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

Explanation

If the sliding contact is in the middle of AC, the wire resistance of AC will be `(R_(0))/(2)` and the circuit can be redrawn as below: Equivalent resistance across AC will be given by
`R_(AC) = ((R_(0))/(2) xx R)/((R_(0))/(2) + R) + (R_(0))/(2) = (R_(0) (R_(0) + 4R))/(2(R_(0) +2R))`
Current, `I= (V)/(R_(AC)) = (2V(R_(0) + 2R))/(R_(0)(R_(0) + 4R))`
Applying Kirchhoff.s loop law in the loop ABA, we get
`-(I-I.) = (R_(0))/(2) + I.R = 0`
`rArr I. = (IR_(0))/((R_(0) + 2R))`
Voltage across resistance `R+ I.R`
`= (IR_(0)R)/((R_(0) + 2R))= (2V (R_(0) + 2R))/(R_(0) (R_(0) + 4R)) xx (R_(0)R)/((R_(0)+2R))`
`:. V_(AB) = (2VR)/(R_(0) + 4R)`