[Coda]

Originally Posted by Potironette

Voltage is either the work it takes or would take to move one coulomb from one place to another.

No, it's the work it would take to move the coulomb upstream, or the work the coulomb can do while moving downstream.

Imagine a weight. Pick it up off the ground; you put work into the weight to lift it against the gravitational field. You've charged the battery.

But if you tie the weight to a rope on a pulley and let go of it, then that weight can itself DO work to lift up something on the other end of the rope. How MUCH work can it do? Disregarding friction in the rope and pulley, EXACTLY the same amount of work it took to lift it up that high. You've discharged the battery and used it to do something.

Work is a convenient thing that relates force and distance to each other.

Work measures a change in energy. Applying a force over a distance is one way to get a change in energy, but it's not the only way -- you are now learning about another way using electricity. Heat is another way.

Hence, work in voltage is how much force and distance it takes to reach a certain point in space.

Voltage measures change in energy (that is, work) per unit charge. Equivalently, for each volt of potential, each coulomb of charge that passes through the circuit can do a joule of work.

[Potironette]

Does that mean that voltage is both the change in potential energy per every positive coulomb of charge and the change in work per every coulomb of charge, because even though work is what has been done and potential energy is what could be done, the work that could be done is equal to work done if the work was done?

I guess in the context of electric fields work is applying a field force over a distance?

[Coda]

Originally Posted by Potironette

Does that mean that voltage is both the change in potential energy per every positive coulomb of charge and the change in work per every coulomb of charge, because even though work is what has been done and potential energy is what could be done, the work that could be done is equal to work done if the work was done?

Yep, exactly, although that's a big statement to try to have in your head all at once. XD

I guess in the context of electric fields work is applying a field force over a distance?

Yes, but don't generalize that too far -- you can't expand W = Fd into W = (ma)d naively; you have to make sure that expansion makes sense in the context.

[Potironette]

I finally understand that there's a difference between potential energy and potential energy per coulomb! Thank-you for explaining things all those times.

I have a circuit diagram with a ground that looks like this in one homework sheet:

What does the ground mean?

[Coda]

We actually discussed that one previously while we were talking about other stuff. The ground is an infinite current sink! It can accept as much charge as you want to push at it.

But if you're working with numbers in a circuit diagram... what does that mean?

Well, it's one of the few times when you can deal with an absolute potential instead of a potential difference! The ground defines a zero potential in the circuit.

[Potironette]

I guess defining a zero potential would be useful when there's more than one battery..?
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What does it mean when when a current shows up negative using Kirchhoff's Laws? I heard something about how it means that the direction of current chosen was wrong in class, and after searching online it seems that it means the current is flowing the other way? But what does that mean?

More specifically, my homework was to find power dissipated through each resistor in this circuit:

It seems that switching the sign of the negative current in the equations works, but how can current across R_2 be going the other way..?

[Coda]

More than one battery, or when you want to make sure current flows in a specific direction, or as a voltage divider, or as a safety mechanism... there are a bunch of reasons.


I THINK that what you're seeing there is that the battery on the right is so much weaker than the one on the left that it's getting charged! The current is not only passing through R2 backwards, it's passing through V2 backwards!

And conveniently, you know what the voltage across the battery is: 1.0V! It's written right there on the paper. So it's acting like a resistor!

That said, I've never worked this kind of problem before and right now I'm not in a great position to do the research, so I can't follow up on this at the moment.

[Potironette]

So..a ground can end a circuit, I suppose?

Thanks for the response! I don't know where to envision the current so I'll ask my teacher tomorrow if I get the chance (if not then on Monday).

[Potironette]

Today I learned that you were right about the battery being charged!

How do you solve for sequences?
I have this homework problem to solve for the nth of a partial sum:

And I'm not sure what I'm supposed to be thinking. So far, I've just been trying to come up with random ideas for everything (except when things alternate from negative to positive, since we pretty much learned how to do that).

[Coda]

Welcome to your first step into calculus! This stuff you're doing right now is the fundamentals that eventually lead into integrals (on the continuous side) and recurrence relations (on the discrete side, which you probably WON'T do in high school).

I can't quite tell what you're trying to do there. Are you trying to find the sum of a_n for n = 1 to ___? If so then your work looks right, although you could have made life a little easier for yourself by simplifying the expression before summing it:

1/(n+1) - 1/(n+2) = ((n+2) - (n+1))/((n+1)(n+2)) = 1/((n+1)(n+2))
n=1 -> 1/(2*3) = 1/6 = 10/60
n=2 -> 1/(3*4) = 1/12 = 5/60
n=3 -> 1/(4*5) = 1/20 = 3/60
n=4 -> 1/(5*6) = 1/30 = 2/60
Σa_n from 1 to 4 = 20/60 = 1/3

Same results!

I'm not sure what S_n there means, though, unless that's supposed to represent the closed form of the sum? I know how to do that but it's... a much more advanced technique than what I would have guessed to you to use at this level. (PROVING that a given formula -- if you already know it -- is the closed form of a sum is much easier.)

[Potironette]

Doing it that way looks a lot cleaner!

Err, I don't know what the closed form of the sum means, but the exact question was: "Find the four partial sums and the nth partial sum of the sequence a_n." Then I was given a_n = 1/(n+1) - 1/(n+2).
I haven't learned techniques for anything complicated though.

I barely know what a "partial sum" is since class ended the moment it was introduced. The previous question had been "Find the four partial sums and the nth partial sum of the sequence a_n" and I was given a_n = 2/3_n. I sort of gave up on finding the partial sum, so I turned to the answer key (which only answers odd questions) and it said "2/3, 8/9, 26/27, 80/81; S_n = 1 - 1/3" so I assumed the "partial sum" was basically an equation for whatever the sequence of partial sums was like.

[Coda]

A closed form is a simple equation that directly describes any element of a sequence without having to reference any other element or evaluate a sum.

For example, the closed form of f(n) = 1 + 2 + ... + n is f(n) = n(n+1)/2.

That answer you got doesn't make sense... Why would it ever write "1 - 1/3" instead of just "2/3"? There has to be an n in there somewhere... Maybe it's "1 - (1/3)ⁿ"?

[Potironette]

Oh woops, I forgot to write the n, it is 1 - (1/3)ⁿ

For example, the closed form of f(n) = 1 + 2 + ... + n is f(n) = n(n+1)/2.

I think that the homework is asking for that? Except without a technique but just by somehow reasoning it..?

[Coda]

Well, the general technique is pretty difficult. XD I studied it my junior year of college, if I recall correctly.

BUT!

Let's take a look at something:

S_n = S_n-1 + 1/(n+1) - 1/(n+2)
S_n = (S_n-2 + 1/(n-1+1) - 1/(n-1+2)) + 1/(n+1) - 1/(n+2)
S_n = S_n-2 + 1/(n) - 1/(n+1) + 1/(n+1) - 1/(n+2)
S_n = S_n-2 + 1/(n) - 1/(n+2)
S_n = (S_n-3 + 1/(n-2+1) - 1/(n-2+2)) + 1/(n) - 1/(n+2)
S_n = S_n-3 + 1/(n-1) - 1/(n+2)

See a pattern there? That might be something you can work with.

[Potironette]

I didn't manage to get it in the end, except for a S_n = S_n + 1/(n+2) - 1/(n+2), which makes 0 = 0.

[Coda]

Huh. Okay, maybe I'm going to have to work that one myself then. XD