I'm exploring voltage dividers and am puzzled by the conclusion made by the YouTuber in this video:

"If you apply Ohm's law, you find that 1 V spread across two equal-value resistors gives you 0.5 V in the middle [. . .] and as you would expect, four equal-value resistors would give you 0.25 V, 0.5 V, and 0.75 V."


I'm trying to work this conclusion out mathematically to try and make sense of it, but am not having much luck. Could someone please explain?

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    \$\begingroup\$ You're much more likely to get an answer if we don't have to watch the video; give us more of his explanation and/or a diagram. Note that 1V across two equal-value resistors will certainly have a current flowing? \$\endgroup\$ – pjc50 Oct 5 '15 at 12:38
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    \$\begingroup\$ Because the actual value of the current doesn't matter. Voltage divider splits voltage in half regardless of the current that goes trough. \$\endgroup\$ – Dmitry Grigoryev Oct 5 '15 at 12:39
  • \$\begingroup\$ @pjc50 The link I provided starts at the exact point I am referencing (13 s into the video) and only lasts for a few seconds. And the quote before the link is the exact quote that I reference in the video. I only provided the link so you could all see the diagrams. \$\endgroup\$ – user88062 Oct 5 '15 at 12:46

Look at this schematic:


simulate this circuit – Schematic created using CircuitLab

Now, V over R1 is given by: V(R1) = I(R1) * R1.
And V over R2 is given by: V(R2) = I(R2) * R2.

Can you see that there is only one loop for the current? That means I(R1) needs to be equal to I(R2), or I(R2) === I(R1).

Thus the equations become:

V(R1) = I(R1) * R1
V(R2) = I(R1) * R2

Since also given that R1 === R2, you can also substitute those.

V(R1) = I(R1) * R1
V(R2) = I(R1) * R1

And presto, the equation for V(R1) and V(R2) are now exactly the same:

V(R1) = I(R1) * R1 = V(R2) --> V(R1) = V(R2)

Thus the voltages need to be the same, so the point at V2 will be exactly half the voltage at V1.

Let's let you try the same trick for 4 resistors, see if you got it.

  • \$\begingroup\$ Excellent answer! Exactly what I was looking for. Thank you very much. \$\endgroup\$ – user88062 Oct 5 '15 at 20:14

As Dmitry noted, there will be a current flowing, but its actual value is not important, as it disappears when you calculate the voltages.

Take 4 resistors in series, each of resistance R. The total resistance is 4*R, and the current is 1/4*R. The voltage across any one of the resistors is U = R.I = R * (1/4*R) = 1/4V. That last expression doesn't contain I (=1/4*R). (And, maybe more important, neither does it contain R.)


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