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
• B. 0.8 V
• C. 0.9 V
• D. 1 V

Answer 1

The correct option is C

0.9 V

Current in the circuit due to 6 V battery is
$l=\frac{6}{1+5+2}=\frac{3}{4}A$
Now emf of cell C =potential difference across AD
$=\frac{3}{4}\times \frac{3\times 60}{100}=0.9V$