# Branch Current Analysis

Branch Current Analysis is a systematic method for determining all of the branch currents in a circuit and uses KVL and KCL.

The steps are as follows:

1. Assign currents to each branch with voltage polarities shown.

2. Apply KVL around each loop of the circuit.

3. Apply KCL at all of the nodes in the circuit.

4. Solve the equations with linear algebra.

This technique is best illustrated through examples and pretty diagrams (An Engineer's best friend)

### Examples:

1. Calculate all of the branch currents in the following circuit:

The first step is to assign currents, voltage polarities and KVL loops:We can then write two equations using KVL around the left and right loops:Applying KCL:From here we have three equations with three unknowns and can subsitute [3] into [1] to eliminate down to 2:Comparing [1'] to [2] if we divide [1'] through by 2 and then subtract [1'] and [2] we can eliminate I3 and solve I2: We can then substitute I2 into [2] to determine I3: Finally we can calculate I1 by substituting I2 and I3 into [3]:

2. Calculate all of the branch currents in the following circuit:

The first step is to assign currents, voltage polarities and KVL loops:

We can then write two equations using KVL around the left and right loops (Note that the direction of I3 in the above has been chosen in the opposite direction that the current will flow, but this has been deliberately chosen to show that a negative current will be found which indicates that the current flows in the opposite direction. Currents can be selected in directions arbitrary, the calculations just have to be consistent. Likewise I1 could very well flow the other way): From here we can make a further simplification by recalling that an ideal current source will ensure that the provided current will flow through it hence I2 will be 10mA giving us: Applying KCL: We can solve for I1 using [3] and then substitute into [1] to eliminate I1: We can solve for I1 using [3] and then substitute into [1] to eliminate I1: We can then substitute this back into [1] to calculate I1: The negative result indicates that I1 actually flows back towards the voltage source. We can then calculate I3 by substituting V2 and I1 into [3]: Once again the negative result informs us that the current I3 flows in the opposite direction as we anticipated :)

### Questions:

1. Calculate all of the currents in the circuit below using branch current analysis:

I1 = 1.176mA, I2=4.412mA, I3=3.235mA

2. Calculate all of the currents in the circuit below using branch current analysis:

I1=3mA, I2=5mA, I3=2mA

3. Calculate all of the currents in the circuit below using branch current analysis:

I1=1.333mA, I2=1.333mA, I3=2.667mA, I4=666.667uA

## Lab:

Draw each of the above circuits for the questions in LTSpice to verify your answers :)