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Calculate the EMF of the cell

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1. Homework Statement
Explain why the PD between the terminals of a cell is not always the same as its EMF.

A cell, a resistor and an ammeter of negligible resistance are connected in series and a current of 0.80 A is observed to flow when the resistor has a value of 2.00 Ω. When a resistor of 5.00 Ω is connected in parallel with the 2.00 Ω resistor, the ammeter reading is 1.00 A. Calculate the EMF of the cell.

Answer: 2.29 V.

2. The attempt at a solution
I did a quick illustration of the problem, as I see it:

b76418c95b69.jpg


So, on the first (I) graph we have V = I R = 0.8 * 2 = 1.6 V. Second (II): we find the resistance 1 / R = 1 / 2 + 1 / 5 → R = 1.43 Ω. So V = I R = 1 * 1.43 = 1.43 V.

I guess it has something to do with the cell and EMF differences. As far as I know, their difference is in the fact that PD is work done to move a charge between two points in a circuit and EMF is the total work done to move charge throught a complete circuit. And the formula is also different: V = I R, (R = total resistance), E = I (R + r), (R + r) = total external and internal resistance.

What am I missing?
 

gneill

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Remember another thread when you were looking into the behavior of a "real" diode that wasn't an ideal diode? What distinguished that "real" diode from an ideal one?
 
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Remember another thread when you were looking into the behavior of a "real" diode that wasn't an ideal diode? What distinguished that "real" diode from an ideal one?
An "ideal" diode has a resistor, right?
 

gneill

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An "ideal" diode has a resistor, right?
No, the opposite. An ideal diode has no resistance at all when it is conducting; That's what makes it ideal: It's a perfect "one-way-valve" for current. A real diode exhibits resistance because it's comprised of real-world, imperfect materials.

But you're on the right track now.
 
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No, the opposite. An ideal diode has no resistance at all when it is conducting; That's what makes it ideal: It's a perfect "one-way-valve" for current. A real diode exhibits resistance because it's comprised of real-world, imperfect materials.

But you're on the right track now.
Hm, but we don't have a diode here, do we? A cell is a basic battery and we have a resistor with an ammeter. Is my quick graph incorrect?
 

gneill

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Hm, but we don't have a diode here, do we? A cell is a basic battery and we have a resistor with an ammeter. Is my quick graph incorrect?
I'm trying to point out that real world components (all of them) have flaws that when examined closely make them non-ideal. Even batteries. Your diagrams are fine.

What you need to do is account for the possibility that the battery is not an ideal component.
 
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I'm trying to point out that real world components (all of them) have flaws that when examined closely make them non-ideal. Even batteries. Your diagrams are fine.

What you need to do is account for the possibility that the battery is not an ideal component.
Hm, I think I got your point. But what in this situation can have flaws? Regarding the ammeter we have this line:
an ammeter of negligible resistance
 

gneill

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Hm, I think I got your point. But what in this situation can have flaws? Regarding the ammeter we have this line:
So you can ignore the ammeter. What's left?
 
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So you can ignore the ammeter. What's left?
As cnh1995 pointed out it's the battery (and as you also suggested with your:
Even batteries.
).

The battery is non-ideal. You should consider its internal resistance.
So the PD of a cell is considered to be an ideal case while the EMF is not ideal. This is the case with:
Explain why the PD between the terminals of a cell is not always the same as its EMF.
?

---

E = I (R + r)

Let's find r, the internal resistance: 0.80 A (2.00 Ω + r) = 1.00 A (1.42857 Ω + r) → r = 0.857 Ω.

Plug it in: E = 0.80 A (2.00 Ω + 0.857 Ω) = 2.29 V.
 

cnh1995

Homework Helper
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So the PD of a cell is considered to be an ideal case while the EMF is not ideal
Emf is the open circuit terminal voltage of the battery. If Pd=Emf for any value of current, that would be an ideal case.
Let's find r, the internal resistance: 0.80 A (2.00 Ω + r) = 1.00 A (1.42857 Ω + r) → r = 0.857 Ω.

Plug it in: E = 0.80 A (2.00 Ω + 0.857 Ω) = 2.29 V.
Right.
 
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So the PD of a cell is considered to be an ideal case...
The voltage of a cell when no current is being drawn is sometimes called the Thevenin voltage. A cell can be conceptualized as an ideal* voltage source in series with a Thevenin resistance.

* Voltage doesn’t drop under load.
 
Last edited:

gneill

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upload_2016-10-2_12-50-43.png
 
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gneill

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The PD is what you would measure at the terminals of the battery under whatever load conditions there might be. It varies with the load (amount of current being drawn).

The EMF is the potential difference that is attributed to the underlying source of electric potential. Typically it is a chemical cell of some sort whose reactions separate charges to some fixed potential difference. It's what you would measure for the PD when there is no load (no current).
 
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So the PD is the EMF + internal resistance, while the EMF is just voltage?
PD = EMF + internal resistance does not pass dimensional analysis muster:
(voltage = voltage + resistance).
All terms in an equation must be the same physical quantity:
PD = EMF – (current x Thevenin resistance)
 

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