Find current (i) in circuit shown in figure.

Above circuit is satisfied, the Wheatstone bridge condition, so no current is flow in the branch BC and it behaves like an open circuit. So, circuit becomes as shown below Now, in the above circuit $$\begin{aligned} & \mathrm{R}_{\mathrm{ABD}}=5+10=15 \Omega \\ & \mathrm{R}_{\mathrm{ACD}}=10+20=30 \Omega\end{aligned}$$ Resistance, $\mathrm{R}_{\mathrm{ABD}}$ and $\mathrm{R}_{\mathrm{ACD}}$ in parallel, so $$\begin{aligned} \mathrm{R}_{\text{net }} & =\frac{15 \times 30}{30+15}=\frac{450}{45}=10 \Omega \\ \therefore \quad \mathrm{i} & =\frac{\mathrm{E}}{\mathrm{R}_{\text{net }}}=\frac{5}{10} \\ & =0.5 \mathrm{ A}\end{aligned}$$