Recent content by NoahCygnus

Yes, O should be positive as conventional current goes from O to A. If I do imagine a voltage source connected to O, could you elaborate how why it should provide zero current?

Thank you. You are very kind. Reading your last statement and Orodruin's comment on how I am treating point O like it has an emf, I had a realization. There's just a wire at point O, there's no potential drop or gain there like in a resistor or a cell whose potential drop/gain we include in the...

Pardon my lack of knowledge, I know EMF is the potential of a cell when current is not drawn from it but I don't quite understand how I am using EMF of point O instead of its potential value.

Oh yes, you used the loop law on AO, BO and CO instead of taking AOC, AOB and BOC like I did. I agree with that! But what's wrong with considering AOC, AOB, and BOC in my equations?

But I did consider ##V_O## in the first equation. I am sure I didn't miss it.

"The algebraic sum of changes in potential around any 'closed path' of an electric circuit is zero." I know the law mentions closed paths but hear me out. I have seen people use the law on open circuits treating the end points of the open path as closed. I will take an example where I have seen...

In this circuit, I have to find the potential at point O. I tried using Kirchhoff's voltage law for the three open loops AOC, AOB and BOC to arrive at potential at O. According to my calculations the potential at O should be 0, but that is not the case according to the source. So I must be using...

So I have been given a uniform electric field ##\vec{E}=20 V/m## in the direction as show in the image. I have been told to calculate the potential difference ##VC - VA##. According to the teacher (on YouTube) the potential difference ##VC - VA = -10\sqrt{2}V##. But I say it's ##-20 V## as...

cyclopenta-1,3-dineyl? 1,3? I don't get how you numbered the ring, you have to number the carbon of the ring attached to the main chain as 1'. See diagram. There is no double bond on 3' carbon of the ring.

Nomenclature rules state that the functional group has a higher priority than a ring or double bonds, so the main chain will be the one marked by blue. I screwed up in the first try, I didn't see that there was a propane chain. I redid the problem and came with the answer...

Well I have two possible answers, it's either 1-(cyclopent - 1', 4'- dienyl) methyl ethan-1-ol or 1-methyl - 1 - (cyclopent-1',4'-dienyl) ethan-1-ol. Unfortunately the answer wasn't provided in the textbook.

So just because it's not given in the problem I can't assume the particle undergoing motion is a car. All right.

Will decrease in magnitude in the negative direction. I guess it wasn't self explanatory that by saying "Velocity will decrease", I meant it will decrease it's magnitude in the negative direction and hit zero and then increase its magnitude again but the direction of the velocity vector will be...

That is what I said in my first post.