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A resistance of Romega draws current from a potentiometer. The potentio - meter has a total resistance R_(0)Omega (Fig). 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 isin 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 between A and B, say, `R_(1)`, will be given by the FOLLOWING expression : `(1)/(R_(1))=(1)/(R)+(1)/((R_(0)//2))RARR 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 flowing through the potentiometer will be `I=(V)/(R_(1)+R_(0)//2)=(2V)/(2R_(1)+R_(0))` The voltage `V_(1)` taken from the potentiometer will be the p roduct of current I and resistance `R_(1)`, `_(1)=IR_(2)=((2V)/(2R_(1)+R_(0)))xxR_(1)` Substituting for `R_(1)`, we have a `V_(1)=(2V)/(2)((R_(0)xxR)/(R_(0)+2R))+R_(0)xx(R_(0)xxR)/(R_(0)+2R)` `V_(1)=(2VR)/(2R+R_(0)+2R) or V_(1)=(2VR)/(R_(0)+4R)` |
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