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# Potential divider watch

1. MS:
http://www.scribd.com/mobile/doc/60324303?width=320
QP:
http://www.scribd.com/mobile/doc/59434847?width=1024

With question 15aii, WHY is the point of contact the potential divider supposed to be at the bottom for the voltmeter reading to equal to 0? Also how do you know which direction the current flows in a circuit?

Thanks so much!
2. Btw for part 15ciii, how does getting rid of component Q mean that resistance of P increases?
Thanks
3. (Original post by krisshP)
MS:
http://www.scribd.com/mobile/doc/60324303?width=320
QP:
http://www.scribd.com/mobile/doc/59434847?width=1024

With question 15aii, WHY is the point of contact the potential divider supposed to be at the bottom for the voltmeter reading to equal to 0? Also how do you know which direction the current flows in a circuit?

Thanks so much!
If the point of contact was at the top you would get the full pd across the divider. (=full pd of the cells on the left) At the bottom you "tap off" none of the divider and get none of the pd across it.

You don't always know the direction of current in a circuit until you do the calculations (Kirchhoff's Laws.) But as a general rule it goes from positive to negative in a conventional circuit diagram.
4. (Original post by krisshP)
Btw for part 15ciii, how does getting rid of component Q mean that resistance of P increases?
Thanks
It doesn't change the resistance of P, but it does change the combined resistance of P and Q in parallel. That's what increases.
If you have two resistors in parallel and remove one, the total combined resistance increases.
Try a quick calculation with 1/R = 1/R1 + 1/R2 and put any values you want for R1 and R2
5. (Original post by krisshP)
MS:
http://www.scribd.com/mobile/doc/60324303?width=320
QP:
http://www.scribd.com/mobile/doc/59434847?width=1024

With question 15aii, WHY is the point of contact the potential divider supposed to be at the bottom for the voltmeter reading to equal to 0? Also how do you know which direction the current flows in a circuit?

Thanks so much!
The voltage or potential difference around parallel circuits/loops is the same so when the contact is at the bottom the voltage of P would be equal to the that of the wire it is parallel with - which is 0.

Imagine it like this. When the circuit is as shown by the diagram. The current starts from the negative terminal of the battery (if there's only one battery don't need to worry about the relative strength of the emf) and somehow has to go to the positive terminal. It likes less resistance routes. It has only one route until it comes to the contact junction. Then the current gets shared proportionally to the resistance. And at the next junction it rejoins and reaches the positive terminal.

But when the contact is at the bottom. When it comes to the junction it has two choices, going through P or through the wire. Now the wire has 0 resistance. Therefore all the current goes through this and not P. Hence no V for P.
6. you're basically short circuiting the divider - by putting the contact at the bottom, you're measuring the pd across two points connected to earth, ie 0 volts.
7. (Original post by Stonebridge)
It doesn't change the resistance of P, but it does change the combined resistance of P and Q in parallel. That's what increases.
If you have two resistors in parallel and remove one, the total combined resistance increases.
Try a quick calculation with 1/R = 1/R1 + 1/R2 and put any values you want for R1 and R2
I tried a quick calculation. It works . Getting rid of the component Q means that the combined overall resistance increases. Therefore the charges have to do more work when encountering P. The voltmeter reading increases.

(Original post by Stonebridge)
If the point of contact was at the top you would get the full pd across the divider. (=full pd of the cells on the left) At the bottom you "tap off" none of the divider and get none of the pd across it.
I still can't understand this.
8. Look at this diagram.

Let's say the cells give 12V.
Then there is 12V across the component AB.
The point is that the contact P can be placed at various points on AB.
In my diagram its half way between the ends. If the voltage changes uniformly from 12 to zero across AB then at half way it would be 6V. This means there is 6V across the other resistor on the right and 6V across the voltmeter.
The nearer you place the contact point to B, the smaller number of volts you get from the component AB. At the top at A you would get all 12 volts.
At the bottom at B you would get zero volts.
By moving the contact point up and down you can vary the number of volts you get. More when you move it up and less when you move it down.
This is how a potentiometer works.

Potentiometer.
http://en.wikipedia.org/wiki/Potentiometer
9. (Original post by Stonebridge)
Look at this diagram.

Let's say the cells give 12V.
Then there is 12V across the component AB.
The point is that the contact P can be placed at various points on AB.
In my diagram its half way between the ends. If the voltage changes uniformly from 12 to zero across AB then at half way it would be 6V. This means there is 6V across the other resistor on the right and 6V across the voltmeter.
The nearer you place the contact point to B, the smaller number of volts you get from the component AB. At the top at A you would get all 12 volts.
At the bottom at B you would get zero volts.
By moving the contact point up and down you can vary the number of volts you get. More when you move it up and less when you move it down.
This is how a potentiometer works.

Potentiometer.
http://en.wikipedia.org/wiki/Potentiometer
So one end of the potentiometer gives you the maximum of 12V while the other end gives you the lowest of 0V. But how how do you know exactly which end would provide you with 0V?

THANKS
10. The voltage you are taking from the component is between P and B in the circuit on the right. The smaller the length PB the smaller the voltage. If PB is zero the pd is zero.
This should make it clear which end.
11. (Original post by Stonebridge)
The voltage you are taking from the component is between P and B in the circuit on the right. The smaller the length PB the smaller the voltage. If PB is zero the pd is zero.
This should make it clear which end.
Thanks

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