Mesh Analysis

The Kirchorff's Voltage Law (KVL) is the main basis of the Mesh Analysis. It is either the current only exist in one mesh or when the current source is in between of two meshes. And these are the two cases in solving the mesh analysis.  


To understand such cases, let's have an example.
Considering this circuit:

Here are the steps in solving:
1. Analyze circuit.
2. Look at the figure. I have assigned I1 for the first current loop and I2 for the second current loop.
3. Since that the second loop has the current source of 2 amperes,therefore,I conclude that:

                              I2 = -2 Amps

4. Solve Mesh 1 to get I1:
---------------------------------------------------------------------------------
12/_0 + 3I1 - j2(I1-I2) = 0
12/_0 + 3I1 - j2I1 - j2I2 = 0
12/_0 + I1(3-j2) - j2I2 = 0
I1(3-j2)-j2I2 = -12/_0
I1(3-j2)-j2(-2/_0) = -12/_0
I1(3-j2)+j4 = -12/_0
I1 (3-j2) = -12/_0 - j4

I1 = 3.508/_-127.87 Amps


Itotal = 4.99/_-146.30 Amps
---------------------------------------------------------------------------------

V = IR
V = (4.99/_-146.30)(j2)
V = 9.98/_-56.307 Volts




Considering this circuit:

Here are the steps in solving:
1. Analyze the circuit.
2. In this case, the current source is in between of two meshes. Then it will be changed by an open circuit. See the figure below.

 3. Solve for Supermesh by using KVL:

-j2I1 - 3I2 = 0 ----------> Equation 1

• Get I2 by KCL: 

I2 = 2/_0 + I1  ----------> Equation 2

• Substitute Eq. 2 in 1:

-j2I1 - 3(2/_0 + I1) = 0

-j2I1 - 6 - 3I1 = 0

I1(-3-j2) = 6

I1 = 1.664/_146.309 A

• for I2:

I2 = 2/_0 + 1.664/_146.309

I2 = 1.109/_56.309 Amps

• Solve for V:

V = IR

V = 1.109/_56.309(3)

V = 3.328/_56.309 Volts

Impedance and its Components

       An impedance of the circuit is the total effective resistance or the opposition measure to the passage of current in an AC voltage source supply. It's unit is ohms and is given V = IZ. Resistance in the form of ohm's law, reamins constant in DC or AC circuits and is different in capacitor and an inductor. The Capacitor in DC circuit, as well as in an inductor is an open circuit,while in AC voltage source, the capacitive resistance is called reactance or Xc, on the other hand, inductive resistance is equal to the reactance or XL.

Nodal Analysis

       A method of determining voltage between the nodes in electric circuit analysis is what it called a Node-Voltage or Nodal Analysis. 

These are the two cases in solving the circuit problem:

1st Case:  When the voltage source is connected to non-reference node and a reference node or ground.

2nd Case: The Supernode. It is when the voltage source is not connected to the ground or reference node.    
    
Let us analyze such cases by these examples:

#1: Consider this circuit. 

Here are the steps in solving a circuit:

1. Analyze the problem.

2. Assign nodes. Assume that I am going to assign V1 as my first node, and V2 as my second node.

3. Since the 12 volts source is given and is connected to the non-reference and a reference node, then V1 is 12V. We have now the exact value of the first node.

4. Solve V2 by adding the quantities of the following:
    
    Since V1 = 12 /_0V; then

    V2(1/3 - 1/j1 + 1/j2) - 12V/3 = 0
    0.33 + j0.5V2 - 6 = 0
    0.33 + j0.5V2 = 6
    V2 = 5.538 - j8.307 or

    V2 = 9.98 /_ -146.309 Volts

And now we get Io value:
    
      Io = V/R
       = 9.98/_ -146.309 / j2

    Io = 4.99/_ -146.309 Amps

#2: Consider this Circuit:

Solve for V1 and V2

Here are the steps in solving a circuit:

1. Analyze the problem.
2. Look at the figure again. As what we saw, the 12 volts source is not connected to the reference node. Then I conclude that it is a supernode. 

3. Assign nodes again. Assume that I am going to assign V1 as my first node, and V2 as my second node. 

4. Join the two nodes as a supernode.

5. Remove the voltage source by changing it in a short circuit. (Take note as well: that if the current source is given and has to be removed, change it in an open circuit.)

6. In getting the value of a supernode, get the equation value by using Kirchoff's Current Law(KCL):

@KCL:

I1 + I2 + I3 = 0

• I1 = V1/R1 = V1/3
• I2 = V2/-j1
• I3 = V2/j2

V1/3 + V2/-j1 + V2/j2 = 0

@KVL:

V1+12/_0 - V2 = 0
V1-V2 = -12/_0
V1 = V2 - 12/_0

Now get the values:

V2 - 12/3 + (V2/-j1) + V2/j2 = 0
V2/3 - 6 +(V2/-j1)+V2/j2 = 0
V2(1/3 + (1/-j1) + 1/j2) = 6
0.33V2 + j0.5 = 6

V2 = 9.98/_ -56.309 V

V1 = 9.98/_ -56.309 - 12
V1 = 10.52/_ -127.89 V

Sinusoids and Phasors

A sine wave

          A curve having a form of a wave or a sine wave is also what we called a sinusoid. It is also termed as a repetitive oscillation. A phasor is represented by a sinusoidal function as a complex number. It's phase difference is about less than or equal to 90 degrees. It is usually get the angle on where the voltage leads the current. Such simple indications are given in order to remember the following: the mnemonic "ELI and ICE man." It is remembered as "ELI" when the voltage leads the current by 90 degrees, with an emf(E),which is ahead of the current(I) and an inductor(L). Otherwise, "ICE" will be remembered when the voltage lags the current.