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

(B) $1.1 \mathrm{~A}, 0.4 \mathrm{~A}, 0.4 \mathrm{~A}$

(A) $1.1 \mathrm{~A},-0.4 \mathrm{~A}, 0.4 \mathrm{~A}$

(C) $0.4 \mathrm{~A}, 1.1 \mathrm{~A}, 0.4 \mathrm{~A}$

(D) $-0.4 \mathrm{~A}, 0.4 \mathrm{~A}, 1.1 \mathrm{~A}$

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Current Electricity

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 currents

I_{2}, I_{3}

and

I_{6}

, respectively are:

Answer: (B)

Apply Kirchhoff's current law junction by junction. Where

I_{3}, I_{4}, I_{5}

meet,

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

gives

I_{3} = 0.8 - 0.4 = 0.4

A. Where

I_{1}, I_{2}, I_{4}

meet,

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

gives

I_{2} = 0.8 - (- 0.3) = 1.1

A. At the remaining junction,

I_{3} + I_{6} = I_{1} + I_{2}

gives

I_{6} = 1.1 + (- 0.3) - 0.4 = 0.4

A. So

I_{2} = 1.1

I_{3} = 0.4

A and

I_{6} = 0.4

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option