Observing and exploring the relationship between current and potential with a USB multimeter

Question

Preface;

Hi, I bought this cheap little USB multimeter/double-adapter accessory on eBay.
Today's question will be focusing on the multimeter aspect of this device.

The white USB cable is for charging the black/red Android smart-phone.
Both of which are relevant to the question.

(Unfortunately the tip of the cable is slightly damaged and may not always be making the absolute best connection.
- Apologies, it was all I had on me at the time).

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If you're curious; the black USB cable is connected to a wireless network interface card.
Said interface is down, and doesn't seem to be actively consuming any resources.
It can probably be safely omitted from the scenario without affecting any other variables.

(It's main purpose here is simply to apply a small amount of physical pressure down on the white cable.
- Ensuring a decent connection, and holding it in place).

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TLDR;

There seems to be an inversely proportional relationship between;
the potential/voltage[V] and current/amperage[A].
(At the node being measured).

##In other words:##

 Min current[A] = Max potential[V];  
More current[A] = Less potential[V];

##And vice versa.##

I recall being told to think of it in terms of a metaphor:

• Where electromotive-force/potential-difference[V]; is like pressure/Pascals-per-square-inch; inside a hose/pipe.
• And Coulombs-per-second/electric-current[A]; is like the flow-of-water/Litres-per-second; through said hose/pipe.

I'm not actually sure how accurate this analogy is.
But it seems to help with creating a mental image and an understanding of the mechanism of action.

• (i.e. Block the tip of a running hose; water stops flowing; you can feel the pressure build up under your thumb; the hose becomes noticeably more tense.
• Unblock the tip of a running hose; water resumes flowing; you can feel the pressure release under your thumb; the hose becomes noticeably more relaxed.)

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The Question;

I'm sure this is a relatively elementary question for seasoned electrical/electronic engineers/technicians, but how accurate is this analogy?
What exactly is the relationship between current and potential (as observed below)?

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FYI: (I noticed that ~4.98V is approximately the maximum voltage produced by this Acer laptops buss.
Apple computers and phone-chargers (just for instance) were observed producing a more ideal maximum of ~5.11V).

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Fully 'open' circuit: Not charging: [4.98V/0.00A]

Fully 'closed' circuit: Charging: [4.89V/0.30A]

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Half 'open' circuit ('loose connection'): Charging ('or trying to'): [4.80V/0.14A]

Fully 'closed' circuit: Charging: [4.88V/0.27A]

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Half 'open' circuit: Not charging: [4.95V/0.00A]

Fully 'closed' circuit: Charging: [4.90V/0.35A]

Answer 1

This phenomenon is voltage drop due to internal resistance of the power supply and its wiring.

Somewhere in the laptop is a voltage regulator. It will adjust the current supplied until its output is at 5V. This is connected both to the internals of the laptop and the USB ports.

However, between the power supply and the voltage measurement point is some wiring, which appears to have a resistance of about 0.4 ohms.

Answer 2

For a resitive element the analogy works. Voltage and current are inversely proportional. A would be like your thumb in the example

But the charger isn't a resistive element. The charger is, ideally, a constant voltage source. It has active elements (transistors, ICs, etc) supported by resistors and capacitors to provide a constant voltage for some range of expected resistance (or inversely expected currents).

To keep with your analogy imagine a pressure regulator where the input is over pressured. Even if you turn on a few faucets, ideally the output side will stay constant since there's enough surplus of pressure on the input to keep up. This is like your voltage regulator

But if you turned on everything, the input to the regulator, in attempting to support demand, would dip below the regulators max value and then both the input and output would drop proportionally to each other. This would be equivalent to like when the lights dim when a compressor kicks on (loosly).

Answer 3

The analogy between water and electrical circuit is pretty accurate, provided that all assumptions are correct. However, to make any observations and derive meaningful conclusions, you need first to calibrate your "cheap USB multimeter". Second, your load (tablet or whatever) is not something simple, it is an intelligent consumer that can change the amount of current depending on the tablet state, internal battery charge level, etc. Therefore, if you have any confusing observations from this setup, don't confuse yourself, and totally disregard what you saw.