In the given circuit, find voltage across $12 \Omega$ resistance.

If $\mathrm{R}_1$ be the equivalent resistance of parallel resistors $8 \Omega, 8 \Omega$ and $4 \Omega$, then $\frac{1}{\mathrm{R}_1}=\frac{1}{8}+\frac{1}{8}+\frac{1}{4}=\frac{4}{8} \Rightarrow \mathrm{R}_1=2 \Omega$ $\therefore$ Total resistance of upper branch, $\mathrm{R}_2=\mathrm{R}_1+4=2+4=6 \Omega$ Now circuit can be redraw of By current division rule, current through $12 \Omega$ resistor $\mathrm{I}_2=\frac{12 \times 6}{12+6}=12 \times \frac{1}{3}=4 \mathrm{~A}$ $\therefore$ Voltage across $12 \Omega$ resistor $\mathrm{V}=\mathrm{I}_2 \mathrm{R}=4 \times 12=48 \mathrm{~V}$