Potentiometer to determine an unknown EMF Watch

originaltitle
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I understand that E1/E2 = the ratio of the lengths of the resistor when the galvanometer reads 0 for each (because then all the EMF of the cell would be used up for the resistor), but I don't understand the purpose of the 'main' loop of the circuit.

Compare these 2 diagrams:
This one is the one we use:

If we eliminated the upper loop, we'd get this:

In this, the current would flow fine (because it's a closed circuit).

Would anybody please explain to me why the upper loop is not eliminated (obviously I'm missing out some piece of info)?
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Stonebridge
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The upper loop provides a constant pd across the wire AB.
The sliding contact J taps off some fraction of this pd. For example, if J was, say two thirds of the way along the wire, you would have a pd equal to two thirds of that across AB.
When you find the point at which there is no current in the galvo, G, you have balanced the emf of the lower cell exactly against the pd across the wire from A to J.
Without the top loop there is nothing to balance against and there would always be a current in the galvo. The whole point of the potentiometer setup is that you arrange that the current in G is zero.
So if two cells balance at the same point on the wire AB they must have the same emf.
If the second cell balances at half the distance the first one did, its emf is half that of the 1st one.

If you know the emf of the 1st cell you can calculate the emf of the 2nd. That's the purpose of a potentiometer circuit.
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originaltitle
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(Original post by Stonebridge)
When you find the point at which there is no current in the galvo, G, you have balanced the emf of the lower cell exactly against the pd across the wire from A to J.
Thanks very much for your answer; I understand most of it now, but if you could just elaborate on this one point, I'd be very grateful indeed. What principle is involved in balancing the 2 EMFs? That current is inversely proportional to voltage? (Then if voltage is the maximum, the current is 0?)
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Stonebridge
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(Original post by originaltitle)
Thanks very much for your answer; I understand most of it now, but if you could just elaborate on this one point, I'd be very grateful indeed. What principle is involved in balancing the 2 EMFs? That current is inversely proportional to voltage? (Then if voltage is the maximum, the current is 0?)
The potentiometer circuit looks like this.
I've just replaced the wire AB with two resistors representing the two sections of the wire to the left and right of J



In the lower loop the emf of the cell E2 is trying to move current clockwise, as shown by the arrows. In the upper loop the cell E1 makes the point A positive, and the pd it creates across AJ tends to make current flow anticlockwise in the lower loop. If the two pds balance then no current flows in the lower loop.
If you have studied Kirchhoff's Laws you can show that the current is zero for particular values of the resistances and/or cell emfs.
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originaltitle
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Thanks very much.
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b.k
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(Original post by Stonebridge)
The potentiometer circuit looks like this.
I've just replaced the wire AB with two resistors representing the two sections of the wire to the left and right of J



In the lower loop the emf of the cell E2 is trying to move current clockwise, as shown by the arrows. In the upper loop the cell E1 makes the point A positive, and the pd it creates across AJ tends to make current flow anticlockwise in the lower loop. If the two pds balance then no current flows in the lower loop.
If you have studied Kirchhoff's Laws you can show that the current is zero for particular values of the resistances and/or cell emfs.
how can you explain interms of kirschoffs laws??
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Stonebridge
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You write down the 2 equations, one for the upper and one for the lower loop.
You put the current in the lower loop equal to zero.
You can then find the value of R (between A and J) that satisfies the equations.
The pd (from V=IR) for that resistor will be found to equal the emf of the cell E2
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b.k
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thnks....that what i thought but wanted to make sure
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