MHT CET · Physics · Current Electricity
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

- A \(0 \cdot 4 \mathrm{~A},-0 \cdot 6 \mathrm{~A}, 1 \cdot 4 \mathrm{~A}\)
- B \(-0 \cdot 6 \mathrm{~A}, 1 \cdot 4 \mathrm{~A}, 0 \cdot 4 \mathrm{~A}\)
- C \(1 \cdot 4 \mathrm{~A}, 0 \cdot 4 \mathrm{~A},-0 \cdot 6 \mathrm{~A}\)
- D \(1 \cdot 4 \mathrm{~A},-0 \cdot 6 \mathrm{~A}, 0 \cdot 4 \mathrm{~A}\)
Answer & Solution
Correct Answer
(D) \(1 \cdot 4 \mathrm{~A},-0 \cdot 6 \mathrm{~A}, 0 \cdot 4 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
\text {Given } & \mathrm{I}_{1}=-0.4 \mathrm{~A}, \mathrm{I}_{4}=1 \mathrm{~A}, \mathrm{I}_{5}=0.4 \mathrm{~A} \\
& \mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}_{4} \\
\therefore \mathrm{I}_{2} &=\mathrm{I}_{4}-\mathrm{I}_{1}=1-(-0.4)=1.4 \\
\mathrm{I}_{5} &=\mathrm{I}_{3}+\mathrm{I}_{4} \\
\therefore \mathrm{I}_{3} &=\mathrm{I}_{5}-\mathrm{I}_{4}=0.4-1=-0.6 \mathrm{~A} \\
\mathrm{I}_{6} &=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3} \\
&=-0.4 \mathrm{~A}+1.4-0.6=0.4 \mathrm{~A}
\end{aligned}
\)
\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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