Question

In the potentiometer circuit shown in figure, the internal resistance of the $6 \,V$ battery is $1\, \Omega$ and the length of the wire $A B$ is $100\, cm$. When $A D=60 \,cm$, the galvanometer shows no deflection. The emf of cell $C$ is (the resistance of wire $A B$ is $1 \,\Omega$ )

• A. 0.7 V
• B. 0.8 V
• C. 0.9 V
• D. 1.0 V

Answer 1

Answer: (C)

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