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Current Electricity

Question
CBSEENPH12039377

A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has a total resistance of Ro. 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 potentiometer wire.

Solution


When the slide is in the middle of the potentiometer, only half of its total resistance i.e. R0/2 (since resistance is directly proportional to length) will be between A and point of contact (C), say R1, will be given by the following expression.
1 over straight R subscript 1 space equals 1 over straight R space plus fraction numerator 1 over denominator straight R subscript 0 divided by 2 end fraction
straight R subscript 1 space equals space fraction numerator RR subscript 0 over denominator 2 straight R plus straight R subscript 0 end fraction

The total resistance between A and B will be sum of the resistance between A & C and C&B i.e R1+R0/2
Current flowing through the potentiometer will be
straight I space equals space fraction numerator straight V over denominator straight R subscript 1 plus begin display style straight R subscript straight o over 2 end style end fraction space equals space fraction numerator 2 straight V over denominator 2 straight R subscript 1 plus straight R subscript 0 end fraction

The voltage V1 taken from the potentiometer will be the product of current I and the resistance R1
straight V subscript 1 space equals space IR subscript 1 space equals space fraction numerator 2 straight V over denominator 2 straight R subscript 1 plus straight R subscript 0 end fraction straight x space straight R subscript 1 space equals space fraction numerator 2 straight V over denominator begin display style fraction numerator 2 RR subscript 0 over denominator 2 straight R plus straight R subscript 0 end fraction end style plus straight R subscript 0 end fraction space straight x fraction numerator RR subscript 0 over denominator 2 straight R plus straight R subscript 0 end fraction space equals space fraction numerator 2 VR over denominator straight R subscript 0 plus 4 straight R end fraction