Effective RMS Values

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I’ve learned that RMS, or Root Mean Square, is the measurement used for any time varying signal's effective value. As they said it is not an "Average" voltage.  Its mathematical relationship to peak voltage varies depending on the type of waveform. By definition, RMS Value, also called the effective or heating value of AC, is equivalent to a DC voltage that would provide the same amount of heat produced in a resistor.

Apparent Power and Power Factor

Complex power

     Complex power       

                                    factor has been studied and  formalized for a very 

                                   long time just to fined the total effect of Parallel loads. 

                                    It is very important in power analysis because it 

                                    contains all the  information pertaining to the 

                                     power absorbed by a given load. 

S = 1/2 VI*   or    S= Vrms Irms

 and

S= Vrms Irms  ∠θv - θi

                                         But complex power maybe expressed in terms of the load impedance Z so we may write

                                       
Example 

                                          complex power contains all the relevant 
                                          power  information in a given load.

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I’ve learned that complex power is very conceptual and essential to understand.

I’ve learned that Complex numbers can help us in acquiring all loads. And these are, resistor absorbs the real power and dissipates in the form of heat and light. Inductor absorbs the reactive power and dissipates in the form of magnetic field. Capacitor absorbs the reactive power and dissipates in the form of electric or electrostatic filed. Using complex power we can easily determine if it is inductive, capacitive or it is resistive loads. For expressing the complex power we have to first consider a single phase network whose voltage and current can be represented in complex form.

Maximum Average Power Transfer

We can only say that we are transferring the maximum average power if the load resistance is equivalent to the thevenin’s resistance where RL = Rth. This is applicable in AC circuit.

In AC circuit the Impedance should also be equal to the output impedance by using thevenin. If this will not achieved, then we cannot transfer the Maximum average power to our load.

Let’s consider circuit below and proved that they must be equal..

                  

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 I’ve learned that we have to inspect carefully the loads if it is match in our circuit. Because if not, maximum power transfer to the load may not happen.

          I’ve learned that the approach in the DC analysis is the same in AC analysis there are some that are not especially in time. But I found out that AC analysis is harder than DC analysis. But, if you fully understand analyzing DC circuits there’s no problem with the next steps.

Instantaneous and Average Power

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I've learned that in solving power, there are many different ways and different formula but comes up with the same answer. I also learned that resistor is the resistive load which also absorbs power from the source. This means that most loads in our home are resistive because we are paying bills for the power we consumed.

I've learned that in AC source capacitors are able to store Power and releases power. They are not like resistive loads that absorb power and dissipate power in the form of heat. 

Thevenin's and Norton Theorem

Thevenin's Theorem- states that any network of voltage sources and resistors can be reduced to a single voltage source and a single resistor which are in SERIES with each other.

Example problem;

             In the circuit above we are required to obtain the thevenin's circuit equivalent.

thevenin's circuit equivalent

                           To achieve this goal, on the circuit given above we must acquire Vth and Zth. To do this, in getting Zth we first kill all sources present in the circuit. We have

So let's apply the series-parallel combination.

* -  means series

//  means parallel

j2 Ω * 6 Ω

Z1 = 6 + j2  Ω

Z1// -j4

Z2 = (6+j2)(-j4)   /    (6+j2)+(-j4)

Z2 = 12/5   -  j16/5   Ω

Z2 * 10 Ω

Zth = 62/5 - j16/5 Ω

 next is to solve for Vth. let's use again our 1st circuit.

                 In solving Vth we can apply any analysis that is fit to the circuit. We will use KVL.

75∠ 20 = 6 I + j2 I + (-j4) I

75∠ 20 = I (6  + j2  + (-j4))

I =  11.86∠ 38.43 A

Vth= I x Z (capacitor)

Vth =  11.86∠ 38.43 x  (-j4)

Vth = 47.43∠ -51.565 

What happen to 10  Ω resistor?

Our circuit on the right side is open. one terminal of our resistor is hanged. Therefore there is no current present across this resistor.

our thevenin equivalent circuit is :

                  I've learned that in solving circuit using thevenin's theorem in AC circuit it has the same application as we solve the dc circuits.The difference is the application of complex numbers. 

Source Transformation

Source Transformation

In source transformation we have to combine the impedances or simply the circuit. But remember that we must leave the branch which is the required parameter is present on it.

Also in source transformation we will transform the voltage source in series with impedance to a current source in parallel with impedance or vise-versa. The transformation of the source with impedance will only use the ohm's Law

Vs = Zs x Is and Is = Vs / Zs.

Illustration:

           

Example:

                               Vx is our unknown in the given circuit. We have to simplify this circuit to obtain Vx.

* means series

3Ω *  j4 Ω

Z1 =  3 + j4   Ω

4 Ω * -j13 Ω

Z2 = 4 + -j13 Ω

As we can see from the circuit, It has a voltage source with series resistor. We will transform it into a current source with a resistor as what illustration above shows. Using ohm's Law we have;

I = (20∠-90)   /   5   =  -j4 A

                          

5 Ω // (3+ j4 Ω)

Z3 = (3+ j4) (5) /   (3+ j4) + (5)

Z3 = 2.5 + j1.25 Ω

V =   (-j4) x (2.5 + j1.25)

V =  5 - j10 V

                           

   (2.5 + j1.25 Ω) *  (4 + -j13 Ω)

  Z4 =  (2.5 + j1.25) + (4 + -j13) 

  Z4 = 6.5 - j11.75 Ω

By this time, we will apply voltage division.

Vx = ( Zx ) (Vs)    /    (Zx) + (Z)

Vx = ( 10 ) (5 - j10)  /   ( 10 )  (6.5 - j11.75)

Vx =  5. 519 ∠ -27.98 V

I've learned that in solving unknown in Ac circuit, we can also use source transformation. For me this is much easier to use than to the other in simplifying the circuit because it lessen your time making solutions. It gives you many illustrations how the circuit changes while solving the problem. Ohm's Law is the most formula we used in here so it is quiet easy. Source transformation can be use for checking your circuit if you are confused.