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)
Options:
A .  0.7 V
B .  0.8 V
C .  0.9 V
D .  1 V
Answer: Option C : C 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=0.9V
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