In the given network,the current through bran | vedclass

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(D) Let the potential at the common upper node $ABCD$ be $0 \, 	ext{V}$ and the potential at the common lower node $HGFE$ be $V$.Using Millman's theo

Vedclass · Accepted Answer

$0.26$

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(D) Let the potential at the common upper node $ABCD$ be $0 \, 	ext{V}$ and the potential at the common lower node $HGFE$ be $V$.Using Millman's theorem, the voltage $V$ is given by:$V = \frac{\sum \frac{E_i}{R_i}}{\sum \frac{1}{R_i}} = \frac{\frac{10}{2} + \frac{15}{2} + \frac{20}{4} + \frac{25}{5}}{\frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{5}} = \frac{5 + 7.5 + 5 + 5}{0.5 + 0.5 + 0.25 + 0.2} = \frac{22.5}{1.45} = \frac{450}{29} \, 	ext{V}$The current $I$ through branch $BG$ is:$I = \frac{V - E_2}{R_2} = \frac{\frac{450}{29} - 15}{2} = \frac{\frac{450 - 435}{29}}{2} = \frac{15}{58} \, 	ext{A} \approx 0.26 \, 	ext{A}$

In the given network,the current through branch $BG$ is almost equal to ............ $A$.

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