# B Potentiometer nulling voltages in a circuit

1. Jul 15, 2017

### esha

this is the potentiometer and the emf of cell E2 is unknown. now to know the emf just finding the null point of the galvanometer is enough. its because at that given length the potentiometer has the same potential as the cell. but why are they equal at null point?

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Last edited by a moderator: Jul 15, 2017
2. Jul 15, 2017

### tech99

They are pushing in opposite directions and balance out at the null point.

3. Jul 15, 2017

### esha

sorry but i dont understand the pushing idea

4. Jul 15, 2017

### Staff: Mentor

@esha -- Currently this looks like a general question, but it may still get moved to the schoolwork forums.

And here is your picture rightside-up, and cleaned up a bit.

5. Jul 15, 2017

### esha

thanks for the edit

6. Jul 15, 2017

### Staff: Mentor

No worries. Is the structure A-J-B supposed to be the potentiometer? If so, it sure is drawn in an unconventional way...

7. Jul 15, 2017

### esha

yea its the potentiometer

8. Jul 15, 2017

### Staff: Mentor

I'm not quite getting the question. The problem is that we do not know the relationship between the resistor in the upper circuit and the resistance of the potentiometer between ends A and B. or is that relationship given? That voltage divider will determine the null point for the potentiometer wiper, and without that relationship I don't think much can be said...

9. Jul 15, 2017

### esha

yea the value of the above resistor is given. but i dont understand why is that resistor required when the potentiometer wire itself has some resistance

10. Jul 15, 2017

### Staff: Mentor

It's not required, but it does complicate the problem. What is the relationship that is given?

11. Jul 15, 2017

### esha

ill be able to figure the value of the potential of the E2 cell... i just need to know why the potential difference AJ is equal to the cell E2 at null point

12. Jul 15, 2017

### Staff: Mentor

When you connect two batteries of the exact same potential in parallel, no current flows between them. That's because there is no potential difference to drive the current.

In the circuit above, start with the wiper not connected to the lower circuit. Using a voltmeter connected between A and J of the potentiometer, move the wiper J until you read the voltage E2 on the voltmeter. Now connect the galvanometer end of the lower circuit to J. This is the same situation as two parallel identical batteries. No current will flow through the bottom loop, because there is no potential difference to drive that extra current (so the galvanometer reads zero). All the voltage drop from A-J is caused by current in the upper loop still.

Hope that helps.

13. Jul 15, 2017

### esha

i got it but why is there no potential difference in the bottom loop? sorry if i m being too stupid

14. Jul 15, 2017

### Staff: Mentor

You're not being stupid. I'm having a conceptual problem with this as well, but I think it's because the circuit is non-physical with zero resistance in the bottom loop. An ideal voltage source and an ideal galvanometer each have zero internal resistance, so the problem is poorly formulated, IMO.

Instead, let's insert a small resistor in series with the galvanometer. Now you can write KCL equations for nodes A and J (make B ground), and you will be able to see how no current flows in the bottom loop when the voltage from A-J is equal to E2 (and the voltage drop across the bottom loop resistor is zero, since there is no current at that nulling point).

15. Jul 15, 2017

### esha

is this the kirchoff's voltage equation you r talking bout?

Last edited by a moderator: Jul 15, 2017
16. Jul 15, 2017

### esha

then if its correct E2 = I2R1.

17. Jul 15, 2017

### Staff: Mentor

Correct.

18. Jul 15, 2017

### esha

so the ultimate conclusion is 2E2 = I1r. this means at null point E2 = 0. okk... so thats y I2R1 must equal E2 at null point. m i right?

19. Jul 15, 2017

### Staff: Mentor

No, E2 is a voltage. The current through the bottom leg is zero at the null point.

20. Jul 15, 2017

### esha

yea but voltage can be altered depending upon the internal resistance.. so its zero at null point... or else current would have flown

21. Jul 15, 2017

### Staff: Mentor

No, at the null point of the potentiometer, E2 is the same voltage it always is, and I1 is zero.

22. Jul 15, 2017

### esha

okk i got it... thanks for the help... you helped me a looooooot today

23. Jul 15, 2017

### sophiecentaur

The null point is where there is no Potential Difference - that's why no current flows through the meter.

I think something very important about using a Potentiometer 'properly' may have been missed out here. The battery voltage ξ1 is not used when a potentiometer is used seriously because it is always supplying current. Three batteries (cells) are used for an accurate measurement of emf. The 'power supply battery' is just chosen as having loads of capacity so it will not change volts (sag) over the period of the measurements. You then compare the ξ2 from a battery under test and ξ3 from a reference battery, making sure that batteries 2 and 3 never draw more than a very small amount of current. This way, it doesn't matter what the power supply PD or its series resistor R are (as long as they are 'appropriate'). The ratio of the lengths of potentiometer wire (at balance) gives the ratio of ξ2 and ξ3 because neither battery is supplying any current.

24. Jul 15, 2017

### esha

but if the batteries are present what prevents them from supplying current?

25. Jul 16, 2017

### cnh1995

You find a point on the potentiometer wire such that the potential difference across that wire segment is equal to the emf of the unknown source. This is why there is zero potential difference across the meter as sophiecentaur said earlier.