining the power source. I did note that the cable and the switch collectively have 20 ohms of resistance, this is what i measured on a 500' roll of 22 awg stranded wire.

If I change the power source in my circuits to 12V then amps would be determined via the I = V/R equation rather than something i am specifying so;

For Circuit A
• Open: V = 12, I = 0
• Closed: V = 12, I = .6
For Circuit B
• Open: V = 12, I = .002 (ie: 2mA)
• Closed: V = 12, I = .6 (ie: 600mA)
Correct so far?

To be correct should I also include a ground in the circuit diagram? I was using only the basic representation i've seen for basic DC circuit diagrams, most of which just loop from com to neg.

#### MrChips

Joined Oct 2, 2009
19,747
If you are trying to simulate the circuit then include a 20-ohm resistor in series with the 12V supply.
That would make a lot more sense.

#### BobTPH

Joined Jun 5, 2013
2,108
I think what you want is a resistor in series with the power source then going through the reed switch. Connect one side of the resistor to the supply +, the other side to the reed switch and the other side of the reed switch to the supply minus. Now see if you can figure out the voltage at the junction of the resistor and the reed switch with the switch open and closed.

Bob

#### wayneh

Joined Sep 9, 2010
16,127
To be correct should I also include a ground in the circuit diagram? I was using only the basic representation i've seen for basic DC circuit diagrams, most of which just loop from com to neg.
Your circuit does not require a ground connection. You could ground either pole of the supply (but not both), and there would be no change in function.

All voltages are relative, and using the lowest voltage as your reference or zero point is fine. If you chose instead to ground your positive pole, all your circuit voltages would be negative relative to ground.

Joined Nov 29, 2018
40
I think what you want is a resistor in series with the power source then going through the reed switch. Connect one side of the resistor to the supply +, the other side to the reed switch and the other side of the reed switch to the supply minus. Now see if you can figure out the voltage at the junction of the resistor and the reed switch with the switch open and closed.

Bob
Hi Bob - is what you're describing what i've read as high side voltage monitoring?

I've added a 10-ohm 20% resistor (don't have a 20) inline with the supply and the supply side of the resistor. I find that for Circuit A (no resistor at switch) I read 116mV when the switch is closed and 0 when the switch is open while for Circuit B (5600 20% resistor at switch) I read 116mV when the switch is closed and 21.8mV when the switch is open. For this I have the V (red) from my multimeter connected to the high side of this 10-ohm resistor and the COM (black) connected to the low side of it

While I don't expect anyone to diag based on it I thought I'd include the setup on my "bench" which shows the bench power supply, breadboard and the 500' spool of 22AWG i'm testing with and collectively the "circuit". Red is +, black is - except for the crimp cables i'm using between my multimeter and the board, the crimp cables between the board and the spool and the white jumper cable on the board which serves as my "switch" allowing me to open/close. Bench could use some cleaning up, i need a little more space. Joined Nov 29, 2018
40
If you are trying to simulate the circuit then include a 20-ohm resistor in series with the 12V supply.
That would make a lot more sense.
Thanks MrChips - I have added a 10-ohm on the high side between my bench power supply and the reed switch. Is this added specifically so that there is some resistance on the circuit even if the switch is open?

#### MrChips

Joined Oct 2, 2009
19,747
Thanks MrChips - I have added a 10-ohm on the high side between my bench power supply and the reed switch. Is this added specifically so that there is some resistance on the circuit even if the switch is open?
No. The resistance is to prevent a dead short across the power source when the switch is closed, in simulation or real life.

1000 feet of 22AWG is 16 ohms.
16 ohms plus 10-ohm resistor make little difference when there is 5600-ohm resistor in the circuit.

(B) Current in 5626-ohm circuit with 12V source = 12/5626 = 0.002A.
Voltage across 10Ω resistor = 10 x 0.002A = 0.02V = 20mV (close to your 16mV).

(A) Current in 10Ω resistor = 0.116/10 = 0.0116A
Total resistance in circuit = 12/0.0116 = 1034Ω (should have been 26Ω)
Something is wrong with your setup and/or measurements.

Edit:
Your PSU is showing 0.3V output, not 12V.
Do not set the PSU current limit to 0.001A. Set the current limit to at least 1A.
You cannot set the power source to V and A at the same time. This was already pointed out in a previous post (post #3). Ohm's Law applies.

Joined Nov 29, 2018
40
Thanks Mr. Chips. The PSU shows 12V when the circuit isn't closed, I'll adjust the A side. I'm a network, SW & systems guy...electronics is a new discipline to me.

The circuits that I'll eventually monitor are pre-existing. Some of them have no resistor, some have a resistor bridging the reed switch and I suspect some will have a resistor bridging the reed switch and a resistor on the supply side near the reed switch which is done to detect when someone try's to bypass the system by shorting the wires near the switch. The resistors in use are 1KΩ, 2KΩ & 5.6KΩ; if a system uses 1KΩ it uses only 1KΩ, same with 2KΩ and 5.6KΩ so it's possible that a single circuit could have as much as 11.2KΩ +/- what i expect is 20% tolerance.

I'm just working through the best way to go about monitoring them and as part of that figuring out these basic circuit parameters.

Greatly appreciate the response and feedback.

#### MrChips

Joined Oct 2, 2009
19,747
Thanks Mr. Chips. The PSU shows 12V when the circuit isn't closed, I'll adjust the A side. I'm a network, SW & systems guy...electronics is a new discipline to me.

The circuits that I'll eventually monitor are pre-existing. Some of them have no resistor, some have a resistor bridging the reed switch and I suspect some will have a resistor bridging the reed switch and a resistor on the supply side near the reed switch which is done to detect when someone try's to bypass the system by shorting the wires near the switch. The resistors in use are 1KΩ, 2KΩ & 5.6KΩ; if a system uses 1KΩ it uses only 1KΩ, same with 2KΩ and 5.6KΩ so it's possible that a single circuit could have as much as 11.2KΩ +/- what i expect is 20% tolerance.

I'm just working through the best way to go about monitoring them and as part of that figuring out these basic circuit parameters.

Greatly appreciate the response and feedback.
Ok. You need to learn how to work with Ohm's Law.

I = V / R
V = I x R
R = V / I

There are three variables. You can fix any two. You have no choice on the third.
If you set V and R (as in your situation) you have no control over I.
Hence you cannot control both V and I on your power supply.
V and I settings are limits. That is the power supply will not exceed those limits.
If the V limit is reached first, then the PSU is in constant voltage mode.
If the I limit is reached first, then the PSU is in constant current mode and will not reach the V limit.
This is why the PSU is showing 0.3V. You have reached the current limit on the PSU.

Set the PSU to 12V. Now short the supply with a piece of wire. Increase the current setting until it reads 1A. Now your PSU is set up for 12V max and 1A max (assuming your PSU can handle that range).

Joined Nov 29, 2018
40
Yep, have been going over that one again and again. The feedback I've received here over the last week has helped a great deal; your guidance implanted that I will be derived from the R along the series circuit rather than something I supply. A few days ago calculating the V drop across the resistors was tripping me up but I think I'm starting to get it. Before long Ohms law & wheel as well as how to actually apply it properly will be permanently seared into my brain the same way the OSI model, a myriad of protocols and standards and countless other things are.

Once I have this piece down then I'll need to circle back to how I actually implement the circuit monitoring.

Thanks for the guidance on the PSU.

Joined Nov 29, 2018
40
One question I had that I don't think I made clear enough in my original post is how does voltage dissipate over the length of a circuit such as the ones i've drawn here. I get that the voltage will be reduced via resistance (resistors, load, cable itself). My question is does voltage dissipate linearly over the length of a segment of cable? It seems as I've been measuring at different points on the low side I see decreasing amounts of voltage the farther i get from the load.

thx

#### crutschow

Joined Mar 14, 2008
23,764
My question is does voltage dissipate linearly over the length of a segment of cable?
Yes.
You can look at it as a an infinite number of infinitely small resistances connected in series.

It's also the way a potentiometer varies the voltage by sliding a wiper across a length of resistive material that has a voltage applied across it.

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