SOURCE TRANSFORMATION AC CIRCUIT

SOURCE TRANSFORMATION

Source transformation in the frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance, or vice versa. As we go from one source type to another, we must keep the following relationship in mind:

Example:

Calculate Vx in the circuit using source transformation

First we transform the voltage source to a current source  where

Is=20angle-90/5

Is=4angle-90 or -j4A

The parallel combinations of 5ohms resistance and (3+j4) impedance gives

z1=5(3+j4)/8+j4

z1=2.5+j1.25ohms

We must convert the current source to a voltage source yields the circuit in the figure where

Vs=IsZ1=-j4(2.5+j1.25)=5-j10v

By voltage division,

Vx=(10/10+2.5+j1.25+4-j13)(5-j10)

Vx=5.519angle-28V

SUPERPOSITION IN AC CIRCUIT

SUPERPOSITION

Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to dc circuits. It becomes important if the circuit has sources operating at different frequencies. In this case, since the impedances depend on frequency, we must have a different frequency domain circuit for each frequency. The total response must be obtained by adding the individual responses in the time domain. It is incorrect to try to add responses in the phasor or frequency domain. Because the exponential factor e^jwt is implicit in sinusoidal analysis, and that factor would change for every angular frequency w. It would therefor not make sense to add responses at different frequencies in the phasor domain. Thus, when a circuit has sources operating at different frequencies, one must add the responses due to the individual frequencies in the time domain.

Example we gather from the book:

Use The superposition theorem to find Io in the circuit in the figure below:

So First we let 

Io=I'o + I''o

Where I'o and I''o are due to the voltage and current sources, respectively. To find I'o consider the circuit in figure a. If we let Z be the parallel combination of -j2 and 8+j10, then

Z=-j2(8+j10)/-2j+8+j10=0.25-j2.25

And current I'o is 

I'o=j20/4-j2+Z

=J20/4.25-j4.25

or I'o= -2.353 + j2.353

To get I''o, consider the circuit in figure b. For mesh 1,

(8+j8)I1-j10I3+j2I2=0

For mesh 2, 

(4-j4)I2+J2I1+J2I3=0

For mesh 3.

I3=5

From equations of mesh 2 and mesh 3

(4-j4)I2+j2I1+j10=0    =>eq 1

Expressing I1 in terms of I2 gives

I1=(2+j2)I2-5     =>eq 2

Substituting equation of mesh 3 and equation 2 into equation of mesh 1 we get

(8+j8)[(2+j2)I2-5]-j50+j2I2=0

or I2=90-40/34 =2.647-j1.176

Current I''o is obtained as

I''o=-I2=-2.647+j1.176

From the quation I'o=-2.353+j2.353 and I''o=-I2 we write

Io=I'o+I''o=-5+j3.529

=6.12angle144.78 A

MESH ANALYSIS IN AC CIRCUIT

MESH ANALYSIS

Kirchhoff's voltage law (KVL) forms the basis of mesh analysis. We must keep in our mind that the very nature of using mesh analysis is that it is to be applied to planar circuits.

EXAMPLE:

Determine current Io in the circuit using mesh analysis

Applying KVL to mesh 1,

(8+j10-j2)I1-(-j2)I2-j10I3=0

Mesh 2,

(4-j2-j2)I2-(-j2)I1-(-j2)I3+20angle90=0

Mesh 3, I3=5 We must substitue this in equation 1 and 2,

(8+j8)I1+j2I2=j50

j2I1+(4-j4)I2=-j20-j10

Put it in matrix form as

   | 8+j8            j2  |     I1=|j50|

         |         j2         4-j4|  I2=|-j30 |

from wiche we obtain the derminants

delta=68

delta 2=416.17angle -35.22

where I2=Delta2/delta=416.17angle-35.22/68

I2=6.12angle-35.22 A      

Supermesh

-supermesh analysis is a better technique instead of using mesh analyis such a complex electric circuit where two meshes have a current source as a common element. This is the same with supernode analysis which we also use when voltage source has a common node in the circuit.

NODAL ANALYSIS IN AC CIRCUIT

NODAL ANALYSIS

The basis of nodal analysis is Kirchoff's current law. Since KCL is valid for phasors, we can also analyze ac circuits using nodal analysis.

This are some steps to analyze AC circuits:

1. Transform the circuit to the phasor or frequency domain.

2.Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc)

3. Transform the resulting phasor to the time domain.

Example:

Find Ix in the circuit using nodal analysis

We first convert the circuit to the frequency domain:

20cos4t    => 20∠0 , Ѡ=4rad/s

1H   =>   JѠL=J4

0.5H  => JѠL=J2

0.1F => 1/JѠC= -j2.5

The frequency domain equivalent circuit is as shown in figure

Applying KCL at node A

20-Va/10= Va/-j2.5 + Va-Vb/J4

or

(1+J1.5)Va+J2.5Vb=20

At node B

2Ix+Va-Vb/J4=Vb/J2

Ix=Va/-j2.5 then substitute

2Va/-j2.5+Va-Vb/J4=Vb/j2

By simplifying, we get

11Va+15Vb=0

Then Equation 1 and equation 2 can be put in matrix form as shown 

|1+j1.5              j2.5| |v1|=|20|

           |11                  11   |  |v2|=|0|             

where delta =15-j5 as i computed using matrix form

and Delta1 equals to -220 as i computed using cramers rule.

where v1=delta1/delta=300/15-j5=18.97angle18.43 V

v2=delta2/delta=-220/15-j5=13.91angle198.3 V

The current Ix is given by

Ix=V1/-j2.5=18.97angle18.43/2.5angle-90

=7.59angle108.4 A

Transforming this to the time domain,

ix=7.59cos(4t+108.4) A

PHASORS

PHASORS

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


Phasors provide a simple means of analyzing linear circuits excited by sinusoidal sources; solutions of such circuits would be intractable otherwise. The notion of solving ac circuits using phasors was first introduced by charles Steinmetz in 1893. Before we completely define phasors and apply them to circuit analysis, we need to be throughly familiar with complex numbers.

A  complex number z can be written in rectangular form as:

                                             z=x+jy

where j=√-1; x is the real part of z; y is the imaginary part of z. In this context, the variables x and y do not represent a location as in two-dimensional vector analysis but rather the real and imaginary parts of z in the complex plane. Nevertheless, we note that there are some resemblances between manipulating complex numbers and manipulating two-dimensional vectors.

The complex number z can also be written in polar or exponential form as:

                                         z=r∠Φ
where r is the magnitude of z, and Φ is the phase of z. We notice that z can be represented in three ways:

                z=x+jy                     z=r∠Φ               z=re^jΦ



the relationship between the rectangular form and the polar form is shown in fig, 9.6, where the x axis represents the real part and the y axis represents the imaginary part of a complex number. Given x and y, we can get r amd Φ as

                              r=√󠄈x^2+y^2,                Φ=tan^-1 (y/x)



On the other hand, if we know r and Φ, we can obtain x and y as

                            x=rcosΦ             y=rsinΦ

Mathematical operations of complex numbers:

ADDITION

Z1+Z2=(X1+X2) + J(y1+y2)

SUBTRACTION

z1-z2=(x1-x2)+(y1-y2)

MULTIPLICATION

z1z2= r1r2∠Φ1+Φ2

DIVISION

z1/z2=r1/r2∠Φ1-Φ2

RECIPROCAL

1/z=1/r∠-Φ

SQUARE ROOT

square root of z=square root of r∠Φ/2

COMPLEX CONJUGATE

z*=x-jy=r∠-Φ

      

                                          

SINUSOIDS AND PHASORS

SINUSOIDS AND PHASORS

Sinusoids

- A Sinusoid is a signal that has the form of the sine and cosine function. Sinusoidal variables are of special importance in electrical and electronic systems, not only because the occur frequently in such systems but also because any periodical signal can be represented as a linear combination of a set of sinusoidal signals of different frequencies, amplitudes, and phase angles.





Sinusoidal variable can be written as:




v(t) = Vm sin(ωt + Ф)

Where;

Vm = the amplitude of the sinusoid

ω = the angular frequency in radians/s

Ф = the phase









Example 1:

Given a sinusoid, i=3sin(3t + 70) , calculate its amplitude, phase, angular frequency

Solution:

Amplitude = 3, phase =70,angular frequency = 3




Example 2:

A sinusoidal current with a frequency of 60 Hz reaches a positive maximum of 20A at . Give the expression of this current as a function of time .
Solution:
We have , , , . As cosine function reaches maximum when (or ), the phase angle should satisfy where and , i.e.,














The current is














Alternatively, the phase angle can be found as shown below:














Solving this we get , same as above.

Example 3:

Find the phase angle between i1=-4sin(377t + 25) and i2=5cos(377t – 40), does i1 lead or lag i2?

Solution:

Add 180 @ i1:

i1=-4sin(377t + 25 + 180)

i1=4cos(377t + 205)

Change: cos to sin (add 90)

Since sin(ωt + 90) = cosωt

i2=5cos(377t – 40+90)

i2=5cos(377t +50)

therefore, i1 leads i2 155o(Phase Angle)
Complex Numbers

The mathematics used in Electrical Engineering to add together resistances, currents or DC voltages uses what are called “real numbers”. But real numbers are not the only kind of numbers we need to use especially when dealing with frequency dependent sinusoidal sources and vectors. As well as using normal or real numbers, Complex Numbers were introduced to allow complex equations to be solved with numbers that are the square roots of negative numbers, √-1.

In electrical engineering this type of number is called an “imaginary number” and to distinguish an imaginary number from a real number the letter “ j ” known commonly in electrical engineering as the j-operator, is used. The letter j is placed in front of a real number to signify its imaginary number operation. Examples of imaginary numbers are: j3, j12, j100 etc. Then a complex number consists of two distinct but very much related parts, a “ Real Number ” plus an “ Imaginary Number ”.

CAPACITOR AND INDUCTOR

CAPACITOR AND INDUCTOR

CAPACITOR
-A capacitor is a two-terminal, electrical component. Along with resistors and inductors, they are one of the most fundamental passive components we use. You would have to look very hard to find a circuit which didn’t have a capacitor in it.

A capacitor, we are familiar with this thing because we use to it into our power supply. We use capacitor to stored energy in our power supply.

INDUCTOR
-An inductor is a passive electronic component that storesenergy in the form of a magnetic field. In its simplest form, an inductor consistsof a wire loop or coil. The inductance is directly proportional to the number ofturns in the coil.

The standard unit of inductance is the henry.

In a capacitor, the formula is expressed as “the current in a circuit is in proportion to the time rate of change of the voltage across it.

written as:

i=c(dv/dt)

In series and parallel capacitors, they are combined in the same way as conductances.

An inductor formula states that the voltage across it is directly equitable to the rate of change of the current through the circuit.

The formula is written like this:

v=L(di/dt)

In solving inductors in series and parallel, it is the same as combining the resistors in the circuit, which is in series or parallel.

FIRST ORDER CIRCUITS: SOURCE FREE RC AND RL CIRCUIT

FIRST ORDER CIRCUITS: SOURCE FREE RC AND RL CIRCUIT

FIRST ORDER CIRCUIT
-A circuit that can be simplified to a Thevenin's or Norton's equivalent connected to either a single equivalent inductor or capacitor.

THE SOURCE-FREE RC CIRCUIT
-an electric circuit composed of resistors and capacitorsdriven by a voltage or current source. A first order RC circuit is composed of one resistor and one capacitor and is the simplest type of RC circuit.

IN SOLVING THE RC CIRCUIT

The time required for the voltage to fall to is called the RC time constant and is given by

SOURCE FREE RL CIRCUIT
-is an electric circuit composed of resistors and inductorsdriven by a voltage or current source. A first order RL circuit is composed of one resistor and one inductor and is the simplest type of RL circuit.

IN SOLVING THE RL CIRCUIT

SAMPLE RL CIRCUIT

MAXIMUM POWER TRANSFER

MAXIMUM POWER TRANSFER THEOREM

maximum power transfer theorem states that, to obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals. 

(RTH=RL)

PROVE THAT RL=RTH

Computing the Power Transfer is equal to V2/4(RL) 

THEVENIN'S AND NORTON'S THEOREM

THEVENIN'S  AND NORTON'S THEOREM

THEVENIN'S THEOREM

-Thevenin’s Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. It is especially useful in analyzing power system and another circuits where one particular resistor in the circuit called the load resistor is subject to change and re-calculation of the circuit is necessary with each trial value of load resistance to determine voltage and current across it.

It is also states that the linear two terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTH in a series resistor RTH where Vth is the open circuit voltage at terminals and Rth is the input or total resistance at the terminals when the independent source are turn off.



There are two cases you need to consider in thevenin's:

Case 1: If the network has no dependent source therefore you need to turn off all independent source. Rth can be computed via parallel or series connection seen from the given.

Case 2: If the network has dependent source turn off all dependent source. Apply V0 at A-b and Also Io at A-b.


Thevenins Equivalent Circuit:

NORTON'S THEOREM

-Any collection of batteries and resistances with two terminals is electrically equivalent to an ideal current source i in parallel with a single resistor r. The value of r is the same as that in the Thevenin equivalent and the current i can be found by dividing the open circuit voltage by r.

Norton's Circuit

Norton's Equivalent Circuit:

SAMPLE PROBLEM: