In the given circuit diagram, a wire is joining points B and D. The current in this wire is :

$\begin{array}{l}Resistors (1 || 4) + (2 || 3) \\ here \left|\right| means parallel and +means series\\ \text{}{R}_{eff}=\frac{\left(1\times 4\right)}{\left(1+4\right)}+\frac{\left(2\times 3\right)}{\left(2+3\right)}=2\Omega \\ Ohm\text{'}s Law, i=\frac{E}{R}=\frac{20}{2}=10A\\ Current in BA=i_1\text{}\\ Current in AD=i-i_1\text{}\\ Current in BC =i_2\text{}\\ Current in BD = i_1-i_2 \\ Kirchhoff Loop in ABCEA,\text{}\\ -\left(i_1\right)\left(1\right)-\left(i_2\right)\left(2\right)+20=0 ------\left(1\right) \\ Kirchhoff Loop in ABDA,\text{}\\ -\left(i_1\right)\left(1\right)+\left(i-i_1\right)\left(4\right)=0\\ \text{}\Rightarrow -\left(i_1\right)\left(1\right)+\left(10-i_1\right)\left(4\right)=0\\ \Rightarrow i_1=8A, and \text{}\\ Put i_1=8 in equation \left(1\right)\\ \Rightarrow i_2=6A\\ \text{}Hence, Current in BD = i_1-i_2=8-6=2A\end{array}$