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A rsisitance of R Omegadraws 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. |
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Answer» Solution :Resistance of WIRE AC of POTENTIOMETER= `R_(0)` Becaue point B is the midpoint of wire AC, resistance of wire AB = `(R_(0))/(2)` If equivalent resistance between points A and B is `R_(1)` then, `(1)/(R_(1)) = (1)/(R) + (1)/((R_(0))/(2))` `therefore (1)/(R_(1)) = (1)/(R) + (2)/(R_(0))` `therefore (1)/(R_(1)) = (R_(0) + 2R)/(R R_(0)) ` ` therefore R_(1) = (R R_(0))/(R_(0) + 2R) "" ` .... (1) Here resistance of wire AC = (resistance of wire AB ) + (resistance of wire BC ) `therefore R. = R_(1) + (R_(0))/(2) = (2R_(1) + R_(0))/(2) "" ` .... (2) Current PASSING through potentiometer wire, I`= (V)/(R.) = (2V)/(2 R_(1) + R_(0)) "" ` [From equation (2) ] Potential difference between points A and B, `V_(1) = I xx R_(1) ` `= ((2V)/(2R_(1) + R_(0))) xx R_(1)` = `(2v)/(2 ((R R_(0))/(R_(0) + 2R) ) + R_(0) ) xx (R R_(0))/(R_(0) + 2R) ` `therefore V_(1) = (2V)/(((2R R_(0) + R_(0)^(2) + 2 R R_(0))/(R_(0) + 2R))) xx (R R_(0))/(R_(0) + 2R)` `= (2V)/(R_(0) (R_(0) + 4R) ) xx R R_(0)` `therefore V_(1) = (2VR)/(R_(0) + 4R)` 
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