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Question
When a resistor of 5Ω is connected across the cell, its terminal potential difference is balanced by 150 cm of potentiometer wire and when a resistance of 10 Ω is connected across the cell, the terminal potential difference is balanced by 175 cm same potentiometer wire. Find the balancing length when the cell is in open circuit and the internal resistance of the cell.
Solution
`r= 5(l_1/150-1)..............(1)`
`r= 10(l_1/175-1)..............(2)`
`5(l_1/150-1)=10(l_1/175-1)`
`l_1/150-1=2l_1/175-2`
`-1+2= 2l_1/175 - l_1/150`
`1= l_1(2/175 -1/150)`
`=l_1((300-175)/(175xx150))`
`=l_1 125/(175xx150)`
`l_1=210 cm`
`r= R (L_1/L_2-1)`
`=5 (210/150-1)`
`r= 2Ω`
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