Question

The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V battery are connected in series with it. What should be the resistance in the box, if it is desired to have a potential drop of 1 µV/mm?

Answer 1

Data: R = 8 Ω, L = 8 m, E = 2 V, K = 1 µV/mm

= 1 × \(\frac{10^{-6} \mathrm{~V}}{10^{-3} \mathrm{~m}}\) = 10^-3 \(\frac{\mathrm{V}}{\mathrm{m}}\)

K = \(\frac{V}{L}=\frac{E R}{\left(R+R_{\mathrm{B}}\right) L^{\prime}}\) where R_B is the resistance in the box.

∴ 10^-3 = \(\frac{2 \times 8}{\left(8+R_{B}\right) 8}\)

∴ 8 + R_B = \(\frac{2}{10^{-3}}\)

= 2 × 10³

∴ R_B = 2000 – 8

= 1992 Ω