Here I will show you how to calculate the different currents in each loop of the figure below using the mesh current method.
This is my favorite approach to a problem like this one:
1) Identify meshes in a planar circuit.
2) Identify currents unknown.
3) Write KVL for each mesh.
4) Simplify and Solve.
Easier said than done I suppose, so let’s do it:
It is important to understand how many equations you are going to need. Here is how you do that:
Using Node-Voltage: #Equations = Ne – 1
Using Mesh-Current: #Equations = Be – (Ne – 1)
Where Ne = Number of Node, Be = Number of circuit components.
Here we have:
Ne = 4
Be = 6
Be – (Ne – 1) = 3 equations.
Now you are ready to start writing your mesh equations. You should only do one loop at a time. This is simply KVL at work here, adding up the voltages around the loop and making sure they sum up to zero or to the source voltage.
Mesh 1: 5 ( i1 – i3) + 26 (i1 – i2) = 80
Mesh 2: 26 ( i2 – i1) + 90 (i2 – i3) + 8 (i2) = 0
Mesh 3: 30 (i3) + 90 (i3 – i2) + 5 (i3 – i1) = 0
We now have 3 equations and 3 unknowns, this is what we want and we should go ahead and move the “80” in the Mesh 1 equation to the left side so that all equations sum to zero.
Confused by this? “( i2 – i1)”
That is normal, and probably the most confusing concept. All 3 of the currents in all 3 loops are heading in the same direction, but in relation to each circuit component (such as 90Ω resistor), they are opposite. If you are in loop number 1, this current (i1) is your reference current for this given loop, and the current through a resistor in that loop is equal to that reference current minus the opposing current on that given component. If the currents were heading in the same direction, they would simply add instead of subtract. This way we have summed up the total current through each component and can therefore calculate the voltage.
You can either use a TI-89 to solve for these unknowns…
Solving the 3×3 system
I1 = 5 A
I2 = 2.5 A
I3 = 2 A