Question

The resistance of a potentiometer wire is
8 ohm 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 microV/mm?​

Answer 1

Answer:

Here, E = 1uV/mm = 10^-6/10^-3 = 10^-3 V/m

Again, Ewire = IRwire/lwire = I 8/8 = I = 2/(8+R) V/m

So, 2/(8+R) = 10^-3

=> R = 1992 ohm = 1.992 mohm