1.

Circuit has been set up for finding the internal resistance of a given cell . The main battery used across the potentimeter wire itself is 4 m long . When the resistance R , connected across the given cell,ha =s values of(i) infinity , (ii)9.5Omegathe balancing lengths on the potentiometer wire are found to be 3 cm and 2.85 m respectively .The value of internal resistance of the cell is

Answer»

`0.25Omega`
`0.95Omega`
`0.5Omega`
`0.75Omega`

Solution :The INTERNAL resistance of the CELL ,
`R=((l_(1))/(l_(2))-1)R=9.5((3)/(2.85)-1)=05Omega`


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