Using Kirchhoff’s rules, calculate the current through the 40 Q and 20 Q resistors in the following circuit:

What is end error in a meter bridge? How is it overcome? The resistances in the two arms of the metre bridge are R = 5Ω and S respectively.
When the resistance S is shunted with an equal resistance, the new balance length found to be 1.5 l₁ where l₁ is the initial balancing length. Calculate the value of S.

Applying Kirchhoff's first law at junction C
$\begin{array}{r}{\mathrm{I}}_1={\mathrm{I}}_2+{\mathrm{I}}_3\\ {\mathrm{I}}_2={\mathrm{I}}_1-{\mathrm{I}}_3\end{array}$                ...(i)
Applying the Kirchhoff's second law in closed loop ABCDA 
$80-20{\mathrm{I}}_1-40{\mathrm{I}}_3=0 ...\left(\mathrm{ii}\right)$
Applying Kirchhoff's second law in closed loop DCFED
$40{\mathrm{I}}_3-10{\mathrm{I}}_2+40=0 ....\left(\mathrm{iii}\right)$
From equations (i) and (iii),
$$\begin{gathered} 40{\mathrm{I}}_3-10\left({\mathrm{I}}_1-{\mathrm{I}}_3\right)-40=0 \\ 50{\mathrm{I}}_3-10{\mathrm{I}}_1+40=0 ....\left(\mathrm{iv}\right) \end{gathered}$$
From equations (ii) and (iv)
$\begin{array}{r}80=20{\mathrm{I}}_1+40{\mathrm{I}}_3\\ 40=10{\mathrm{I}}_1+50{\mathrm{I}}_3\end{array}$
We get,  I₃=0
and  ${\mathrm{I}}_1=4\mathrm{A}$
${\mathrm{I}}_1=4\mathrm{A}$
Therefore, current through 40$\mathrm{\Omega }$ resistor is 0 and current through 20$\mathrm{\Omega }$ resistor is 42 A.
OR
The shifting of zero of the scale at different points give rise to the end error in metre bridge wire.
This can be overcome by,
1. By trying to obtain the balance point in the middle of the metre bridge.
2. By taking metre bridge wire of uniform cross-section
3. By taking the copper strips thick, so that their resistance can be ignored.
As we know that,
$\frac{\mathrm{R}}{\mathrm{S}}=\frac{{\mathrm{l}}_1}{100-{\mathrm{l}}_1} ....\left(\mathrm{i}\right)$
When resistance S is shunted with equal resistance. Then, ${\mathrm{S}}^{\text{'}}=\frac{\mathrm{S}}{2}$
Therefore, new balancing point is, $\Rightarrow \frac{\mathrm{R}}{\mathrm{S}/2}=\frac{1.5{\mathrm{l}}_1}{100-1.5{\mathrm{l}}_1}$
$\Rightarrow \frac{\mathrm{R}}{\mathrm{S}}=\frac{1.5{\mathrm{l}}_1}{2\left(100-1.5{\mathrm{l}}_1\right)} .....\left(\mathrm{ii}\right)$
On comparing equations (i) and (ii)
$\begin{array}{l}\frac{{\mathrm{l}}_1}{100-{\mathrm{l}}_1}=\frac{1.5{\mathrm{l}}_1}{2\left(100-1.5{\mathrm{l}}_1\right)}\\ 200-3{\mathrm{l}}_1=150-1.5{\mathrm{l}}_1\\ 1.5{\mathrm{l}}_1=50\\ {\mathrm{l}}_1=\frac{50}{1.5}\end{array}$
From equation (i),
$\begin{array}{l}\therefore \frac{5}{\mathrm{S}}=\frac{50}{1.50}\left(100-\frac{50}{1.5}\right)\\ \Rightarrow \frac{5}{\mathrm{S}}=\frac{50\times 1.5}{1.5\left(100\right)}(\because \mathrm{R}=\mathrm{S})\\ \Rightarrow \mathrm{S}=\frac{500}{50}=10\mathrm{\Omega }\end{array}$