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In the following network, $\mathrm{I}_{1}=-0 \cdot 4 \mathrm{~A}, \mathrm{I}_{4}=1 \mathrm{~A}$ and $\mathrm{I}_{5}=0 \cdot 4 \mathrm{~A}$. The values of $\mathrm{I}_{2}, \mathrm{I}_{3}$
and $\mathrm{I}_{6}$ respectively are
Options:
and $\mathrm{I}_{6}$ respectively are
Solution:
1745 Upvotes
Verified Answer
The correct answer is:
$1 \cdot 4 \mathrm{~A},-0 \cdot 6 \mathrm{~A}, 0 \cdot 4 \mathrm{~A}$
$$
\begin{aligned}
\text { Given } & \mathrm{I}_{1}=-0.4 \mathrm{~A}, \mathrm{I}_{4}=1 \mathrm{~A}, \mathrm{I}_{5}=0.4 \mathrm{~A} \\
& \mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}_{4} \\
\therefore \mathrm{I}_{2} &=\mathrm{I}_{4}-\mathrm{I}_{1}=1-(-0.4)=1.4 \\
\mathrm{I}_{5} &=\mathrm{I}_{3}+\mathrm{I}_{4} \\
\therefore \mathrm{I}_{3} &=\mathrm{I}_{5}-\mathrm{I}_{4}=0.4-1=-0.6 \mathrm{~A} \\
\mathrm{I}_{6} &=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3} \\
&=-0.4 \mathrm{~A}+1.4-0.6=0.4 \mathrm{~A}
\end{aligned}
$$
\begin{aligned}
\text { Given } & \mathrm{I}_{1}=-0.4 \mathrm{~A}, \mathrm{I}_{4}=1 \mathrm{~A}, \mathrm{I}_{5}=0.4 \mathrm{~A} \\
& \mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}_{4} \\
\therefore \mathrm{I}_{2} &=\mathrm{I}_{4}-\mathrm{I}_{1}=1-(-0.4)=1.4 \\
\mathrm{I}_{5} &=\mathrm{I}_{3}+\mathrm{I}_{4} \\
\therefore \mathrm{I}_{3} &=\mathrm{I}_{5}-\mathrm{I}_{4}=0.4-1=-0.6 \mathrm{~A} \\
\mathrm{I}_{6} &=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3} \\
&=-0.4 \mathrm{~A}+1.4-0.6=0.4 \mathrm{~A}
\end{aligned}
$$
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