I'm facing a problem I don't exactly know how to resolve.
I have this circuit
The 2k resistance is a potentiometer. Basically we have a wheatstone bridge. Here's the picture
The question is : how do you adjust the potentiometer so that the amp meter shows a 0 current.
Well, the answer should be that - and only with reasoning on how the resistances on the two vertical branches (the lower/upper part of the potentiometer) act as a tension divider, or at least that's how they did it in my solution book - the upper part (say X) and lower part (say Y) of the potentiometer should make it so that X/Y = 8,2/3.
Seriously I don't get it. I have tried methods using superposition principle (considering for example X in serial with Req(Y+3k), and 8,2 in serial with Req(Y+3k)), but I end up with X = 8,2k which is non sense since the potentiometer is 2k.
Can someone explain? I know since its an amp meter it's just like if it was not there. I don't get how you can use the tension divider law if the the middle of the two vertical branches are connected with a wire (since I consider the amp meter as a wire, "ideal amp meter").