Complex Power

COMPLEX POWER

Complex power is the product of voltage peak (Vm) and the conjugate(*) of the current divided by two.

or Vrms multiply by Irms.

symbol is bold S or “S“ and the unit is VA.

S = V . I*

V is the phasor representation of voltage and I* is the conjugate of current phasor.

So if V is the reference phasor then V can be written as |V| ∠0.
(Usually one phasor is taken reference which makes zero degrees with real axis. It eliminates the necessity of introducing a non zero phase angle for voltage)

Let current lags voltage by an angle φ, so

I = | I | ∠-φ

(current phasor makes -φ degrees with real axis)

I*= | I | ∠φ

So,
S = |V| | I | ∠(0+φ) = |V| | I | ∠φ

(For multiplication of phasors we have considered polar form to facilitate calculation)

Writting the above formula for S in rectangular form we get

S = |V| | I | cos φ + j |V| | I | sin φ

The real part of complex power S is |V| | I | cos φ which is the real power or average power and the imaginary part |V| | I | sin φ is the reactive power.

So,

S = P + j Q

Where

P = |V| | I | cos φ and Q = |V| | I | sin φ

In this topic, I've learned that resistors absorbs the real power and dissipates in the form of heat. Capacitors absorbs the reactive power and dissipates in the form of electric. Inductor absorbs the reactive power and dissipates in the form of magnetic field. Using this topic complex power, we can determine if it is inductive, resistive or capacitive. In complex power, we have to consider the phase network which voltage and current represented in complex form.

Apparent and Power Factor

APPARENT POWER
(active + reactive)

Apparent power is a power which is easy to identify, easy to see or to know. Not literally that we see but we can measure the power.
Apparent power is a measure of alternating current (AC) power that is computed by multiplying the root-mean-square (rms) current by the root-mean-square voltage. In a direct current (DC) circuit, or in an AC circuit whose impedance is a pure resistance, the voltage and current are in phase.
Apparent power is measured in units’ volt-ampere (VA).
Apparent power symbol by/represented by “S”.

POWER FACTOR

Power factor is defined as the ratio of real power (P) to apparent power (S). This definition is often mathematically represented as kW/kVA, where the numerator is the active (real) power and the denominator is the (active+ reactive) apparent power or simply PF = P/S

PF = cos angle of PF= cos angle of(angle of voltage -angle of current)

In sinusoid, PF is cosine of the phase difference between the voltage and current.

In this topic, I find apparent power easy to solve for a reason that it is easy to found/measure. Apparent will be divided into two power which is the reactive and real power that has their own function.

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 Z_Th:

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

b.) The average power:

P_max = 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 * X_C = - 50*2 = - 100 mvar

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

V_L = 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.