in first case I1 = 40cm
\(\frac{R}{S}=\frac{l_1}{100-l_1}\) \(\Rightarrow \frac{R}{S}=\frac{40}{60}=\frac{2}{3}\) ....(i)
in second case when S and 30Ω are in parallel balancing length I2 = 50 cm, so

From (i) \(S=\frac{3}{2}R\)
Subtituing this value in (ii), we get


Also from equation (iii), S' = R
