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 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 betweeen 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) + 2 R )`
The total resistance between A and C will be the sum of resistance between A and B,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 ) = ( 2 V)/( 2 R _(1) + R _(0))`
The voltage `V_(1)` taken from the potentiometer will be product of current I and resistance `R_(1),`
`V _(1)=IR _(1) = ((2 V)/( 2 R_(1) + R _(0))) xx R _(1)`
Substituting for `R _(1),` we have a
`V _(1) = ( 2V )/( (( R _(0) xx R )/( R _() + 2 R))xx R _(0))xx (R _(0) xx R )/( R _(0) + 2 R)`
` V _(1) = ( 2 VR)/( 2 R + R _(0) + 2 R)`
` or V _(1) = ( 2 VR)/(R_(0) + 4 R)`