Current Electricity and Electrical InstrumentMedium

Question

In a potentiometer wire experiment the emf of a battery in the primary circuit is 20V and its internal resistances is 5Ω. There is a resistance box in series with the battery and the potentiometer wire, whose resistance can be varied from 120Ω to 170Ω. Resistance of the potentiometer wire is 75Ω. The following potential differences can be measured using this potentiometer.

Options

A.5 V
B.6 V
C.7 V
D.8 V

Solution


imin = A
imax =Amp
Potential = iminRPM = × 75 = 7.5V
Across potentiometer V = imaxRPM = ×  75 = 6V


 

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