AC POWER ANALYSIS

INSTANTANEOUS AND AVERAGE POWER

Instantaneous Power is a power at any instant of time. It is measured as watts.

p(t)=v(t)i(t) or

p(t)= VmImcos(ωt + θv)cos(ωt + θi)

Vm and Im is the amplitudes (or peak values) and the θv, θi are the phase

Using trigonometric Identity

p(t)= VmImcos(ωt + θv)cos(ωt + θv)= 1/2(VmImcos(θv-θi)+ VmIm(2ωt +θv+θi)

The instantaneous power entering the circuit

The instantaneous power is changing with respect to time so it is difficult to measure the power. The other method to solve the power is used the AVERAGE POWER formula, in fact the instrument used to measured the power is the wattmeter the used average power.

P = 1/2 VmImcos(θv-θi)

Thevenin’s and Norton Theorem

Any combination of batteries and resistances with two terminals can be replaced by a single voltage source e and a single series resistor r. The value of e is the open circuit voltage at the terminals, and the value of r is e divided by the current with the terminals short circuited.

 

STEPS

First is to solve for the Zth, thevenin’s impedance. The things to do is to turn off the sources ( short the voltage and open the current).

Second is to solve the Vth, thevenin’s voltage.  In the sample given above, Just solve the voltage on the R3, the R2 is cannot be used because it is open circuit, not connected to the circuit. The voltage across R3 is equal to the Vth.

 

Source Transformation AC ANALYSIS

Since ac circuit are linear, the Source Transformation theorem applies to ac circuits the same way it applies on the dc circuit. Source transformation is to transform the circuit into a simpler form. The formula used to transform the circuit is the Ohm’s law. 

STEPS:

First is to analyse the circuit how to attack it. 

To transform voltage to current source, the source must be series on the impedance and used ohm’s law. If it is current to voltage source the current source must be parallel to the impedance. (see below)  

 

 

After transforming, there is a time that the impedance can combine to the others (parallel or series) or the sources can be combine as one.   

Repeating the steps above until the circuit can be solved easily using (current or voltage division)

Just like below:

 

Circuit A the original circuit, Is is parallel to R so it transform the voltage source. After that i can be solved easily the V1 and V2 using voltage division formula.

 

 

          I learned that it is not all sources (voltage and current) be converted in the circuit. The first thing to do is to know what is the unknown or to understand the problem. Second is how to make the circuit simplify and where to used the source transformation. If the unknown is the current or voltage drop, do not touch that part or there is a dependent variable.