Question

In the secondary circuit of a potentiometer a cell of internal resistance 1.5$\Omega$ gives balancing length of 52 cm. To get a balancing length of 40 cm, how much resistance is to be connected across the cell?

• A. 5Ω
• B. 6 Ω
• C. 7 Ω
• D. 4 Ω

Answer 1

The correct option is A

5Ω

$r=1,5\Omega ,I_1=52cm,I_2=40cm\frac{r}{R}=\frac{I_1-I_2}{I_2}\text{Resistance R}=\frac{r{I}_2}{(I_1-I_2)}=\frac{1.5\times 40}{(52-40)}=\frac{1.5\times 40}{12}=5\Omega$