Question

please help

Answer 1

http://screencast.com/t/jhfxLoQeXDHE

Answer 2

@theEric

Answer 3

I'm going to be going to bed in just a moment.

But here's this:

As the variable resistor resists more, it will start to drop the voltage over it. That means there is less for the voltage drop across the resistor with the voltmeter.
So, the p.d. will decrease.

Answer 4

Also, as you add more resistance, the current through all of it (is the same since it's series) and decreases.

Answer 5

So they both decrease....
Now I have to think where to go from there...

Answer 6

I guess use math... Find V as a function of I, then see what it looks like... maybe?

Answer 7

hmm

Answer 8

The voltage across the \(2\ \Omega\) resistor will have only a part of the voltage drop across it, according to the other resistor, as you've seen before!

Answer 9

If the variable resistor is at \(x\ \Omega\)...
Voltage is split proportionately in series (and not at all in parallel).
It makes sense, when you think about it for a bit...
\(V_{p.d.}=12{\rm V}\times \rm\dfrac{2\Omega}{(2+x)\Omega}\)
Well, that's the p.d... Maybe it'll help to look at current?
\(I=\dfrac {V_{source}} R =\dfrac{12\rm V}{(2+x)\ \Omega}\)
They both have a \(2+x\) in there... So, they have something in common... So I can solve for it now from the electrical current equation...
\(I =\dfrac{12\rm V}{(2+x)\ \Omega}\)
\(\implies(2+x)\ \Omega=\dfrac{12\rm V}{I}\)
And then use it to substitute it into the voltage equation...
\(\implies V_{p.d.}=12{\rm V}\times \rm\dfrac{2}{2+x}=12{\rm V}\times\dfrac{2\Omega\times I}{12\ \rm V}=2\Omega\times I\)
And so
\(V_{p.d.}=2\Omega\times I\)

There was probably a simpler, and maybe more conceptual way... But I think I'm too tired to understand it. Do you see what that last equation implies, though?

Answer 10

Oh.... I did all that to arrive at \(V=I\ R\). I'm my own killjoy!

Answer 11

thanks got it...so the answer shuld be either B or A

Answer 12

Well, it is know what happens when \(I=0\). So examine that point in A and B.

Answer 13

V=0?

Answer 14

Right!

Answer 15

but the correct answer is B o.O

Answer 16

I should've gotten that one faster... It's proof that I'm tired!
Oh, sorry.. You're right, it is B.... Because you will never hit 0V. There will always be some voltage across it, even when the variable resistor resists a lot. Sorry!

Answer 17

I'm sorry if you lost points because I mislead you. Do you understand why it is what it is now?

Answer 18

is that because the volt meter is connected at 2ohms resistor?

Answer 19

Right. The voltage across it can't be 0. There will always be a voltage across it, and some current through it, because it is always connected in a closed loop with the battery.

Answer 20

check this one
http://screencast.com/t/jhfxLoQeXDHE

Answer 21

That looks like the same thing...

Answer 22

nope the answer is wrong

Answer 23

It's not B? And did it ask the same question twice?

Answer 24

it is |D for the second question

Answer 25

The link is exactly the same.

Answer 26

Did you maybe not copy the new link, so, when you pasted, you pasted the old link?

Answer 27

yeah sorry my bad
http://screencast.com/t/u8YfTPLyA1P

Answer 28

Okay. So Look at this circuit as you have others.
When the resistance increases, the voltage drop across it increases since it's a larger portion of the resistance from positive terminal to negative terminal.
At the same time, current decreases, since there is more total resistance.
So you want to see voltage increasing as current is decreasing. Or the opposite - voltage decreasing as current is increasing (as if the resistance was lowered).

Now, the resistance CAN be turned down to zero, implying the voltage drop is also zero. So, you can see a p.d. of zero. At which point, the current will be at its maximum...

Answer 29

il brb go have ur sleep

Answer 30

Okay! I'm headed to bed! Good luck with any more assignments!