Sunday, October 13, 2013

Nodal Analysis and Mesh Analysis

Nodal Analysis

nodal analysis is one of the most powerful in doing circuit analysis in the sense that it is systematic and you are solving it in the right manner. Lets take a look of some example.

                                      suppose we want to know the Vo across the 4k ohms Resistor.


Current flows from a higher potential to a lower potential in a resistor.




Nodal analysis with voltage source

Mesh Analysis

 is formed by enclosing a voltage source connected between two non reference nodes and elements connected in parallel with it.
     


        

Solve for the current through the 5 ohm resistor and the current through the 4V source using Node-Voltage Analysis.
Now write KCL at each node (except the reference):
KCL at V1:
-5A + V1/5 + (V1-V2)/10 + [V1-(V2+4)]/10 = 0
Note that there are four terms in the equation, one for each branch leaving the node. The terms list the current leaving right, down, left, and up.
KCL at V2:
(V2-V1)/10 + V2/2 - 2A + [V2-(V1-4)]/10 = 0
Note that there are four terms in the equation, one for each branch leaving the node. The terms list the current leaving right, down, left, and up.
Now gather terms (multiplying through by 10 to clear up the fractions):
4V1 - 2V2 = 54
-2V1 + 7V2 = 16
Now solve the set of 2 equations with 2 unknowns.
V1 = 17.08V
V2 = 7.17V
We can now determine the current through the 5 ohm by Ohm's law:
I = V1/5 = 3.41A
The current through the 4V source can be found as:
I = [V1-(V2+4)]/10 = 0.59A





Mesh Analysis

   Mesh analysis (or the mesh current method) is a method that is used to solve planar circuits for the currents (and indirectly the voltages) at any place in the circuit.

sample problem
                                 


Label each mesh  with a mesh current. Then write the KVL equations for each pane. Note that we were forced to label the voltage over the current source (Vx) in order to write the voltage term there:

We now have an extra unknown (Vx), so we need another equation. It is found be relating the two mesh currents to the current source.

Note that i1 is positive because it is in the same direction of the source. I2 is negative because it is in the opposite direction as the source.
Now solve the three equations in three unknowns. I1 is found to be -320mA. Since ix is in the opposite direction of i1, then ix = 320mA.

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