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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."

https://www.youtube.com/watch?v=XxLKfAZrhbM&feature=youtu.be&t=13s

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
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Look at this schematic:

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.

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