Physics - Current Electricity Question with Solution | TestHub

The circuit given resembles the balanced Wheatstone Bridge as $\frac{4}{6}=\frac{2}{3}$. Thus, middle arm containing $4\Omega$ resistance will be ineffective and no current flows through it. The equivalent circuit is shown as below :Net resistance of $AB$ and $BC$, $R^{\prime }=4+2=6\Omega$ Net resistance of $AD$ and $DC$, $R^{\prime \prime }=6+3=9\Omega$ Thus, parallel combination of $R^{\prime }$ and $R^{\prime \prime }$ gives $\begin{array}{c}R=\frac{R^{\prime }\times R^{\prime \prime }}{R^{\prime }+R^{\prime \prime }}\\ =\frac{6\times 9}{6+9}=\frac{54}{15}=\frac{18}{5}\Omega \end{array}$ Hence, current $i=\frac{V}{R}=\frac{V}{18/5}=\frac{5V}{18}$