A rsisitance of `R Omega ` draws current from a potentiometer. Te potentiometer has a total resistance `R _(0) Omega` A voltage V is supplited to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potenttometer.

Explanation

While the slide is in the middle of the potentiometer only half of its resistance`((R_(0))/(2))`will be between the points A and B. Hence, the total resistance between A and B, say, `R_(1)` will be given by the following expression:
`(I)/(R_(1))=(I)/(R)+(I)/((R_(0))/(2)), " "R_(1)=(R_(0)R)/(R_(0)+2R)`
The total resistance between A and C will be sum of resistance between A and B and B and C,i.e.,`R_(1)+(R_(0))/(2)`
`therefore`The current flowing through the potentiometer will be `I=(V)/(R_(1)+(R_(0))/(2))=(2V)/(2R_(1)+R_(0))`
The voltage `V_(1)`, taken from the potentiometer will be the product of current I and resistance `R_(1)`
`V_(1)=IR_(1)=((2V)/(2R_(1)+R_(0)))xxR_(1)`
Substitution for `R_(1)` we have `V_(1)=(2V)/(2((R_(0)xxR)/(R_(0)+2R))+R_(0))xx(R_(0)xxR)/(R_(0)+2R)=(2VR)/(2R+R_(0)+2R)orV_(1)=(2VR)/(R_(0)+4R)`