Data: L = 4 m,
R = 4 Ω,
E = 2 V,
r = 2 Ω
The required potential drop per unit length of the wire is
10-4 V/cm = \(\cfrac{10^{-3}V}{10^{-2}m}\) = 0.1 V/m
Let Rs be the series resistance for which the desired potential drop is obtained. The current in the circuit is
I = \(\cfrac E{R+r+R_s}\) = \(\cfrac 2{4+2+R_s}\)
The potential difference across the wire is
V = IR = \(\left(\cfrac2{6+R_s}\right)\) x 4 = \(\cfrac8{6+R_s}\)
The potential drop per metre of the wire
= \(\cfrac VL\) = \(\cfrac8{4(6+R_s)}\) = \(\cfrac2{6+R_s}\)
\(\therefore\) 0.1 = \(\cfrac2{6+R_s}\)
\(\therefore\) 6 + Rs = 20
\(\therefore\) Rs = 14 Ω
