The current passing through the battery in the given circuit, is

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Professor PC Thomas Classes · Answer

Circuit $A B C D E F$ form wheatstone bridge, So, Resistance across arm $A D, R_{A D}=6 \Omega$ can be eliminated. From circuit $A B C D E F$ : Resistance $5 \Omega$ and $3 \Omega$ are in series and $2.5 \Omega$ and $1.5 \Omega$ are in series. Then both the series combination are in parallel with each other. Thus equivalent resistance is \ $ \begin{aligned} & \frac{1}{R_{e q}}=\frac{1}{8}+\frac{1}{4}=\frac{3}{8} \ 	herefore & R_{e q}=\frac{8}{3} \Omega \end{aligned} $ \ Now, the given circuit can be redrawn as Current through battery, $ I=\frac{5}{10}=0.5 \mathrm{~A} $