Instantaneous & Average Power

Instantaneous & Average Power

Instantaneous power is the power at any instant of time. In Ac circuit the instantaneous electric power is given by

P= VI

but these quantities are continuously varying. Instantaneous Power p(t) is the power, p(t)= u(t)*i(t). It is the product of the time functions of the voltage and current. 

This definition of instantaneous power is valid for signals of any waveform. The unit for instantaneous power is VA. 

Where; P(t) = 1/2 VmIm cos(ɸv-ɸi) + 1/2 VmIm cos(2wt + ɸv + ɸi)

Instantaneous power is at positive values if power is absorbed by the circuit. It will be on the negative values if power is absorbed by the source. 

Average power is the average of instantaneous power over one period. Defined as;

P = 1/2 VmIm cos(ɸv-ɸi) 

Characteristics

1. P IS NOT TIME DEPENDENT

2. ɸv=ɸi, purely resistive load

3. ɸv-ɸi = 90, purely reactive load. 

THEVENIN EQUIVALENT CIRCUITS WITH DEPENDENT SOURCES

Thevenin's Equivalent circuits with dependent sources

Procedures in Solving Thevenin's Theorem with dependent sources:

1. Determine Vth the usual way

2. Determine Zth

3. Remove all independent sources

4. Place an assumed value of voltage at the open circuit

5. Determine the current supplied by the assumed voltage source

6. Solve for the Zth using the formula where Zth=Vo/Io

Example:

Find the thevenin equivalent of the circuit in the figure below as seen from therminals a-b.

To find Vth we apply KCL at node 1 in the figure below

15=Io+ O.5Io  => Io=10A

Applying KVL to the loop on the right-hand side in the figure above we obtain

-Io(2-j4)+0.5Io(4+j3) + Vth=0

or

Vth=10(2-j4) - 5(4+j3) = -j55

Vth= 55angle -90 V

To Obtain Zth, we remove the independent source. Due to the presence of the dependent current source, we connect a 3-A current source (3 is an arbitrary value chosen for convenience here, a number divisible by the sum of currents leaving the node) to terminals a-b as shown in figure b at the node, KCL gives

3=Io+0.5Io  =>  Io=2A

Applying KVL to the outer loop in figure b gives

Vs=Io(4+j3+2-j4)=2(6-j)

The Thevenin impedance is

Zth= Vs/Is =2(6-j)/3 =4-j0.6667 ohms

Thevenin and Norton Equivalent Circuits AC circuit

Thevenin and Norton Equivalent Circuits

Thevenin's and Norton's theorems are applied to ac circuits in the same way as they are to dc circuits. The only additional effort arises from the need to manipulate complex numbers. The frequency domain version of a thevenin equivalent circuit is depicted where a linear circuit is replace by a voltage source in series with an impedance. The norton equivalent is a linear circuit replaced by a current source in parallel with an impedance. Two equivalent circuits are related as

The thevenin or Norton equivalent circuit must be determined at each frequency. This leads to entirely different equivalent circuits, one for each frequency, not one equivalent circuit with equivalent sources and equivalent impedances.

Example:

Obtain the thevenin equivalent at terminals a-b of the circuit 

We must find Zth by setting the voltage source to zero. As shown in the figure below, the 8ohm resistance is now in parallel with the -j6 reactance, so that their combination gives

Z1=-j6ll8=-j6(8)/8-j6=2.88-j3.84ohms

Similarly, the 4ohm resistance is in parallel with the j12 reactance and their combination gives

z2=4llj12=j12(4)/4+j12=3.6+j1.2ohms

The thevenin impedance is the series combination of Z1 and Z2 that is:

Zth=Z1+Z2=6.48-j2.64ohms

To find Vth consider the circuit in the figure below. Currents I1 and I2 are obtained as

I1=120angle75/8-j6 A

I2=120angle75/4+j12 A

Applying KVL around loob gives

Vth-4I2+(-j6)I1=0

or

Vth=4I2+j6I10=(480angle75)/4+j12)+(720angle75+90)/8-j6

=37.95angle3.43+72angle201.87

Vth=37.95angle220.31 V

Example:

Obtain current Io using Norton's Theorem

Our first objective is to find the Norton equivalent at terminals a-b. Zn is found in the same way as Zth. We set the sources to zero as shown in the figure above. As evident from the figure, the (8-j2) and (10+j4) impedances are short-circuited, so that 

Zn=5ohms

To get In, we short-circuit terminals a-b as in figure below and apply mesh analysis. Notice that meshes 2 and 3 form a supermesh because of the current source linking them. 

For Mesh 1,

-j40+(18+j2)I1-(8-j2)I2-(10+j4)I3=0

(a) finding Zn,   (b) finding Vn,  (c) Calculating Io

For the supermesh,

(13-j2)I2+(10+j4)I3-(18+j2)I1=0

at node a, due to the current source between meshes 2 and 3,

I3=I2+3

Adding Equations gives

-j40+5I2=0  => I2=j8

From eq where 

I3=I2+3=3+j8

The norton current is 

IN=I3=(3+j8) A

Io= 5/5+20+j15 IN = 3+j8/5+j3=1.465angle 38.48 A

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