# BRANCH CURRENT METHOD

In the branch current method Kirchhoff's voltage and current laws are used to solve for the current in each branch of a circuit. Once the branch currents are known, voltages can be determined.

STEPS

The following are the general steps used in applying the branch current method.

**Step 1:** Assign a current in each circuit it branch in an direction.

**Step 2:** Show the polarities of the resistor voltages according to the assigned branch current direction.

**Step 3:** Apply Kirchhoff's voltage law around each closed loop (Sum of voltages equal to zero).

**Step 4:** Apply Kirchhoff's current law at the minimum number of nodes so that all branch currents are included (Sum of currents at a node equals zero).

**Step 5:** Solve the equations resulting from step 3 and 4 for the branch current values.

CIRCUIT FOR DEMONSTRATING BRANCH CURRENT ANALYSES

First |
The branch current I |

Second |
The polarities of the voltage drops across R |

Third |
Kirchhoff's voltage law applied to the two loops gives the following equations. Equation 1: R Equation 2: R |

Fourth |
Kirchhoff's current law is applied to node A, including all branch currents as follows. Equation 3: I the negative signs indicates that I |

Fifth |
The three equations must be solved for the three unknown currents, I |

The three equations in the above steps are called Simultaneous Equations and can be solved by substitution.

## BRANCH CURRENT METHOD EXAMPLE

Finding Current in each branch using branch current method.

Step 1: |
Assign branch current keep in mind that you can assume any current direction at this point and that the final solution will have a negative sign if the actual current is opposite to the assigned current. |

Step 2: |
Mark the polarities of the resistor voltage drops as shown in the figure. |

Step 3: |
Applying Kirchhoff's voltage law around the left loop gives 47I Around the right loop gives 22I |

Step 4: |
At node A, the current equation is I |

Step 5: |
The equation are solved by substitution as follows. First find I, in terms of I I Now substitute I 47(I 47I 69I Next, take the right loop equation and solve for I I Substituting this expression for I2 into 69I 69(5-68I (345 - 4692I -260.27I I Now, substitute this value of I 22I Solve for I I Substituting I I I |