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 I1, I2, and I3 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 R1, R2 and R3 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: R1I1 + R2I2 ----VS1 = 0 for loop 1 Equation 2: R2I2 + R3I3 -----VS2 = 0 for loop 2
Fourth	Kirchhoff's current law is applied to node A, including all branch currents as follows. Equation 3: I1 - I2 + I3 = 0 the negative signs indicates that I2 is out of the junction.
Fifth	The three equations must be solved for the three unknown currents, I1, I2 and I3.

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:

47I1 + 22I2 - 10 = 0 Around the right loop gives 22I2 + 68I3 - 5 = 0
Step 4:	At node A, the current equation is I1 - I2 + I3 = 0
Step 5:	The equation are solved by substitution as follows. First find I, in terms of I2 and I3 I1 = I2 - I3 Now substitute I2 - I3 for I1 in the left loop equation. 47(I2 - I3) + 22I2 = 10 47I2 - 47I3 + 22I2 = 10 69I2 - 47I3 = 10 Next, take the right loop equation and solve for I2 in terms of I3 22I2= 5 - 68I3 I2 = (5 - 68I3)/22 Substituting this expression for I2 into 69I2 - 47I3 = 10 you get 69(5-68I3/22) - 47I3 = 10 (345 - 4692I3)/22 - 47I3 = 10 -260.27I3 = -5.68 I3 = 5.68/260.27 = 0.0218 A = 21.8 mA Now, substitute this value of I3 into the right loop equation. 22I2 + 68(0.0218) = 5 Solve for I2 I2 = (5 - 68(0.0218))/22 = 3.52/22 = 0.16 A = 160 mA Substituting I2 and I3 values into the current equation at node A, you obtain I1 - 0.16 + 0.0218 = 0 I1 = 0.16 - 0.0218 = 0.138 A = 138 mA