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

Question

Aicharya · Accepted Answer

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 $R^*=R_3+R_4=6+3$ $=9 \Omega$
Now $R^{\prime}$ and $R^{\prime \prime}$ are in parallel
$\begin{array}{l}
\therefore \frac{1}{R_{e q}} =\frac{1}{R^{\prime}}+\frac{1}{R^n} \\
=\frac{1}{6}+\frac{1}{9}=\frac{5}{18} \\
\Rightarrow R_{e q} =\frac{18}{5} \\
\text {Now from } V =I r_{e q} \\
\Rightarrow I =\frac{V}{R_{e q}}=\frac{5}{18} \mathrm{~V}
\end{array}$

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

Solution