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Originally posted by MattPI can only assume that the formulas for potential dividers do not work in respect to internal resistance then? Otherwise it would have worked....
No, the answer is 1ohm.
The working goes like this:
when the switch is closed you have the cell in series with a load resistance, R = 4ohms. Model the internal resistance as a resistance, r, in series with the cell.
Now you have a voltage source, Vs, wired in series with an internal resistance, r, and a load resistance, R.
Using Kirchhoff's voltag ...[text shortened]... 3A.
Now r = (1.5 - 1.2)/0.3 = 1ohm.
I hope this is helpful, sorry if it isnt very clear.
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Originally posted by Bad wolfThe cell resistance is in series with the load so you have to use the potential differance between the off voltage and the on voltage, in this case off is 1.5 V and on is 1.2 V and the differance of course is 0.3V. That is the effective voltage across the internal resistance of the cell. BTW, unless they are throwing you a curve, the internal resistance of real world devices should be a LOT lower than the load resistance. The open voltage represents the unloaded voltage of the cell but the closed voltage is the load and internal resistance in series. Internal resistance is the ionic resistance of the individual cells.
I can only assume that the formulas for potential dividers do not work in respect to internal resistance then? Otherwise it would have worked....
I think when you say potential dividers, you mean two or more external resistors in series?
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Originally posted by sonhouseI think when you say potential dividers, you mean two or more external resistors in series?
The cell resistance is in series with the load so you have to use the potential differance between the off voltage and the on voltage, in this case off is 1.5 V and on is 1.2 V and the differance of course is 0.3V. That is the effective voltage across the internal resistance of the cell. BTW, unless they are throwing you a curve, the internal resistance of ...[text shortened]... ls.
I think when you say potential dividers, you mean two or more external resistors in series?
- I would assume. 😕
http://tinyurl.com/2k27tx
I worked the question with internal resistance as R1 and it didn't work, lol.
(Ionic resistance? 😞)
Originally posted by Bad wolfBad Wolf, I assume you know your circuit formulas? I always remember the system by RIVB which stands for Eq. Resistance, I (current), V (Voltage Drops), and B (Branch Currents). From here you can solve most circuit problems, Matt and Sonhouse have answered the problem for you though, so I don't think there's anything else I can offer.
[b]I think when you say potential dividers, you mean two or more external resistors in series?
- I would assume. 😕
http://tinyurl.com/2k27tx
I worked the question with internal resistance as R1 and it didn't work, lol.
(Ionic resistance? 😞)[/b]
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Originally posted by cmsMasterMost of them, they're not very easy to learn, on a mock exam some time ago, I did very poorly. 🙁
Bad Wolf, I assume you know your circuit formulas? I always remember the system by [b]RIVB which stands for Eq. Resistance, I (current), V (Voltage Drops), and B (Branch Currents). From here you can solve most circuit problems, Matt and Sonhouse have answered the problem for you though, so I don't think there's anything else I can offer.[/b]
I have the real test on this pretty soon, so I need to work on it...
You'll have to be less vague with RIVB, I am not sure how you would use it...
Originally posted by Bad wolfIts fairly easy to explain when you have a sit and think about it.
I just looked at my textbook, its in there alright, but I've not had anyone explain it to me, nor had any questions on it.
🙁
The tricky part is to model the cell as a voltage source in series with an internal resistance, r, and a load resistance, R.
You seem to have been taught that the p.d = EMF - lost volts.
Using ohm's law the p.d is given by IR and the lost volts are given by Ir. (As they are in series the same current will flow though both resistors.)
Therefore, IR = EMF -Ir.
So EMF - IR - Ir = 0.
But the EMF is Vs. So we get Vs - IR - Ir = 0 as given in your text book.
A perhaps more rigerous way to explain the problem is, as mentioned before, using Kirchhoff's voltage Law. Kirchhoff's voltage law states that the sum of electrical potential differences around a closed loop must be zero. This is straight forward to prove by considering energy conservation.
The upshot of this is that the voltage across the source must be equal and opposite to the sum of the voltage drops across the two resistors, making the total sum of the potential differences around the circuit zero. So we again arrive at Vs = IR + Ir, or Vs - Ir - IR = 0.
I hope this has explained where the equation comes from.
NB: the equation p.d = Vs/(1+(r/R)) can be derived by treating the system as a podential divider with the load as R2 in the standard formular for potential dividers. This relationship is useful when considering the power output into the load. It is easy to show, by differentiating with respect to R, that the maximum power in the load is achived when R = r. However this is very inefficent.
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Originally posted by MattPAs in the p.d. across the cells, is countered by p.d. lost across the resistors.
A perhaps more rigerous way to explain the problem is, as mentioned before, using Kirchhoff's voltage Law. Kirchhoff's voltage law states that the sum of electrical potential differences around a closed loop must be zero. This is straight forward to prove by considering energy conservation.
I see.
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Originally posted by MattPWell, AS, if you want to get specific. lol 🙂
Correct! 🙂 So what test are you studying for? A-levels? best of luck by the way, im sure ul do better then you think.
Maths, that's split into pure and mechanics, physics, accounts and government&politics.
Too much work...
🙁