Sum
The emf of a cell is balanced by a length of 120 cm of a potentiometer wire. When the cell is shunted by a resistance of 10 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell.
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Solution
Data: R = 10 Ω, l1 =120 cm,l2 = 120 - 20 = 100 cm
r = `"R"(("l"_1 - "l"_2)/"l"_2)`
`= 10 ((120 - 100)/100)`
= 2 Ω
The internal resistance of the cell is 2 Ω.
Concept: Potentiometer
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