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

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Professor PC Thomas Classes · Answer

(d) : Since given circuit is in the form of Wheatstone bridge, $$ \begin{aligned} & \frac{1}{R_{e q}}=\frac{1}{(4+2)}+\frac{1}{(6+3)} ; R_{e q}=18 / 5 \Omega \ & V=i R_{e q} \Rightarrow \quad i=\frac{V}{R_{e q}}=\frac{5 V}{18} . \end{aligned} $$