Physics - Current Electricity Question with Solution | TestHub

Since the network is balanced, wheatstone's bridge. Therefore no current flows through the arm $\mathrm{AC}$.Here $R_1$ and $R_2$ are in series $\therefore R_1{}^{\prime }=R_1+R_2=4+2=6\Omega R_3$ and $R_4$ are in series $R^{\ast }=R_3+R_4=6+3$ $=9\Omega$ Now $R^{\prime }$ and $R^{\prime \prime }$ are in parallel $\begin{array}{l}\therefore \frac{1}{R_{eq}}=\frac{1}{R^{\prime }}+\frac{1}{R^n}\\ =\frac{1}{6}+\frac{1}{9}=\frac{5}{18}\\ \Rightarrow R_{eq}=\frac{16}{5}\\ \text{Now from }V=I{r}_{eq}\\ \Rightarrow I=\frac{V}{R_{eq}}=\frac{5}{18}\mathrm{V}\end{array}$