Here, we have to calculate the internal resistence of the unknown cell.
So, we have to use the following formulae:
Internal resistance of the unknown cell is
r=(l2l1−1)R=(2.853−1)(9.5Ω)=0.5Ω
Hence the internal resistance will be 0.5 Ohm
A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0V and a negligible internal resistance. The potentiometer wire itself is 4m long. When the resistance, R, connected across the given cell, has values of.
(i) infinity (ii)9.5Ω
The 'balancing lengths, on the potentiometer wire are found to be 3m and 2.85m, respectively. The value of internal resistance of the cell is:
Held on 30 Apr 2014 · Verified 9 Jul 2026.
0.25Ω
9.5Ω
0.5Ω
0.75Ω
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