Question

For the network shown in the figure, the value of the current $i$ is

• A. $\frac{9V}{35}$
• B. $\frac{5V}{18}$
• C. $\frac{5V}{9}$
• D. $\frac{18V}{5}$

Answer 1

Answer: (B)

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 $R$ " gives
$R=\frac{R^{\prime }\times R^{\prime \prime }}{R^{\prime }+R^{\prime \prime }}$
$=\frac{6\times 9}{6+9}=\frac{54}{15}=\frac{18}{5}\Omega$
Hence, current $i=\frac{V}{R}=\frac{V}{18/5}=\frac{5V}{18}$