In the potentiometer circuit shown in figure, the internal resistance of the 6 V battery is 1 Ω 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 Ω )

Question

In the potentiometer circuit shown in figure, the internal resistance of the 6 V battery is 1 Ω 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 Ω ) (A) 0.7 V (B) 0.8 V (C) 0.9 V (D) 1.0 V

Tardigrade Admin · Accepted Answer

Current in the circuit due to 6 V battery is I=(6/1+5+2)=(3/4) A Now emf of cell C= potential difference across A D (3/4) × (2 × 60/100)=0.9 V

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

0.7 V

0.8 V

0.9 V

1.0 V

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 $A D$
$\frac{3}{4} \times \frac{2 \times 60}{100} = 0.9 V$

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