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### Transcript of INC 111 Basic Circuit Analysis Week 12 Complex Power Complex Frequency

• INC 111 Basic Circuit AnalysisWeek 12 Complex PowerComplex Frequency

• ExampleFind i(t) = 2

AC

50cos(2t)

+

-

2.5H

3

i(t)

5

2H

50sin(2t)

-

+

AC

5090

+

-

j5

3

i(t)

5

j4

-

+

500

• Use superpositioni(t) can be found from current divider

AC

5090

+

-

j5

3

i(t)

5

j4

i1(t)

• i(t) can be found from current divider

AC

j5

3

i(t)

5

j4

-

+

500

i2(t)

• Average Power in AC circuitsIn AC circuits, voltage and current values oscillate.This makes the power (instantaneous power) oscillate as well.

However, electric power is best represented as one value.Therefore, we will use an average power.

Average power can be computed by integration ofinstantaneous power in a periodic signal divided by time.From instantaneous power

• Let v(t) in the formChange variable of integration to We gotThen find the instantaneous powerintegrate from 0 to 2 (1 period)

• Compare with power from DC voltage sourceDCAC

AC

Asin(t+)

+

-

R

i(t)

AC

A

+

-

R

i(t)

• Root Mean Square Value (RMS) In DC circuits,In AC, we define Vrms and Irms for convenience to calculate power.Vrms and Irms are defined such that,Note: Vrms and Irms are constant all the time

• V (volts)t (sec)311VV peak (Vp) = 311 VV peak-to-peak (Vp-p) = 622VV rms = 220V 3 ways to tell voltage0

• Reactive PowerCapacitors and inductors have average power = 0 because they have voltage and current with 90 degree phase difference.Change variable of integration to Then integrate from 0 to 2 (1 period)

• Capacitors and Inductors do not have average power although there are voltage and current.

Therefore, reactive power (Q) is defined

• Complex PowerPower can be divided into two parts: real and imaginaryComplex power S = P + jQP = real powerQ = Reactive powerInductor has no real power P =0But it has complex power, computed by V, I that are perpendicularto each other.

• Complex power = S is a vectorwherefor sine wavesSPQ is the angle difference of (V-I)or (S-P)

• Phasor Diagram of an inductorviPower = (vi cos)/2 = 0Phasor Diagram of a resistorviPower = (vi cos)/2 = vi/2Note: No power consumed in inductorsi lags v

• Phasor Diagram of a capacitorviPower = (vi cos)/2 = 0Phasor Diagram of a resistorviPower = (vi cos)/2 = vi/2Note: No power consumed in capacitorsi leads v

• Power Conservation TheoremThe sum of real power and reactive power in a circuit is equal to zeroThis is because power cannot disappear. It can only change form.

• ExampleFind i(t), vL(t)

AC

5sin(3t+/3)

+

-

3H

i(t)

2

+ vR(t) -

+vL(t)-

AC

560

+

-

j9

i(t)

2

• VIVRVLPhasor Diagram1. Resistor consumes power2. Inductor consumes no real power P = 0 but it has reactive power = 77.47

• 3. Voltage source supplies powerConservation of energy holdsResistor 2 ohmInductor 3HVoltage source

• Power FactorPower factor of a source is the ratio of real power to the complex powerSPQ = 77.47Complex Power DiagramConsider the voltage source,The voltage source supplies 0.292W real power and 1.318VAR reactive power.Power factor = 0.217

• Complex Frequencyis a fundamental waveform of electrical engineeringwhere s can be a complex number in a rectangular formIt is composed of real and imaginary parts.The theory of phasor can be extended to supportsignal

• Defines is called complex frequencyIf = 0

The voltage source will be in the format that we already know.If is not 0, we get a new form of signal.

AC

sint

+

-

AC

etsint

+

-

• Summary of Procedures Change voltage/current sources in to phasor form

Change R, L, C value into phasor form Use DC circuit analysis techniques normally, but the value of voltage, current, and resistance can be complex numbers

Change back to the time-domain form if the problem asks.

R

L

C

R

sL

1/sC

• ExampleFind i(t), ic(t)

AC

20e-2tcos(4t+/2)

+

-

1H

0.1f

i(t)

5

ic(t)

• We then substitute s = -2+j4 and got

AC

2090

+

-

s

1/0.1s

i(t)

5

ic(t)

• ic(t) can be computed from current divider

• Complex Frequency Characteristics111