Nodal Analysis And Mesh Analysis In AC circuit

Since we know that Kirchhoff's Law is applicable to the AC circuit. We will also apply the Nodal Analysis and Mesh Analysis in analyzing an ac circuits.

STEPS TO ANALYZE AC CIRCUITS:
1. Transform the circuit to the phasor or frequency domain.
2. Solve the problem using circuit techniques(Nodal/Mesh analysis)
3. Transform the resulting phasor to the time domain.


~We've been through about Nodal Analysis and Mesh Analysis in my past blogs. We will just recall it.

Nodal Analysis provide a general procedure for analyzing circuits using node voltages as the circuit variables.

Steps to Determine Node Voltages:

1. Select a node as the reference node, Assign voltages v1, v2, . . . . . , 

vn-1 to the remaining n-1 nodes. The voltages are referenced with respect to the reference node.

2.Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law to express currents in terms of node voltages.

3. Solve the resulting simultaneous equations to obtain the unknown node voltages.

Nodal Analysis with Voltage Sources

Case 1: If the voltage source (dependent or independent) is connected between two non-reference nodes, the two non-reference nodes form a generalized node or super node, we apply both KCL and KVL to determine the node voltages.

Case 2: if a voltage source is connected between the reference node and a non-reference node, we simply set the voltage at the non-reference node equal to the voltage of the voltage source.

~In this case we will solve a problem in which the three(capacitor,conductor,resistor) elements are included. Also with the rectangular form and polar form since the phasor and frequency domain was being applied. 


Example:

The unknown for this problem is Io, in order to get Io we must transform first the inductor and capacitor into the impedance. 

@node Vo,

But Io= (25-Vo) / 2000, so substitute to get the Vo and also the answer must be in polar form.

We can get now the Io using Ohm's Law.

MESH ANALYSIS;

A Mesh is a loop that does not contain any other loop within it.

STEPS TO DETERMINE MESH CURRENTS:
1. Assign mesh currents I1, I2,... In to the n meshes.
2. Apply KVL to each  of the n meshes. Use ohm's law to express the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh currents.

~We  will apply mesh to ac;

Example: 

 

Source:

- Fundamental of Electric Circuits by Alexander and Sadiku

- Google

- Youtube

Phasor Relationship For Circuit Elements

Impedance and Admittance

Impedance Z – of a circuit is the ratio of the phasor voltage to the phasor current, measured in Ohms(Ω).

               Z=V/I                    or           V=ZI

Admittance Y – is the reciprocal of impedance, measured in siemens(s).

               Y=1/Z = I/V

                              Impedances and admittances of passive elements.

Elements	Impedance	Admittance
R (Ω)	Z=R	Y=1/R
L (H)	Z= jωL	Y=1/jωL
C (F)	Z=1/jωC	Y=jωC

That three elements(R,L,C) only the Resistances are real and the rest were imaginary and the reactances. The Capacitor is the positive reactance while the Inductor is the negative reactance.

Example:

We can transform the imaginary elements into an impedance by the use of shown equations.

1H

Z= jωL = j1x10 = j10

1F

Z=1/jωC = 1/j10x1 = -j0.1

In order to proceed and get the unknown values, we must first identify if what concept should be done in this problem. Since, circuit analysis was all about analyzing the different character of a circuit in any ways, we will use the basic law which we’ve been through in a few months “The Kirchhoff’s Law” was already discuss here in the past topics. We will just recall some important matters.

Kirchhoff’s Law has two parts, the Kirchhoff’s voltage Law and Kirchhoff’s current Law.

Kirchhoff's voltage law, states that the algebraic sum of all the voltages around a closed circuit equals zero. 

Kirchhoff's current law, states that the algebraic sum of all the currents entering and leaving a node is equal to zero. 

Looking at the circuit, the KVL must be applied since the sum of all the voltage drops in a closed circuit will equal the voltage source if only we’ll combine all of the impedances in order to have a single loop. And here we can apply the time domain converted into phasor domain.

Z= 1 + (1/j10 + 1/-j0.1 + 1/1)^-1 = 1.01010 – j0.1 = 1.015∠ -5.653

We can solve now the unknown which is the current using Ohm’s Law;

I=V/R = 2∠0/ 1.015∠ -5.653

I = 1.9704∠5.653 = 1.9704cos(10t+5.65) A

SINUSOIDS AND PHASORS

A Sinusoid is a signal that has the form of the sine or cosine function.

                              There are two parts of sinusoid, the Sinusoidal Current and Sinusoidal Voltage. Sinusoidal current is usually referred to as alternating current. Such a current reverses at regular time intervals and has alternately positive and negative values. Circuits driven by sinusoidal current or voltage sources are called ac circuit. 

Sinusoidal voltage,

v(t) = Vmsinωt

where;

Vm= the amplitude of the sinusoid

ω = the angular frequency in radian/s

ωt = the argument of the sinusoid

Sample equation to determine it's label;

  6cos(200t + 15° )

Amplitude- 6

Phase angle- 15°

Angular Frequency- 200t

A Phasor – is a complex number that represents the amplitude and phase of a sinusoid.

2 PHASES

* IN PHASE

*OUT OF PHASE

IN PHASE,

The same;

*Time

*Period

*Frequency

OUT OF PHASE,

It's either have the same amplitude or not.

~Sinusoids are easily expressed in terms of phasors, in which are more convenient to work with than sine and cosine function.

Sinusoid-Phasor Transformation 

Time Domain representation	Phasor Domain Representation
Vmcos(ωt + ɸ )	Vm ∠ ɸ
Vmsin(ωt + ɸ )	Vm ∠ ɸ - 90 °
Imcos(ωt + 0 )	Im∠0
Imsin(ωt + 0 )	Im∠ 0 - 90 °

To transform the Time domain into the Phasor domain, the time domain is in the rectangular form of

 z = x + jy,

where in x is the real part of z and y is the imaginary part.

The equation going to polar form for the amplitude;

Square root of x squared plus y squared

For the Phase angle;

arctan(y divided by x) 

Example; 

5 + j2
= 5.39∠ 21.80°

Since all of these was all about the currents and voltages, in getting the value of each of them we go through graphing in a sinusoidal form. So between that two, there must be leading and lagging. 

Looking at the figure, the voltage leads the current since leading is when a sinusoid peaks first in time and it is closer to the reference axis. And the current here is lagging.