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Figure 6.13 shows a 2.0 V potentiometer ...

Figure `6.13` shows a `2.0 V` potentiometer used for the determination of internal resistance of a `1.5 V` cell. The balance point of the cell in open circuit is `76.3 cm`. Whan a resistor of `9.5 Omega` is used in the external circuit of the cell, the balance point shifts to `64.8 cm`, length of the potentiometer. Dentermine the internal resistance of the cell.

Text Solution

Verified by Experts

Here, `l_(1) = 76.3 cm`, `l_(2) = 64.8 cm`, `r =` ?, `r = 95 Omega`
Now, `r = ((l_(1) - l_(2))/(l_(2))) R = ((76.3 - 64.8)/(64.8)) xx 9.5= 1.68 Omega`
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