In the given circuit diagram, the currents, $I_{1} = - 0.3 A, I_{4} = 0.8 A$ and $I_{5} = 0.4 A$ , are flowing as shown. The current $I_{2}, I_{3}$ and $I_{6}$ , respectively, are:

1.1 A, 0.4 A, 0.4 A

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-0.4 A, 0.4 A, 1.1 A

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0.4 A, 1.1 A, 0.4 A

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1.1 A, -0.4 A, 0.4 A

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Solution

The correct option is A $1.1 A, 0.4 A, 0.4 A$

Given that,
$I_{1} = - 0.3 A, I_{4} = 0.8 A, I_{5} = 0.4 A$

Applying Kirchhoff's current law at $R$

$I_{1} + I_{2} = I_{4}$

$\Rightarrow I_{2} = 1.1 A$

Applying Kirchhoff's current law at $S$

$I_{4} = I_{3} + I_{5}$

$\Rightarrow I_{3} = 0.4 A$

From the given figure,

$I_{5} = I_{6} = 0.4 A$

Hence, option $(B)$ is correct.

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Similar questions

I_{1} = - 0.3 A, I_{4} = 0.8 A

I_{5} = 0.4 A

I_{2} I_{3}

I_{6}

I_{n} = \frac{π}{2} \int \frac{π}{4} {cot}^{n} x d x

Identify correct relation for the circuit mentioned below, assuming the current $I_{3}$ flows through a conductor with zero resistance:

I_{n} = \int_{0}^{π / 4} t a n^{n} x d x

I_{2} + I_{4}, I_{3} + I_{5}, I_{4} + I_{6}, I_{5} + I_{7}

I_{n} = \int_{0}^{π / 4} {tan}^{n} (x) d x

I_{2} + I_{4}, + I_{3} + I_{5}, + I_{4} + I_{6}, . . .

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