Source Transformation (AC Analysis)

Source transformation is simplifying a circuit solution, especially with mixed sources, by transforming a voltage into a current source, and vice versa. Finding a solution to a circuit can be difficult without using methods such as this to make the circuit appear simpler. Source transformation is an application of Thévenin's theorem and Norton's theorem.


Performing a source transformation consists of using Ohm's law to take an existing voltage source in series with a resistance, and replace it with a current source in parallel with the same resistance. Remember that Ohm's law states that a voltage on a material is equal to the material's resistance times the amount of current through it (V=IR). Since source transformations are bilateral, one can be derived from the other. [2] Source transformations are not limited to resistive circuits however.

They can be performed on a circuit involving capacitors and inductors, as long as the circuit is first put into the frequency domain. In general, the concept of source transformation is an application of Thévenin's theorem to a current source, or Norton's theorem to avoltage source.

Specifically, source transformations are used to exploit the equivalence of a real current source and a real voltage source, such as a battery. Application of Thévenin's theorem and Norton's theorem gives the quantities associated with the equivalence. Specifically, suppose we have a real current source I, which is an ideal current source in parallel with an impedance. If the ideal current source is rated at I amperes, and the parallel resistor has an impedance Z, then applying a source transformation gives an equivalent real voltage source, which is ideal, and in series with the impedance. This new voltage source V, has a value equal to the ideal current source's value times the resistance contained in the real current source. The impedance component of the real voltage source retains its real current source value.



In general, source transformations can be summarized by keeping two things in mind:
· Ohm's Law
· Impedance's remain the same

Source transformation also is one of the easiest way to solve for the wanted values. However, it is also not applicable to every circuit, and also it requires a lot of redrawing of circuit. Being unable to have the talent of good drawing capabilities, it is quite a downfall. 

Superposition (AC Analysis)

The superposition theorem for electrical circuits states that for a linear system the response (voltage or current) in any branch of a bilateral linear circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, where all the other independent sources are replaced by their internal impedances.

To ascertain the contribution of each individual source, all of the other sources first must be "turned off" (set to zero) by:

1. Replacing all other independent voltage sources with a short circuit (thereby eliminating difference of potential i.e. V=0; internal impedance of ideal voltage source is zero (short circuit)).

2. Replacing all other independent current sources with an open circuit (thereby eliminating current i.e. I=0; internal impedance of ideal current source is infinite (open circuit)).

This procedure is followed for each source in turn, and then the resultant responses are added to determine the true operation of the circuit. The resultant circuit operation is the superposition of the various voltage and current sources.

The superposition theorem is very important in circuit analysis. It is used in converting any circuit into its Norton equivalent or Thevenin equivalent.

The theorem is applicable to linear networks (time varying or time invariant) consisting of independent sources, linear dependent sources, linear passive elements (resistors, inductors, capacitors) and linear transformers.


Another point that should be considered is that superposition only works for voltage and current but not power. In other words the sum of the powers of each source with the other sources turned off is not the real consumed power. To calculate power we should first use superposition to find both current and voltage of each linear element and then calculate the sum of the multiplied voltages and currents.

It is one of the easiest process in order to calculate the wanted value. However, it is not applicable to all circuits. That is saddening but we should just accept such. That is life, we have to accept and move on.

Mesh Analysis (AC Analysis)

MESH ANALYSIS:

Another method of analyzing circuits and the counterpart of nodal analysis is the mesh analysis. Also known as loop analysis or the mesh-current method. A mesh is a loop which does not contain any other loops within it. And we can recall that a loop in an electronic circuit is a closed path with no node passed more than once. The current through a mesh is known as mesh current.

Mesh analysis makes use of Kirchhoff’s Voltage Law (KVL) in forming equations to solve for unknown currents in a given circuit. If it’s Vx for nodal analysis, it’s Ix for mesh analysis. Don’t you find it ironic that nodal analysis applies KCL to find unknown voltages, while mesh analysis applies KVL to find unknown current? Though mesh analysis isn’t quite as general as nodal analysis because it is only applicable to planar circuits**.

**one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar.

Here are the steps in determining mesh currents:

Assign the meshes found in the circuit as mesh currents i1, i2, … , in.
Apply KVL to each of the assigned meshes to form equations. Use Ohm’s law (V=IR) to express the voltages in terms of the mesh currents.
Solve the formed equations from step 2 to get the value of the unknown mesh currents.
Let’s take this very familiar circuit for this tutorial:

Assuming that all the values for the elements are already given. And that we are supposed to determine the value of the voltage across the capacitor Z3 (Vz3).

Again, as mentioned above, a mesh is a loop which does not contain any other loops within it. By looking at this simple circuit, we can already say that there are two meshes present in the circuit. The bigger loop outside the circuit cannot be considered as a mesh, because it contains loops within it. The direction of the mesh current can either be clockwise or counterclockwise and it will not affect the validity of the solution, as long as you follow the rules in mesh analysis. And we will encounter those rules as we proceed in solving this circuit.

I have assumed that both meshes i1 and i2 loop at a clockwise direction. You can have meshes at different directions (one goes clockwise, another goes counter) but for ethicality purposes, it’s better to have them at the same direction.

To determine Vz3, we have to first get the current that passes through the capacitor Z3. Then we apply Ohm’s law to solve for Vz3. If you look at Z3, you can observe that there are two currents that are passing through it but not in the same direction as each other: i1 and i2. We can also see that i2 passes through a current source Is but in a different position. This means we can conclude that i2 = -Is.

In nodal analysis, the voltage of a non-reference node directly connected to a voltage source is already equal to that source. Unless the node is connected to another voltage source, then that’d be a supernode. In mesh analysis, the current of a mesh that passes through a current source is already equal to that source (positive if they go the same direction, otherwise negative). Unless the mesh shares the current source with another mesh, then that’d be a supermesh.

So, now that we know mesh i2 = -Is, we’ll only need the current of mesh i1. And we can form its equation using KVL this way:

@mesh i1:

+Vs = 0

In nodal analysis, a current source is considered a constant* and its sign depends on its direction. In mesh analysis, the voltage source is the one considered as a constant* and its sign depends on the direction of the mesh that enters it.

*unless it’s a dependent source.

In this case, Vs is positive since mesh i1 came out of the positive terminal of the source (this is one of the rules I was talking about earlier). Moving on as mesh i1 passes through the resistor Z1..

+Vs – (i1 · Z1) = 0

You might think that the i1 · Z1 is quite familiar. That is actually just Ohm’s law, where V = I · R. So, it’s Vz1 = i1 · Z1. See the resemblance? Now as mesh i1 passes through the capacitor Z3..

+Vs – (i1 · Z1) – Z3(i1 – i2) = 0

As you can see from the last expression of the equation, mesh i1 is subtracted by mesh i2 and is multiplied to Z3. This is because both meshes go through the capacitor but with a difference in direction. Maybe you’d add the meshes if they go to the same direction, but i’ve never tried it before since i’m ethical in solving my circuits.

..and since i2 = -Is..

+Vs – (i1 · Z1) – Z3(i1 – (-Is)) = 0

..or..

+Vs – (i1 · Z1) – Z3(i1 + Is) = 0

..simplifying further..

Vs – (i1 · Z1 ) – (i1 · Z3) – (Is · Z3) = 0

Vs – (Is · Z3) = (i1 · Z1) + (i1 · Z3)

Vs – (Is · Z3) = i1(Z1 + Z3)

[Vs – (Is · Z3)] / (Z1 + Z3) = i1

i1 = [Vs – (Is · Z3)] / (Z1 + Z3)

And now that we have solved i1, we substitute it to:

Vz3 = Z3 · (i1 – i2)

..or..

Vz3 = Z3 · (i1 – Is)

Tadaa! And that is how you do mesh analysis, without the presence of a supermesh. I hope you were able to comprehend all that. If you have some questions, leave them below!