In the following network, $\mathrm{I}_{1}=-0 \cdot 4 \mathrm{~A}, \mathrm{I}_{4}=1 \mathrm{~A}$ and…

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
  1. $0 \cdot 4 \mathrm{~A},-0 \cdot 6 \mathrm{~A}, 1 \cdot 4 \mathrm{~A}$
  2. $-0 \cdot 6 \mathrm{~A}, 1 \cdot 4 \mathrm{~A}, 0 \cdot 4 \mathrm{~A}$
  3. $1 \cdot 4 \mathrm{~A}, 0 \cdot 4 \mathrm{~A},-0 \cdot 6 \mathrm{~A}$
  4. $1 \cdot 4 \mathrm{~A},-0 \cdot 6 \mathrm{~A}, 0 \cdot 4 \mathrm{~A}$

Solution

$\begin{aligned} \text { Given } & \mathrm{I}_{1}=-0.4 \mathrm{~A}, \mathrm{I}_{4}=1 \mathrm{~A}, \mathrm{I}_{5}=0.4 \mathrm{~A} \\ & \mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}_{4} \\ \therefore \mathrm{I}_{2} &=\mathrm{I}_{4}-\mathrm{I}_{1}=1-(-0.4)=1.4 \\ \mathrm{I}_{5} &=\mathrm{I}_{3}+\mathrm{I}_{4} \\ \therefore \mathrm{I}_{3} &=\mathrm{I}_{5}-\mathrm{I}_{4}=0.4-1=-0.6 \mathrm{~A} \\ \mathrm{I}_{6} &=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3} \\ &=-0.4 \mathrm{~A}+1.4-0.6=0.4 \mathrm{~A} \end{aligned}$ .

Asked in: MHT CET 2020 (13 Oct Shift 1)

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