In this video by Khan academy, Mahesh Shenoy (The narrator) says around 7:14 of the video that if increase the length of the red wire we take, then the potential supplied by the battery increases (By increase length, means to move that contact of wire connected to galvanometer further to the right)

enter image description here

I couldn't understand why he said this, so I try to figure out on my own by labelling potentials on the circuit, assuming the negative terminal as 0V , we gain 9V when we pass through battery from negative to positive enter image description here

In this diagram it seems left end of the red wire (The blue dot) has highest potential and the PD should drop as we move to the right. However, this is inconsistent with the Mahesh's result, where have I made the mistake?

  • \$\begingroup\$ maybe try to translate your pretty picture in a schematic made of discrete resistors. The correct result should jump right at you. \$\endgroup\$ – Marcus Müller Feb 21 at 11:46
  • \$\begingroup\$ If you connect both lamp wires to the blue dot you will have zero potential difference across the lamp so it will turn off. Logically if you move them further away from each other there will be a greater potential difference. \$\endgroup\$ – HandyHowie Feb 21 at 11:56
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    \$\begingroup\$ I'll be honest, I think a video that looks like that is not a great introduction to linear electrical networks. Maybe read or watch someone who's not making money of Youtube preferring the most colorful, bright and flashy explanations... \$\endgroup\$ – Marcus Müller Feb 21 at 11:58
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    \$\begingroup\$ Hey, Buraian, maybe get one of the free video lectures from real universities. You can't go very wrong with MIT's lectures: Here's the intro course you should be watching: a link to the video lectures, and it even comes with exercises with which you can check whether you correctly understood. It's not harder than anywhere else, but it's correct, and you can say you've learned electronics at an MIT B.Sc. EE level. Not from "Khan academy" (which sounds like a scam, tbh). \$\endgroup\$ – Marcus Müller Feb 21 at 12:03
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    \$\begingroup\$ @MarcusMüller Khan Academy is usually a lot better than this, honestly. They're not a scam, I can at least tell you that. \$\endgroup\$ – Hearth Feb 21 at 13:01

In this diagram it seems left end of the red wire (The blue dot) has highest potential and the PD should drop as we move to the right.

That would be correct if you took battery negative as your reference or "ground" point. In the illustration the author isn't measuring voltage from a ground point, s/he's just measuring difference between two points and the battery positive is the reference.


simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) The author's circuit. (b) A more conventional circuit.

Imagine the resistors R1 and R2 as ladders with nine steps. The wiper is on the sixth step.

  • In (a) we're measuring the difference between the wiper and the top step so VM1 reads 9 - 6 = 3 V.
  • In (b) we're measuring the difference between ground and the wiper, 6 V.

The author is correct. The voltage reading will increase as the wiper / contact is moved to the right.

  • \$\begingroup\$ Ohh, it makes sense now, the more of the resistor wire we take, the larger the potential drop we will get Thank you!! \$\endgroup\$ – Buraian Feb 21 at 12:16

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