Question

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

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

Answer 1

The correct option is D $\frac{5V}{18}$

This arrangement can be modified as shown in figure.


We can clearly see that the given circuit is balanced Wheatstone bridge and no current flows through $4\Omega$ resistance connected across the diagonal.

Now circuit reduces to


So, equivalent resistance of circuit will be

$R_{eq}=\frac{6\times 9}{6+9}=\frac{18}{5}\Omega$

$\therefore$ Current will be

$i=\frac{V}{R}=\frac{V}{18/5}=\frac{5V}{18}$

Hence, option $\left(b\right)$ is correct.