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schematic

simulate this circuit – Schematic created using CircuitLab

Hi. I'm testing a very simple circuit with a multimeter. In Ohm's Law, a resistor should affect the voltage, but in my circuit it doesn't happen.

I'm using a Raspberry and on PIN 1 I have 3.3v at 0.076A. What I've did is very simple, I've connected a Resistor to my Raspberry voltage source, and I've connected the other side of the resistor to the multimeter and the multimeter to Raspberry GND pin.

By measuring the voltage I've obtained same result as circuit without resistor (3.3v) and by measuring the current I've obtained 0.026A.

Why?

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  • \$\begingroup\$ Because some voltage regulators are better than others. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 18 '17 at 6:02
  • \$\begingroup\$ That is strange, 3.3 V over 82 ohms should give 40 mA, not 26mA, unless there is a 1.17V voltage drop over the 3.3V's internal resistance. Check your measurements. \$\endgroup\$ – Bart Aug 18 '17 at 8:18
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The Raspberry Pi has an onboard voltage regulator. This will maintain 3.3 V on its output when the load draws from 0 mA up to its rated output. e.g., A 100 mA regulator will maintain the voltage at 3.3 V from 0 mA to 100 mA but above 100 mA you can expect to see the voltage droop as the regulator will enter current limiting mode.

I have 3.3v at 0.076A.

If you edit your question to explain how you got 0.076 A we can explain further. If you just connected an ammeter between 3.3 V and GND then that is the short-circuit current - the maximum that it will supply but note that you had close to zero volts available then and the Pi would have shut down.

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  • \$\begingroup\$ I've tried to do the same thing by using an AA battery and it's the same (3.3v in output). How can I change the frequency? I did that because I need to run an old PC fan that draws 0.5v at 0.35A. \$\endgroup\$ – Syncro Aug 18 '17 at 6:31
  • \$\begingroup\$ AA batteries are 1.5 V nominal so there's something wrong with your measurement. PC fans don't run at 0.5 V - usually 12 V. Batteries output DC - direct current - with a frequence of zero. \$\endgroup\$ – Transistor Aug 18 '17 at 6:33
  • \$\begingroup\$ In measuring voltage between + and - of my battery I've obtained 1.5V, so I've placed a resistor between + and - and I still obtaining 1.5V with different current. \$\endgroup\$ – Syncro Aug 18 '17 at 6:35
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    \$\begingroup\$ If you can find a second multimeter and measure the voltage and current simultaneously you will find that the voltage drops when you draw current. The more current you draw the more the voltage drops. There is some internal resistance in your battery. \$\endgroup\$ – Transistor Aug 18 '17 at 6:37
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I'm using a Raspberry and on PIN 1 I have 3.3v at 0.076A. What I've did is very simple, I've connected a Resistor to my Raspberry voltage source, and I've connected the other side of the resistor to the multimeter and the multimeter to Raspberry GND pin.

So you have this:

schematic

simulate this circuit – Schematic created using CircuitLab

That is not how you measure a voltage, I mean you can, but usually you place the voltmeter in parallel to a component you want to measure the voltage across. What your approach does is increase your measurement error for measuring the voltage of Pin 1. With the configuration you used not by much.

The voltmeter is basically a big resistor (in the order of 1 MegaOhm or more), so in this configuration you have almost no current flowing. If you just switch to measuring ampere, the amperemeter is a very low resistance, so your circuit changes completely between the two modes.

So you might measure the voltage like this:

schematic

simulate this circuit

Now this will not teach you anything about Ohm's law in a useful way, just like your way didn't. Here everything is in parallel and the resistor doesn't have an influence on the voltage because the whole voltage will drop across the resistor. (in an ideal world, in the real world you will get a small drop of voltage because of the internal resistance inside your voltage source when you connect the resistor)

To study Ohm's law I suggest something like this:

schematic

simulate this circuit

Because now, the R3 will have an measurable impact on the voltage over R2. And you can change the values of R2 and R3 easily to check if your calculations are correct.

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  • \$\begingroup\$ I've tried to measure voltage following your instructions, but in your second circuit I still obtain 3.3 volt, while in third I obtain 3.14 for the second resistor and 0.15 for the first resistor, can you explain me why this happens? \$\endgroup\$ – Syncro Aug 18 '17 at 12:33
  • \$\begingroup\$ That you measure 3.3 V in the second circuit is completely expected. You measure in parallel to the voltage source, so you measure just what that source is giving you, the resistor has no influence. In the third, if you'd use two 82 Ohm resistors, you should end up with 1.65 V over each resistor. Seems like the first resistor is a smaller one. So you measure 3.14 V over 82 Ohm, following Ohm's law you have a current of 3.14 V / 82 Ohm = 38.3 mA flowing in your circuit. Again using Ohm's law I conclude that the first resistor is 0.15 V / 0.0383 A = 3.92 Ohm. You also note that 3.14 V + 0.15 V \$\endgroup\$ – Arsenal Aug 18 '17 at 12:48
  • \$\begingroup\$ ... is 3.3 V, so the total what your pin is providing. This is a basic circuit known as voltage divider, because it divides the input voltages according to the resistances. \$\endgroup\$ – Arsenal Aug 18 '17 at 12:50
  • \$\begingroup\$ How you calculated 1.65V? \$\endgroup\$ – Syncro Aug 18 '17 at 12:54
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    \$\begingroup\$ @Syncro well in the meantime I know that by heart, but you could go like this: you have 3.3 V and two 82 Ohm resistors in series, so their total resistance is 164 Ohm. So the current in the circuit is I = U / R = 3.3 V / 164 Ohm = 20.1 mA. And over one resistor this current will produce a voltage of: U = R * I = 82 Ohm * 20.1 mA = 1.6482 V (=1.65 if you do not round). \$\endgroup\$ – Arsenal Aug 18 '17 at 13:01
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Because your voltmeter has an almost infinite resistance (1Meg-10Meg, see the manual for your voltmeter), so when you measured the voltage you got a divider - R1+Rvoltmeter in series. The current is very tiny but non zero I = 3.3/(R1+Rvoltmeter), and the voltage drop across R1: VR1 = I * R1 is almost 0V on Rvolt: V(Rvolt)= 3.3-V(R1) == 3.3V - I*Rvolt.

And the numbers you see on the voltmeter is actually the V(Rvolt) i.e. ~3.3V

schematic

simulate this circuit – Schematic created using CircuitLab

The same situation is when you measure the current, your ampermeter has the almost zero resistance (1R--0.01R). So there is two resistance in series R1+Ramp. And the current is I = 3.3/(R1+Ramp) ~ 3.3/R1

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