- #1
moenste
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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:
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?