The value of current $i_{1}$ in the given circuit is:

\frac{i}{5}

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\frac{2 i}{5}

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\frac{3 i}{5}

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\frac{4 i}{5}

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Solution

The correct option is C $\frac{3 i}{5}$
By Kirchhoff's law,
$i_{1} r + i_{3} r = 2 r (i - i_{1}) \Rightarrow i_{1} + i_{3} = 2 i - 2 i_{1} \Rightarrow i_{3} + 3 i_{1} = 2 i \Rightarrow i_{3} = 2 i - 3 i_{1} ⟶ (1) A l s o, {2 i}_{1} + 2 i_{3} = 2 i_{3} + i - i_{1} \Rightarrow 4 i_{3} = 3 i_{1} - i \Rightarrow i_{3} = \frac{3}{4} i_{1} - \frac{i}{4} ⟶ (2)$
Equating $(1)$ and $(2)$
$2 i - 3 i_{1} = \frac{3}{4} i_{1} - \frac{i}{4} \Rightarrow i_{1} = \frac{3}{5} i$

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