A resistance of `R Omega` draws current from a potentiometer. The potentiometer has a total resistance `R_(0) Omega`. A voltage V is supplied to the potentiometer. Dervie can expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

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, says `R_(1)` will be given by the following expression
`(1)/(R_(1)) = (1)/(R ) + (1)/((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., `R_(1) + R_(0) //2`
`:.` 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))) xx R_(1)`
Substituting for `R_(1)` we have a
`V_(1) = (2V)/(2((R_(0) xx R)/(R_(0) + 2R)) + R_(0)) xx (R_(0) x R)/(R_(0) + 2R)`
`V_(1) = (2VR)/(2R + R_(0) + 2R)`
(or) `V_(1) = (2VR)/(R_(0) + 4R)`