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

When slide is in the middle of the potentiometer wire, only half of the resistance of potentiometer wire `( = R_(0)//2)` will be between the points A and B. Hence effective resistance `(R_(1))` between A and B is
`1/R_(1) = 1/R + 1/(R_(0)//2) or R_(1) = (R_(0)R)/(R_(0) + 2R)`
Total resistance between A and C `= R_(1) + R_(0)/2`
Current through the potentiometer wire will be `I=V/(R_(1) + R_(0)//2) = 2V/(2 R_(1) +R_(0))`
The voltage `V_(1)` taken from the potentiometer will be the product of current I and resistance `R_(1)` ,
i.e., `V_(1) = IR_(1) = ((2V)/(2R_(1) + R_(0))) xx R_(1)`
`=(2V)/((2(R_(0)R)/(R_(0)+2R))+R_(0)) xx (R_(0)R)/(R_(0)+2R)`
` = ( 2 VR)/( 2R + R_(0) + 2R) = (2VR)/(R_(0) +4R)`