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

$\begin{array}{l}1\mathrm{\Omega } \mathrm{and} 4\mathrm{\Omega } \mathrm{are} \mathrm{in} \mathrm{parallel} \frac{1\left(4\right)}{1+4}=\frac{4}{5}\mathrm{\Omega }\\ 2\mathrm{\Omega } \mathrm{and} 3\mathrm{\Omega } \mathrm{are} \mathrm{in} \mathrm{parallel} \frac{2\left(3\right)}{2+3}=\frac{6}{5}\mathrm{\Omega }\\ \frac{4}{5}\mathrm{\Omega } \mathrm{and} \mathrm{are} \mathrm{in} \mathrm{series} \\ {\mathrm{R}}_{\text{eff}}=\frac{4}{5}+\frac{6}{5}=\frac{10}{5}=2\Omega \\ \\ \text{i}=\frac{V}{{\mathrm{R}}_{\text{eff}}}=\frac{10}{2}=5\text{\hspace{0.17em}A}\end{array}$

Apply junction rule at point B

$Let R_1=4\Omega ,R_2=1\Omega ,R_3=3\Omega ad R_4=2\Omega$

$$\begin{gathered} current through R_1=4\Omega is i_1 \\ current through R_2=1\Omega is i_2 \\ current through R_3=3\Omega is i_3 \\ current through R_4=2\Omega is i_4 \end{gathered}$$

${\text{Current across 4Ω,I}}_1=\frac{R_2\text{i}}{R_1+R_2}==\frac{1(\text{5)}}{1+4}=1A$

${\text{Current across 3Ω,I}}_3=\frac{R_4\text{i}}{R_3+R_4}==\frac{2(\text{5)}}{2+3}=2A$

$\mathrm{current} \mathrm{through} \mathrm{BC}=\mathrm{i}={\mathrm{I}}_3-{\mathrm{I}}_1=2-1=1\mathrm{A}$