1.

A battery of emf 6 V and negligible internal resistance is connected to the terminals of apotentiometer wire of length 4 m. The wire is of uniform cross-section and its resistance is 100 Omega . The difference of potential between two points separated by 40 cm on the wire will be

Answer»

0.4 V
0.6 V
1.0 V
2V

Solution :`V = (EPSI l)/(L) = (6 xx (40 CM))/((4M)) = (6 xx (40 cm))/((400 cm)) = 0.6 V`


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