State Kirchhoff’s law for an electrical network. Using these laws deduce the condition for balancing in a Wheatstone bridge?

a) Kirchhoff’s current law (or) first law :
“The sum of electric currents flowing into a junction is equal to sum of electric currents flowing out of same junction”. (or)
“The algebraic sum of all electric currents meeting at any junction of an electrical network is zero.”

From Kirchhoff’s current law
${\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3={\mathrm{I}}_4+{\mathrm{I}}_5$
${\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3-{\mathrm{I}}_4-{\mathrm{I}}_5=0$. 

This is based on ‘law of conservation of charge’.
b) Kirchhoff’s voltage law (KVL) or loop theorem: “In any closed circuit the algebraic sum of the potential differences is equal to zero”.
Mathematically $\sum _{ }^{ }$ V = 0
This law is based on ‘law of conservation of energy’. 

From the figure shown using kirchoff’s voltage law

we get $\mathrm{E}-{\mathrm{IR}}_1-{\mathrm{IR}}_2-{\mathrm{IR}}_3=0$
$\mathrm{I}=\frac{\mathrm{E}}{{\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3}$
Wheatstone’s bridge: Wheat stone’s bridge network is connected as shown below

1) Applying Kirchhoff’s I^stlaw, at junction
${{ }^{\text{'}}\mathrm{C}}^{\mathrm{\prime }},{\mathrm{i}}_1={\mathrm{i}}_{\mathrm{g}}+{\mathrm{i}}_3\to \text{, (1), }$
$\text{ at junction 'D' }{\mathrm{i}}_2+{\mathrm{i}}_{\mathrm{g}}={\mathrm{i}}_4\to \text{, (2) }$
2) Applying Kirchhoff’s II^nd law for the closed mesh ACDA, $-{\mathrm{i}}_1\mathrm{P}-{\mathrm{i}}_{\mathrm{g}}\mathrm{G}+{\mathrm{i}}_2\mathrm{R}=0\to \left(3\right)$
For the closed mesh CBDC, 
 $-{\mathrm{i}}_3\mathrm{Q}+{\mathrm{i}}_4\mathrm{S}+{\mathrm{i}}_{\mathrm{g}}\mathrm{G}=0\to \left(4\right)$
3) When no current passes through the galvanometer i_g = 0, the bridge is said to be balanced.
4) If i_g = 0, the above four equations become,
$\begin{array}{l}{\mathrm{i}}_1={\mathrm{i}}_3\mathrm{\&}{\mathrm{i}}_2={\mathrm{i}}_4\\ {\mathrm{i}}_1\mathrm{P}={\mathrm{i}}_2\mathrm{R}\to \left(5\right){\mathrm{i}}_3\mathrm{Q}={\mathrm{i}}_4\mathrm{S}\to \left(6\right)\\ \frac{\left(5\right)}{\left(6\right)}\Rightarrow \frac{{\mathrm{i}}_1\mathrm{P}}{{\mathrm{i}}_3\mathrm{Q}}=\frac{{\mathrm{i}}_2\mathrm{R}}{{\mathrm{i}}_4\mathrm{S}}\Rightarrow \frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}\end{array}$
This is the principle of wheat stone’s bridge.