In the following network,I_{1} = -0.4 text{ A | vedclass

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(D) Applying Kirchhoff's Current Law $(KCL)$ at the junctions:At the right junction: $I_{1} + I_{2} = I_{4}$Given $I_{1} = -0.4 	ext{ A}$ and $I_{4}

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$1.4 	ext{ A}, -0.6 	ext{ A}, 0.4 	ext{ A}$

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(D) Applying Kirchhoff's Current Law $(KCL)$ at the junctions:At the right junction: $I_{1} + I_{2} = I_{4}$Given $I_{1} = -0.4 	ext{ A}$ and $I_{4} = 1 	ext{ A}$,we have:$-0.4 + I_{2} = 1 \implies I_{2} = 1.4 	ext{ A}$At the bottom-left junction: $I_{5} = I_{3} + I_{4}$Given $I_{5} = 0.4 	ext{ A}$ and $I_{4} = 1 	ext{ A}$,we have:$0.4 = I_{3} + 1 \implies I_{3} = -0.6 	ext{ A}$At the top-left junction: $I_{6} = I_{5} + I_{3}$ (Wait,looking at the diagram,$I_{6}$ is the current entering the top-right junction from the left branch,so $I_{6} = I_{5}$ is not correct. Let's re-evaluate the circuit).Actually,applying $KCL$ at the top-right junction: $I_{6} = I_{1} + I_{2} + I_{3}$.Substituting the values: $I_{6} = -0.4 + 1.4 + (-0.6) = 0.4 	ext{ A}$.Thus,$I_{2} = 1.4 	ext{ A}$,$I_{3} = -0.6 	ext{ A}$,and $I_{6} = 0.4 	ext{ A}$.

In the following network,$I_{1} = -0.4 \text{ A}$,$I_{4} = 1 \text{ A}$ and $I_{5} = 0.4 \text{ A}$. The values of $I_{2}$,$I_{3}$ and $I_{6}$ respectively are:

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