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Figure 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 76.8 cm. when 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 wire. Determine the internal resistance of the cell. |
Answer» SOLUTION : `R=((l_(o)-l_(E))/(l_(C)))R=(76.3-64.8)/(64.8)xx9.5` `r=1.68Omega` |
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