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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.

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The correct Answer is:
`((70)/(60)-1)xx9.5=(9.5)/(6)0hm`
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RESONANCE ENGLISH-CURRENT ELECTRICITY-Exercise-1 (Part-1)
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