Question

Consider a circuit with a main wire that branches into two other wires. If the current is 10 A in the main wire and 4 A in one of the branches, how much current is in the other branch? Express your answer as an integer and include the appropriate units.

Answer 1

Answer:

6A

Explanation:

Kirchhoff's Current Law (KCL) states that for a parallel path circuit with current flow the sum of the current entering the circuit's junction is equal to the sum of the current moving away from it.

That is 10A enters the junction

and 4A is observed in one of the two branches, that means

10A - 4A = 6A the current in the other branch

Answer 2

Answer:

6A

Explanation:

Given that the current that enters the junction is 10A, According to Kirchoff's current law which states that the total amount of current entering a junction must be equal to the total amount of current leaving the junction, then the total current that also leaves the junction must be equal to 10A, and since one of the two branches by which current leaves the junction has a current of 4A, and the total current leaving the junction must be equal to 10A. So, current passing through the second wire can be given as;

I(total) = I1 + I2

I1 = 4A

I(total) = 10A

I2 = I(total) - I1 = 10A - 4A = 6A

Therefore, the amount of current leaving through the other branch is 6A.