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. Derive an expression for the voltage fed into the circuit when the slide contact is in the middle of 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, say, `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.e., `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))`
Substituting for `R_(1)`, we have a
`V_(1)=(2V)/(((R_(0)xxR)/(R_(0)+2R))+R_(0))xx(R_(0)xxR)/(R_(0)+2R)`
`V_(1)=(2VR)/(2R+R_(0)+2R)`
or `V_(1)=(2VR)/(R_(0)+4R)`.