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Question
When a resistance of 12 ohm is connected across a cell, its terminal potential difference is balanced by 120 cm length of potentiometer wire. When the resistance of 18 ohm is connected across the same cell, the balancing length is 150 cm. Find the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.
Solution
Solution:
Given:
R1 = 12Ω, l2 = 120 cm,
R2= 18 Ω,` l_2^'` = 150 cm
To find:
Balancing length (l1),
Internal resistance (r)
Solution:
`r=R((l_1-l_2)/l_2)`
`r=12((l_1-120)/120)`.......................(1)
From second condition
`r=18((l_1-150)/150)`......................(2)
From equation 1 and 2
`12((l_1-120)/120)=18((l_1-150)/150)`
`5(l_1-120)=6(l_1-150)`
`l_1=300 cm`
Using equation (i) we get Internal resistance,
`r=R((l_1-l_2)/l_2)`
`r=12((300-120)/120)`
`r=18Omega`
i. The balancing length is 300 cm.
ii. The internal resistance of the cell is 18 Ω.
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