Question

The following currents are measured in the same direction in a three-branch parallel circuit: 250 mA, 300 mA, and 800 mA. What is the value of the current into the junction of these three branches?

Answer 1

The current in the junction is 1350 mA or 1.350 A.

Explanation:

As per Kirchoff's first law, the algebraic sum of current meeting at any junction should be equal to the algebraic sum of current leaving the junction. As in the present case, three parallel branch circuit is given the current in 250 mA, 300 mA and 800 mA, respectively, the sum of these three current will be equal to the current in the junction.

So,

I₁+I₂+I₃ = I₄

So I₁,I₂ and I₃ are the current passed in the three parallel branches and I₄ is the current in the junction.

250 + 300 + 800 = 1350 mA

So the current in the junction is 1350 mA or 1.350 A.