Saturday, February 8, 2014

Maximum Average Power Transfer

Maximum Average Power Transfer

The 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.


RL= Re{ZTh} and XL = - Im{ZTh}

Where V2Th and I2N represent the square of the sinusoidal peak values.

We’ll next illustrate the theorem with some examples.

Example 1

R1 = 5 kohm, L = 2 H, vS(t) = 100V cos wt, w = 1 krad/s.

a) Find C and R2 so that the average power of the R2-C two-pole will be maximum











b) Find the maximum average power and the reactive power in this case.

c) Find v(t) in this case.

The solution by the theorem using V, mA, mW, kohm, mS, krad/s, ms, H, m F units:v

a.) The network is already in Thévenin form, so we can use the conjugate form and determine the real and imaginary components of ZTh:

R2 = R1 = 5 kohm; wL = 1/w C = 2 ® C = 1/w2L = 0.5 mF = 500 nF. 

b.) The average power: 

Pmax = V2/(4*R1) = 1002/(2*4*5) = 250 mW

 The reactive power: first the current: 

I = V / (R1 + R2 + j(wL – 1/wC)) = 100/10 = 10 mA
Q = - I2/2 * XC = - 50*2 = - 100 mvar

c.) The load voltage in the case of maximum power transfer: 

VL = I*(R2 + 1/ (j w C ) = 10*(5-j/(1*0.5)) =50 – j 20 = 53.852 e -j 21.8° V
and the time function: v(t) = 53.853 cos (wt – 21.8°) V




In this topic, I noticed that the way we solve DC Circuits compared to AC Circuits is the same but there are some that is not.  In solving problems with regards to Maximum average Power Transfer is that we really have to inspect carefully the loads so that we can get correctly the needed unknown and Maximum Power transfer will occur.

EFFECTIVE or RMS VALUE

EFFECTIVE or RMS VALUE
  • We have a few different ways to specify the size of an ac current or voltage.
  • We can give either
  • the peak value, or
  • the peak-to-peak value, or
  • Something called the effective value (also called rms value).
  • These distinctions apply only to ac, not to dc.

In the previous lesson/experiment I learned how to use the multimeter to measure voltages and currents in dc circuits and how to use the oscilloscope to measure the peak voltage or peak-to-peak voltage of an AC waveform. When we used to measure AC voltages or currents, the multimeter gives us something called the effective value, or rms value.
The root-mean-square (rms) value or effective value of an AC waveform is a measure of how effective the waveform is in producing heat in a resistance.
The rms value is am constant itself which depending on the shape of the function i(t).
The Effective value or rms value of an AC waveform is an equivalent DC value.
Example: If you connect a 5 Vrms source across a resistor, it will produce the same amount of heat as you would get if you connected a 5 V dc source across that same resistor. On the other hand, if you connect a 5 V peak source or a 5 V peak-to-peak source across that resistor, it will
Not produce the same amount of heat as a 5 V dc source.
That’s why rms (or effective) values are useful: they give us a way to compare ac voltages to dc voltages.
To show that a voltage or current is an rms value, we write rms after the unit: for example, Vrms = 25 V rms.
P=1/2 VmIm cos(angle of voltage – angle of current)or= Vrms I rms cos(angle of voltage – angle of current)
Resistive load only:
True power, reactive power, and apparent power for a purely resistive load.                                                                                                                        
In this topic, I've learned that RMS (Root Mean Square) is used to measure varying signals effective value and the mathematical relationship to peak voltage do varies depending on the type of waveform.

Superposition Theorem

Superposition Theorem
The superposition theorem eliminates the need for solving simultaneous linear equations by considering the
effect on each source independently.

To consider the effects of each source we remove the remaining sources; by setting the voltage sources to
zero (short-circuit representation) and current sources to zero (open-circuit representation).


The current through, or voltage across, a portion of the network produced by each source is then added 
algebraically to find the total solution for current or voltage.

The only variation in applying the superposition theorem to AC networks with independent sources is that we
will be working with impedances and phasorsinstead of just resistors and real numbers.

The superposition theorem is not applicable to power effects in AC networks since we are still dealing with a 
nonlinear relationship.

It can be applied to  networks with sources of different frequencies only if the total response for each
frequency is found independently and the results are expanded in a nonsinusoidal expression .

One of the most frequent applications of the superposition theorem is to electronic systems in which the DC and AC analyses are treated separately and the total solution is the sum of the two.

When a circuit has sources operating at different frequencies,
•The separate phasor circuit for each frequency must be solved independently, and
•The total response is the sum of time-domain responses of all the individual phasor circuits.  

Sample Problem using Superposition.
















I learned that superposition Theorem  to AC Networks with independent sources is that we are dealing with
impedances  and phasors instead of having real numbers. 

Saturday, January 11, 2014

Thevenin and Norton Equivalent Circuit

Thevenin's & Norton Equivalent Circuit

Thevenin's Equivalent Theorem

- Thévenin’s theorem, as stated for sinusoidal AC circuits, is changed only to include the term impedance instead of resistance.

- Any two-terminal linear ac network can be replaced with an equivalent circuit consisting of a voltage source and an impedance in series.

- VTh is the Open circuit voltage between the terminals a-b.

- ZTh is the impedance seen from the terminals when the independent sources are
set to zero.

**Take note that: VTh is the Open circuit voltage between the terminals a-b while ZTh is the impedance seen from the terminals when the independent sources are set to zero**






Norton's Equivalent Theorem

 - The linear circuit is replaced by a current source in parallel with an impedance. IN is the Short circuit current flowing between the terminals a-b when the terminals are short circuited. 

 - In Norton’s theorem, it states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source Iin parallel with a resistor RN, where Iis the short-circuit current through the terminals and RNis the input or equivalent resistance at the terminals when the independent sources are turned off.



 









Thevenin and Norton equivalents are related by:







In Thevenin's and Norton's Equivalent Theorem, This is good if we are going to put concentration on a particular part of a circuit so that in other part in other way around, The circuit can be replaced by using Thevenin's equivalent.

Source Transformation

Source Transformation

Transform a voltage source in series with an impedance to a current source in parallel with an impedance for simplification or vice versa.













Practice Problem : Calculate the current Io.











If we transform the current source to a voltage source, we obtain the circuit





















In source transformation, I observed and i learned that using this theorem will simplify the circuit given for us to be able to convert it into more easier form that it can be, In such way, The circuit will appear simple. 


Friday, December 6, 2013

Nodal and Mesh Analysis of Phasor Circuit


Nodal Analysis
  • Since KCL is valid for phasors, we can analyze AC circuits by NODAL analysis.
  • Determine the number of nodes within the network.
  • Pick a reference node and label each remaining node with a subscripted value of voltage: V1, V2 and so on.
  • Apply Kirchhoff’s current law at each node except the reference. Assume that all unknown currents leave the node for each application of Kirhhoff’s current law.
  • Solve the resulting equations for the nodal voltages.
  • For dependent current sources: Treat each dependent current source like an independent source when Kirchhoff’s current law is applied to each defined node. However, once the equations are established, substitute the equation for the controlling quantity to ensure that the unknowns are limited solely to the chosen nodal voltages.
Since KCL is valid for phasors, we can analyze AC circuits by NODAL analysis.

Sample Problem no.1 :
Find v1 and v2 using nodal analysis
















- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


Mesh Analysis

Since KVL is valid for phasors, we can analyze AC circuits by MESH analysis.


Sample Problem 2:
Calculate the current Io




















Saturday, November 16, 2013

Phasors

Phasor Diagrams

In AC electrical theory every power source supplies a voltage that is either a sine wave of one particular frequency or can be considered as a sum of sine waves of differing frequencies. The neat thing about a sine wave such as V(t) = Asin(ωt + δ) is that it can be considered to be directly related to a vector of length A revolving in a circle with angular velocity ω - in fact just the y component of the vector. The phase constant δ is the starting angle at t = 0.



Since a pen and paper drawing cannot be animated so easily, a 2D drawing of a rotating vector shows the vector inscribed in the centre of a circle as indicated below. The angular frequency ω may or may not be indicated.





When two sine waves are produced on the same display, one wave is often said to be leading or lagging the other. This terminology makes sense in the revolving vector picture as shown in Figure 3. The blue vector is said to be leading the red vector or conversely the red vector is lagging the blue vector.



Considering sine waves as vertical components of vectors has more important properties. For instance, adding or subtracting two sine waves directly requires a great deal of algebraic manipulation and the use of trigonometric identities. However if we consider the sine waves as vectors, we have a simple problem of vector addition if we ignore ω. For example consider

Asin(ωt + φ) = 5sin(ωt + 30°) + 4sin(ωt + 140°) ;
the corresponding vector addition is:





Ax = 5cos(30° + 4cos(140°) = 1.26595Ay = 5sin(30° + 4sin(140°) = 5.07115
Thus

and




So the Pythagorean theorem and simple trigonometry produces the result
5.23sin(ωt + 76.0°) .
Make a note not to forget to put ωt back in!

Phasors and Resistors, Capacitors, and Inductors
The basic relationship in electrical circuits is between the current through an element and the voltage across it. For resistors, the famous Ohm's Law gives

VR = IR .                     (1)

For capacitors
VC = q/C .                 (2)

For inductors
VL = LdI/dt .             (3)

These three equations also provide a phase relationship between the current entering the element and the voltage over it. For the resistor, the voltage and current will be in phase. That means if I has the form Imaxsin(ωt + φ) then VR has the identical form Vmaxsin(ωt + φ) where Vmax = ImaxR. For capacitors and inductors it is a little more complicated. Consider the capacitor. Imagine the current entering the capacitor has the form Imaxsin(ωt + φ). The voltage, however, depends on the charge on the plates as indicated in Equation (2). The current and charge are related by I = dq/dt. Since we know the form of I simple calculus tells us that q should have the form − (Imax/ω)cos(ωt + φ) or (Imax/ω)sin(ωt + φ - 90°). Thus VC has the form (Imax/ωC)sin(ωt + φ - 90°) = Vmaxsin(ωt + φ - 90°). The capacitor current leads the capacitor voltage by 90°. Also note that Vmax = Imax/ωC. The quantity 1/ωC is called the capacitive reactance XC and has the unit of Ohms. For the inductor, we again assume that the current entering the capacitor has the form Imaxsin(ωt + φ). The voltage, however, depends on the time derivative of the current as seen in Equation (3). Since we assumed the form of I, then the voltage over the inductor will have the form ωLImaxcos(ωt + φ) or Vmaxsin(ωt + φ + 90°). The inductor current lags the inductor voltage. Here note that the quantity ωL is called the inductive reactance XL. It also has units of Ohms.
The phase relationship of the three elements is summed up in the following diagram,




Note that in all three cases, resistor, capacitor, and inductor, the relationship between the maximum voltage and the maximum current was of the form


Vmax = ImaxZ .                     (4)


We call Z the impedance of the circuit element.  Equation (4) is just an extension of Ohm's Law to AC circuits.  For circuits containing any combination of circuit elements, we can define a unique equivalent impedance and phase angle that will allow us to find the current leaving the battery.  We show how to do so in the next section.


Phasors and AC Circuit Problems
Phasors reduce AC Circuit problems to simple, if often tedious, vector addition and subtraction problems and provide a nice graphical way of thinking of the solution. In these problems, a power supply is connected to a circuit containing some combination of resistors, capacitors, and inductors. It is common for the characteristics of the power supply, Vmax and frequency ω, to be given. The unknown quantity would be the characteristics of the current leaving the power supply, Imax and the phase angle φ relative to the power supply. To solve one needs only to follow the rules:
  1. Circuit elements in parallel share the same voltage.
  2. Circuit elements in series share the same current.
  3. Do one branch of the circuit at a time.
  4. Maintain the phase relationships given in Figure 5.
  5. Use Ohm's Law V = IZ where Z is the equivalent impedance of any combination of circuit elements being considered.