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I want to utilize 2.5V, from a 5V source.

This is a diagram:

enter image description here

My question is, where do I draw the 2.5 volts from?

Is it from the wire I painted pink, or from the wire I painted green?

Or it doesn't matter since the voltage in the sides of a resistor is the same?

I have to make sure.

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    \$\begingroup\$ Green is ground, what voltage will be there? \$\endgroup\$
    – Eugene Sh.
    Oct 26, 2020 at 16:07
  • \$\begingroup\$ If voltage on each side of resistor is the same like you say, and if you expand that to both resistors, then your are saying 5V is 0V and that makes no sense. So the assumption is wrong, and you can use Ohm's law to solve the circuit. \$\endgroup\$
    – Justme
    Oct 26, 2020 at 16:37
  • \$\begingroup\$ Voltage is the potential difference between TWO nodes ! \$\endgroup\$
    – LvW
    Oct 26, 2020 at 16:53
  • \$\begingroup\$ How much current do you plan to draw from the voltage divider? What load do you intend to give the 2.5V to? \$\endgroup\$
    – JRE
    Oct 26, 2020 at 17:15

1 Answer 1

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. An adjustable potential divider and the fixed equivalent potential divider when the wiper is in the centre position.

In both cases the voltage at the bottom is 0 V, at the top it's 5 V and in the middle it's 2.5 V.

The difference with the potentiometer is that it's adjustable. Turn the wiper to the bottom and you get zero volts. Slide it up and you can get any voltage you want between 0 and 5 V. It adjusts the "potential" (voltage) between the wiper and the terminals, hence "potentiometer".

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  • \$\begingroup\$ So the pink line in my diagram is where i will get my power. Thank you! (BTW, 127k rep, 4k answers... You are an EE god, thanks for your contribution) \$\endgroup\$ Oct 26, 2020 at 22:57

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