__Branch Current Calculator:__

__Branch Current Calculator:__

Enter the values of current through resistor, I_{r1(A)}, I_{r2(A)} and I_{r3(A)} to determine the value of Branch current, I_{b(A)}.

__Branch Current Formula:__

__Branch Current Formula:__

Branch current refers to the flow of electrical current through different pathways or branches in an electrical circuit.

Branch current analysis helps in identifying the distribution of currents in circuits, which is necessary for optimizing circuit performance and preventing overload.

It adheres to Kirchhoff’s Current Law (KCL), which states that the total current entering a junction or node is equal to the total current leaving the junction.

This law, which states that the total current entering a junction must equal the total current leaving the junction, is fundamental in analyzing branch currents.

In circuits with multiple paths (branches), each branch can carry a different amount of current depending on the circuit configuration and component values.

The amount of current in each branch depends on the electrical resistance or impedance within that branch relative to others in the circuit.

Branch current, I_{b(A)} in amperes is calculated by the sum of current through resistor, I_{r1(A)}, I_{r2(A)} and I_{r3(A)} in amperes.

Branch current, I_{b(A)} = I_{r1(A) }+ I_{r2(A)} + I_{r3(A)}

I_{b(A)} = branch current in amperes, A.

I_{r1(A) }= current through resistor 1 in amperes, A.

I_{r2(A)} = current through resistor 2 in amperes, A.

I_{r3(A)} = current through resistor 3 in amperes, A.

__Branch Current Calculation:__

__Branch Current Calculation:__

**Calculate the total branch current at a node where three currents of 2 A, 3 A, and 4 A are flowing into the node:**

Given: I_{r1(A) }= 2A, I_{r2(A)} = 3A, I_{r3(A)} = 4A.

Branch current, I_{b(A)} = I_{r1(A) }+ I_{r2(A)} + I_{r3(A)}

I_{b(A)} = 2 + 3 + 4

I_{b(A)} = 9A.

**Determine one of the branch currents if the total branch current at a node is 15 A, and the other two currents are 6 A and 7 A respectively:**

Given: I_{r1(A) }= 6A, I_{r2(A)} = 7A, I_{b(A)} = 15A.

Branch current, I_{b(A)} = I_{r1(A) }+ I_{r2(A)} + I_{r3(A)}

I_{r3(A)} = I_{b(A)} – I_{r1(A) }– I_{r2(A)}

I_{r3(A)} = 15 – 6 – 7

I_{r3(A)} = 2A.