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)
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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:
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
Was this answer helpful ?
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