1. INC 111 Basic Circuit Analysis
Week 12
Complex Power
Complex Frequency

2. Example
Find i(t)
ω = 2

3. Use superposition
i(t) can be found from current divider

4. i(t) can be found from current divider

6. Average Power in AC circuits
From instantaneous power
In 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 of
instantaneous power in a periodic signal divided by time.

7. Let v(t) in the form
Change variable of integration to θ
We got
Then find the instantaneous power
integrate from 0 to 2π (1 period)

8. Compare with power from DC voltage sourceAC

9. 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
For sine wave,

10. 3 ways to tell voltage
V (volts)
311V
t (sec)
V peak (Vp) = 311 V
V peak-to-peak (Vp-p) = 622V
V rms = 220V

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

12. Capacitors and Inductors do not have average power although there are voltage and current.
Therefore, reactive power (Q) is defined

13. Complex Power Power can be divided into two parts: real and imaginary
Complex power S = P + jQ
P = real power Q = Reactive power
Inductor has no real power P =0
But it has complex power, computed by V, I that are perpendicular
to each other.

14. Q S
Complex power = S is a vector θ P
where
θ is the angle difference of (V-I)
or (S-P)
for sine waves

15. Phasor Diagram of an inductor
Phasor Diagram of a resistor v v i i
Power = (vi cosθ)/2 = 0
Power = (vi cosθ)/2 = vi/2
Note: No power consumed in inductors
i lags v

16. Phasor Diagram of a capacitor
Phasor Diagram of a resistor i v v i
Power = (vi cosθ)/2 = 0
Power = (vi cosθ)/2 = vi/2
Note: No power consumed in capacitors
i leads v

17. Power Conservation Theorem
“The sum of real power and reactive power in a circuit is equal to zero”
This is because power cannot disappear. It can only change form.

18. Example
Find i(t), vL(t)

19. Phasor Diagram VL V
θ = 77.47 I VR
1. Resistor consumes power
2. Inductor consumes no real power P = 0 but it has reactive power

20. 3. Voltage source supplies power
Conservation of energy holds
Resistor 2 ohm
Inductor 3H
Voltage source

21. Power Factor
Power factor of a source is the ratio of real power to the complex power
Consider the voltage source, Q S
Complex Power Diagram
Power factor = 0.217
θ = 77.47 P
The voltage source supplies 0.292W real power and 1.318VAR reactive power.

22. Complex Frequency The theory of phasor can be extended to support
signal
is a fundamental waveform of electrical engineering
where s can be a complex number in a rectangular form
It is composed of real and imaginary parts.
Consider only real part

23. If σ = 0
The voltage source will be in the format that we already know.
If σ is not 0, we get a new form of signal.
Define
s is called “complex frequency”

24. 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.

25. Example
Find i(t), ic(t)

26. We then substitute s = -2+j4 and got

27. ic(t) can be computed from current divider

28. Complex Frequency Characteristics1 1 1