A potential difference of 220 V is maintained across a 12000 ohm rheostat AB. The voltmeter V has a resistance of 6000 ohm and point C is at one fourth of the distance from A to B. What is the reading in the voltmeter ?

Explanation

As in case of linear rheostat
`RpropL,(R_(AC))/(R_(AB))=(AC)/(AB)`
Here, as `R_(AB)=12000Omega` and `AC=1/4AB`
So, `R_(AC)=12000xx1/4=3000Omega`
Now as the resistance `R_(AC) ( = 3000 Omega)` is in parallel with voltmeter of resistance of `6000Omega`. So the effective resistance between points A and C will be
`R_(AC).=(3000xx6000)/((3000+6000))=2000Omega`
Now as `R_(BC)=(R_(AB)-R_(AC))`
`=12000-3000)=9000Omega`
is in series with `R_(AC). (= 2000Omega)` and in series potential divides in proportion to resistance, potential difference across AC,
i.e., voltmeter reading will be,
`V_(AC)=(R_(AC).V_(AB))/((R_(BC)+R_(AC).))=(2000)/((9000+2000))xx220=40V`