Question regarding resistors and potentiometer

[Sanosuke Sagara]

I have spend many hours trying to solve this question but still can't figure out the answer.I hope that somebody will help me with this question.Thanks for anybody that spend some time on this question.


I have my question and solution in the attachment followed.

 

[Sanosuke Sagara]

Please , I really need somebody to help me figure out this question and I thanks for anybody that spend some time on this question.

 

[roger]

Sanosuke

consider the 4 volt battery, the 1 ohm rheostat and the 2 ohm wire .

We are measuring the midpoint of the wire so we can assume R to be approx. 1 ohm x2

So far we have the 4 volt battery and a 2ohm resistor and a 1 ohm resistor all in series . ( add the individual resistances )

We are now measuring the pd across the 1 ohm resistor .
But the voltmeter has a series resistance of 20 ohm .
So calculate the total parallel resistance of the 20 ohm and 1 ohm resistance...
Now we just find the voltage drop across the combined resistance

The circuit is a potential divider so voltage drop is :

R / TOTAL resistance X V

(20/21) / (62/21) X 4 volts

= 1.29 volts

 

[Sanosuke Sagara]

Thanks for your help, Roger.With your help,I finally know how and why you used this way to solve this question.Thanks !