(a) Writing two loop equations
(b) Calculation of currents through $40\mathrm{\Omega }$ and $20\mathrm{\Omega }$ resistors

In loop $\mathit{A}\mathit{B}\mathit{C}\mathit{D}\mathit{A}$
$+80-20{\mathit{I}}_2+40{\mathit{I}}_1=0$
$4={\mathit{I}}_2-2{\mathit{I}}_1$
In loop FCDEF
$-40{\mathit{I}}_1-10\left({\mathit{I}}_1+{\mathit{I}}_2\right)+40=0$
$-50{\mathit{I}}_1-10{\mathit{I}}_2+40=0$
$5{\mathit{I}}_1+{\mathit{I}}_2=4$
Solving these two equations
${\mathit{I}}_1=0\mathrm{A}$
${\mathit{I}}_2=4\mathrm{A}$
OR
End error, overcoming
Formula for metre bridge
Calculation of value of $\mathit{S}$
The end error, in a metre bridge, is the error arising due to
(i) Ends of the wire not coinciding with the $0\mathrm{cm}∕100\mathrm{cm}$ marks on the metre scale.
(ii) Presence of contact resistance at the joints of the meter bridge wire with the metallic strips.
It can be reduced/overcome by finding balance length with two interchanged positions of $\mathit{R}$ and $\mathit{S}$ and taking the average value of ' $\mathit{S}$ ' from two readings.
For a metre bridge
$\frac{\mathit{R}}{\mathit{S}}=\frac{\mathit{l}}{100-\mathit{l}}$
For the two given conditions
$\frac{5}{\mathit{S}}=\frac{{\mathit{l}}_1}{100-{\mathit{l}}_1}$
$\frac{5}{\mathit{S}∕2}=\frac{1.5{\mathit{l}}_1}{100-1.5{\mathit{l}}_1}$
Dividing the two equations
$2=\frac{1.5{\mathit{l}}_1}{\left(100-1.5{\mathit{l}}_1\right)}\times \frac{\left(100-{\mathit{l}}_1\right)}{{\mathit{l}}_1}$
$200-3{\mathit{l}}_1=150-1.5{\mathit{l}}_1$
${\mathit{l}}_1=\frac{100}{3}\mathrm{cm}$
Putting the value of ${\mathit{l}}_1$ in any one of the two given conditions.
$\mathit{S}=10\mathit{\Omega }$