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Q.

A potentiometer has a wire of 100 cm length and its resistance is 10 ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2 V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40 cm length of potentiometer wire, the value of ‘E’ is

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a

0.8 V

b

0.4 V

c

0.08 V

d

0.16 V

answer is D.

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Detailed Solution

Length L=100 cm Resistance R=10 Ω Series resistance RS=40 Ω Battery of emf E=2 V Balancing length l=40 cm emf of the cell in secondary circuit,E1=(iρ)l E1=(ER+RS) RLl =(210+40) 10100×40 =2×10×4050×100 =850 ∴ E1=0.16 V

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