# 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_{1}, I_{2}, and I_{3} are assigned in the direction as shown in figure-1. Do not worry about the actual current directions at this point. |

Second | The polarities of the voltage drops across R_{1}, R_{2} and R_{3} are indicated in the figure-1 according to the current directions. |

Third | Kirchhoff's voltage law applied to the two loops gives the following equations. Equation 1: R _{1}I_{1} + R_{2}I_{2} ----V_{S1} = 0 for loop 1Equation 2: R _{2}I_{2} + R_{3}I_{3} -----V_{S2} = 0 for loop 2 |

Fourth | Kirchhoff's current law is applied to node A, including all branch currents as follows. Equation 3: I _{1} - I_{2} + I_{3} = 0the negative signs indicates that I _{2} is out of the junction. |

Fifth | The three equations must be solved for the three unknown currents, I_{1}, I_{2} and I_{3}. |

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: |

47I_{1} + 22I_{2} - 10 = 0Around the right loop gives 22I _{2} + 68I_{3} - 5 = 0 | |

Step 4: | At node A, the current equation is I _{1} - I_{2} + I_{3} = 0 |

Step 5: | The equation are solved by substitution as follows. First find I, in terms of I_{2} and I_{3}I _{1} = I_{2} - I_{3}Now substitute I _{2} - I_{3} for I_{1} in the left loop equation.47(I _{2} - I_{3}) + 22I_{2} = 1047I _{2} - 47I_{3} + 22I_{2} = 1069I _{2} - 47I_{3} = 10Next, take the right loop equation and solve for I _{2} in terms of I_{3} 22I_{2}= 5 - 68I_{3}I _{2} = (5 - 68I_{3})/22Substituting this expression for I2 into 69I _{2} - 47I_{3} = 10 you get69(5-68I _{3}/22) - 47I_{3} = 10(345 - 4692I _{3})/22 - 47I_{3} = 10-260.27I _{3} = -5.68I _{3} = 5.68/260.27 = 0.0218 A = 21.8 mANow, substitute this value of I _{3} into the right loop equation.22I _{2} + 68(0.0218) = 5Solve for I _{2}I _{2} = (5 - 68(0.0218))/22 = 3.52/22 = 0.16 A = 160 mASubstituting I _{2} and I_{3} values into the current equation at node A, you obtainI _{1} - 0.16 + 0.0218 = 0I _{1} = 0.16 - 0.0218 = 0.138 A = 138 mA |

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