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Question

In the potentiometer circuit shown in figure the internal resistance of the 6 V battery is 1 ohm and the length of the wire AB is 100 cm. When AD = 60 cm the galvanometer shows no deflection. The emf of cell C is (the resistance of wire AB is 2 ohm)


A
0.7 V
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B
0.8 V
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C
0.9 V
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D
1 V
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Solution

The correct option is C 0.9 V

Current in the circuit due to 6 V battery is
l=61+5+2=34A
Now emf of cell C =potential difference across AD
=34× 3 × 60100=0.9 V


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