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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. |
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Answer» Solution :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)` |
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