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A rsisitance of R Omega draws current f...

A rsisitance of `R Omega ` draws current from a potentiometer. Te potentiometer has a total resistance `R _(0) Omega` A voltage V is supplited to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potenttometer.

Text Solution

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While the silde 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 toal resictance 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`
`therefore` The current following through the potentiometer 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),`
`V _(1) = I R _(1) ((2V)/(2R_(1) + R _(0))) xx R_(1)`
`V _(1) = (2 V)/(2 ((R _(0) xx R)/( R _(0) + 2R )) + R) xx (R_(0) xx R)/( R _(0) +2R)`
`V _(1) = (2VR)/( 2R+ R_(0) + 2R)`
or ` V _(1) = (2VR)/( R _(0) + 4R)`
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