1
$\begingroup$

I've come across a past paper question which asks to describe the inside of a dial pot potentiometer. I've been reading through this page which describes it very well, but I still have a few unanswered questions that I can't seem to find the solutions to.

The linked page provides this diagram to show the inside of the variable resistor, and from what I understand, current flows through either the top or bottom terminal. Then as current travels along the coil of wire which makes the resistance track, it travels along the wiper as soon as it comes into contact with it, and then leaves through the central terminal and back into the rest of the circuit. Therefor the closer the wiper is to the terminal from which current entered, the lower the resistance, as current spends less time on the resistance track.

What I still have yet to understand is the purpose of the third terminal. From the second picture below it shows that the third terminal is not connected to the power supply, so is it connected to anything at all, or is it used as a means to trap electrons to increase the resistance.

Also the page describes how a fixed resistor is used inside the variable resistor to provide an upper cap on the resistance which the variable resistor can provide, but were would this component be in the first diagram.

enter image description here

Here's the question in the past paper: enter image description here

$\endgroup$
3
$\begingroup$

The third terminal is used when you have to divide the potential for two separate or sub circuits. So in that case you can use the middle terminal as the power supply, and connect the other two terminals to other circuits. This will only be useful when you have similar entities on each sub circuits so that varying potential across both is linearly dependant on the angle wiper makes. Also, there is no such thing as electrons capture going on here. What this is, is simple ohms law, where the resistance increases linearly with length. Hope that helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.