Question

In a potentiometer circuit. E = 2 V, r = 2 Ω, R_wire = 10 Ω, R_ext = 1988 Ω and L = 4 m (in the usual notation). What is the potential gradient along the wire ?

Answer 1

K = \(\cfrac{E\,R_{wire}}{(R_{ext}+R_{wire}+r)L}\) = \(\cfrac{2\times10}{(1988+10+2)\times4}\)

= \(\cfrac{5}{2000}\) = 2.5 x 10^-3 V/m

The potential gradient along the wire is 2.5 × 10^-3 V/m