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MHTCET 20 Apr 2026 Shift 2 Paper

Question : 30 of 150

Marks: +1, -0

i

Solution:

Concept:

This problem is solved using Kirchhoff's Current Law (KCL).

At any junction, the total current entering equals the total current leaving.

Explanation:

At the first junction, currents of

2\,\text{A}

and

3\,\text{A}

enter.

So the current leaving this junction is:

2\,\text{A} + 3\,\text{A} = 5\,\text{A}

At the second junction, the incoming current is

5\,\text{A}

One branch takes away

1\,\text{A}

, so the current moving toward the third junction is:

5\,\text{A} - 1\,\text{A} = 4\,\text{A}

At the third junction, the incoming current is

4\,\text{A}

It divides into three outgoing branches:

0.6\,\text{A}

1.3\,\text{A}

, and

i

Applying KCL:

4\,\text{A} = 0.6\,\text{A} + 1.3\,\text{A} + i

4\,\text{A} = 1.9\,\text{A} + i

i = 4\,\text{A} - 1.9\,\text{A} = 2.1\,\text{A}

Answer:

The value of current

i

2.1\,\text{A}

Hence, the correct option is C.

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