14: Power in AC Circuits - Electrical Engineering

14: Power in AC Circuits

• Average Power • Cosine Wave RMS • Power Factor + • Complex Power • Power in R, L, C • Tellegen’s Theorem • Power Factor Correction • Ideal Transformer • Transformer Applications • Summary


E1.1 Analysis of Circuits (2017-10213)

AC Power: 14 – 1 / 11

Average Power 14: Power in AC Circuits



AC Power: 14 – 2 / 11



Intantaneous Power dissipated in R: p(t) =


v 2 (t) R

AC Power: 14 – 2 / 11





v 2 (t) R

AC Power: 14 – 2 / 11




v 2 (t) R

Average Power dissipated in R:

P =


1 T

RT 0

p(t)dt

AC Power: 14 – 2 / 11



Intantaneous Power dissipated in R: p(t) = Average Power dissipated in R:

P =


1 T

RT 0

p(t)dt =

1 R

×

1 T

RT 0

v 2 (t) R

v 2 (t)dt

AC Power: 14 – 2 / 11




P =


1 T

RT 0

p(t)dt =

1 R

×

1 T

RT 0

2

v 2 (t) R

v (t)dt =

hv2 (t)i R

AC Power: 14 – 2 / 11




R 1 T T 0

1 R

1 T

RT

2

v 2 (t) R

hv2 (t)i

P p(t)dt = × v (t)dt = R 0

2= v (t) is the value of v 2 (t) averaged over time


AC Power: 14 – 2 / 11




R 1 T T 0

1 R

1 T

RT

2

v 2 (t) R

hv2 (t)i

P p(t)dt = × v (t)dt = R 0


We define the RMS Voltage (Root Mean Square): Vrms


p , hv 2 (t)i

AC Power: 14 – 2 / 11




R 1 T T 0

1 R

1 T

RT

2

v 2 (t) R

hv2 (t)i

P p(t)dt = × v (t)dt = R 0



hv2 (t)i

p , hv 2 (t)i

(Vrms )2 R

= The average power dissipated in R is P = R Vrms is the DC voltage that would cause R to dissipate the same power.


AC Power: 14 – 2 / 11




R 1 T T 0

1 R

1 T

RT

2

v 2 (t) R

hv2 (t)i

P p(t)dt = × v (t)dt = R 0



hv2 (t)i

p , hv 2 (t)i

(Vrms )2 R

= The average power dissipated in R is P = R Vrms is the DC voltage that would cause R to dissipate the same power. We use small letters for time-varying voltages and capital letters for time-invariant values.


AC Power: 14 – 2 / 11

Cosine Wave RMS 14: Power in AC Circuits


Cosine Wave: v(t) = 5 cos ωt. Amplitude is V = 5 V.


AC Power: 14 – 3 / 11



Cosine Wave: v(t) = 5 cos ωt. Amplitude is V = 5 V. 2

2

2

Squared Voltage: v (t) = V cos ωt = V


2

1 2

+

1 2

cos 2ωt

AC Power: 14 – 3 / 11




2

2


2 Mean Square Voltage: v =


V2 2

2

1 2

+

1 2

cos 2ωt

since cos 2ωt averages to zero.

AC Power: 14 – 3 / 11




2

2


2 Mean Square Voltage: v =


V2 2

2

1 2

+

1 2

cos 2ωt

since cos 2ωt averages to zero.

AC Power: 14 – 3 / 11




2

2


2

1 2

+

1 2

cos 2ωt

2 V 2 Mean Square Voltage: v = 2 since cos 2ωt averages to zero. p RMS Voltage: Vrms = hv 2 i = √1 V = 3.54 V 2


AC Power: 14 – 3 / 11




2

2


2

1 2

+

1 2

cos 2ωt

2 V 2 Mean Square Voltage: v = 2 since cos 2ωt averages to zero. p RMS Voltage: Vrms = hv 2 i = √1 V = 3.54 V 2 Note: Power engineers always use RMS voltages and currents exclusively and omit the “rms” subscript. For example UK Mains voltage = 230 V rms = 325 V peak.


AC Power: 14 – 3 / 11




2

2


2

1 2

+

1 2

cos 2ωt

2 V 2 Mean Square Voltage: v = 2 since cos 2ωt averages to zero. p e RMS Voltage: Vrms = hv 2 i = √1 V = 3.54 V = V 2 Note: Power engineers always use RMS voltages and currents exclusively and omit the “rms” subscript. For example UK Mains voltage = 230 V rms = 325 V peak.

√ e = In this lecture course only, a ~ overbar means ÷ 2: thus V E1.1 Analysis of Circuits (2017-10213)

√1 V 2

.

AC Power: 14 – 3 / 11

Power Factor

+



Suppose voltage and current phasors are:

V = |V | ejθV I = |I| ejθI


AC Power: 14 – 4 / 11

Power Factor

+






⇔ ⇔

v(t) = |V | cos (ωt + θV ) i(t) = |I| cos (ωt + θI )

AC Power: 14 – 4 / 11

Power Factor

+






⇔ ⇔


AC Power: 14 – 4 / 11

Power Factor

+





⇔ ⇔


Power dissipated in load Z is

p(t) = v(t)i(t) = |V | |I| cos (ωt + θV ) cos (ωt + θI )


AC Power: 14 – 4 / 11

Power Factor

+





⇔ ⇔



p(t) = v(t)i(t) = |V | |I| cos (ωt + θV ) cos (ωt + θI ) = |V | |I|


1 2

cos (2ωt + θV + θI ) +

1 2

cos (θV − θI )

AC Power: 14 – 4 / 11

Power Factor

+





⇔ ⇔



p(t) = v(t)i(t) = |V | |I| cos (ωt + θV ) cos (ωt + θI ) 1 2

1 2

= |V | |I| cos (2ωt + θV + θI ) + cos (θV − θI ) = 21 |V | |I| cos (θV − θI ) + 12 |V | |I| cos (2ωt + θV + θI )


AC Power: 14 – 4 / 11

Power Factor

+





⇔ ⇔




1 2



AC Power: 14 – 4 / 11

Power Factor

+





⇔ ⇔




1 2


Average power: P = 12 |V | |I| cos (φ)


where

φ = θV − θI

AC Power: 14 – 4 / 11

Power Factor

+





⇔ ⇔




1 2



= Ve Ie cos (φ)


where

φ = θV − θI

AC Power: 14 – 4 / 11

Power Factor

+





⇔ ⇔




1 2



= Ve Ie cos (φ)


where

φ = θV − θI

cos φ is the power factor

AC Power: 14 – 4 / 11

Power Factor

+





⇔ ⇔




1 2



= Ve Ie cos (φ)

where

φ = θV − θI

cos φ is the power factor

φ > 0 ⇔ a lagging power factor (normal case: Current lags Voltage) φ < 0 ⇔ a leading power factor (rare case: Current leads Voltage) E1.1 Analysis of Circuits (2017-10213)

AC Power: 14 – 4 / 11

Complex Power 14: Power in AC Circuits


e = √1 |V | ejθV and If V 2

I˜ =

√1 2

|I| ejθI

e × Ie∗ The complex power absorbed by Z is S , V where * means complex conjugate.


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI

e × Ie∗ The complex power absorbed by Z is S , V where * means complex conjugate. Ve × Ie∗ = Ve ejθV × Ie e−jθI


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI


Ve × Ie∗ = Ve ejθV × Ie e−jθI = Ve Ie ej(θV −θI )


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI


Ve × Ie∗ = Ve ejθV × Ie e−jθI = Ve Ie ej(θV −θI ) = Ve Ie ejφ


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI


Ve × Ie∗ = Ve ejθV × Ie e−jθI = Ve Ie ej(θV −θI ) e e jφ e e = V I e = V I cos φ + j Ve Ie sin φ


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI


Ve × Ie∗ = Ve ejθV × Ie e−jθI = Ve Ie ej(θV −θI ) e e jφ e e = V I e = V I cos φ + j Ve Ie sin φ = P + jQ


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI




AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI



Complex Power:


S , Ve Ie∗ = P + jQ measured in Volt-Amps (VA)

AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI



S , Ve Ie ∗ = P + jQ measured in Volt-Amps (VA) e e Apparent Power: |S| , V I measured in Volt-Amps (VA) Complex Power:


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI




Average Power:


P , ℜ (S) measured in Watts (W)

AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI




Average Power: P , ℜ (S) measured in Watts (W) Reactive Power: Q , ℑ (S) Measured in Volt-Amps Reactive (VAR)


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI




Average Power: P , ℜ (S) measured in Watts (W) Reactive Power: Q , ℑ (S) Measuredin Volt-Amps Reactive (VAR)

e − ∠Ie = P Power Factor: cos φ , cos ∠V |S|


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI






Machines and transformers have capacity limits and power losses that are independent of cos φ; their ratings are always given in apparent power.


AC Power: 14 – 5 / 11




I˜ =

√1 2

|I| ejθI






Machines and transformers have capacity limits and power losses that are independent of cos φ; their ratings are always given in apparent power. Power Company: Costs ∝ apparent power, Revenue ∝ average power.


AC Power: 14 – 5 / 11

Power in R, L, C 14: Power in AC Circuits


e Ie∗ = P + jQ For any impedance, Z , complex power absorbed: S = V


AC Power: 14 – 6 / 11



e Ie∗ = P + jQ For any impedance, Z , complex power absorbed: S = V 2 2 e |2 V | e = IZ e (b) Ie × Ie∗ = Ie we get S = Ie Z = ∗ Using (a) V Z


AC Power: 14 – 6 / 11




2 2 |Ve | e Resistor: S = I R = R


φ=0

Absorbs average power, no VARs (Q = 0)

AC Power: 14 – 6 / 11





φ=0


2 2 |Ve | e Inductor: S = j I ωL = j ωL

φ = +90◦

No average power, Absorbs VARs (Q > 0)


AC Power: 14 – 6 / 11





φ=0



φ = +90◦


|Ie|

2

Capacitor: S = −j ωC

2 = −j Ve ωC

φ = −90◦

No average power, Generates VARs (Q < 0)


AC Power: 14 – 6 / 11





φ=0



φ = +90◦


|Ie|

2


2 = −j Ve ωC

φ = −90◦

No average power, Generates VARs (Q < 0) VARs are generated by capacitors and absorbed by inductors


AC Power: 14 – 6 / 11





φ=0



φ = +90◦


|Ie|

2


2 = −j Ve ωC

φ = −90◦

No average power, Generates VARs (Q < 0) VARs are generated by capacitors and absorbed by inductors The phase, φ, of the absorbed power, S , equals the phase of Z E1.1 Analysis of Circuits (2017-10213)

AC Power: 14 – 6 / 11

Tellegen’s Theorem 14: Power in AC Circuits


Tellegen’s Theorem: The complex power, S , dissipated in any circuit’s components sums to zero.

xn = voltage at node n Vb , Ib = voltage/current in branch b


(obeying passive sign convention)

AC Power: 14 – 7 / 11




xn = voltage at node n Vb , Ib = voltage/current in branch b (obeying passive sign convention)

abn

  −1 if Vb starts from node n , +1 if Vb ends at node n   0 else


AC Power: 14 – 7 / 11





abn


e.g. branch 4 goes from 2 to 3 ⇒ a4∗ = [0, −1, 1]


AC Power: 14 – 7 / 11





abn


e.g. branch 4 goes from 2 to 3 ⇒ a4∗ = [0, −1, 1] Branch voltages: Vb =


P

n

abn xn (e.g. V4 = x3 − x2 )

AC Power: 14 – 7 / 11





abn



P

abn xn (e.g. V4 = x3 − x2 ) P P ∗ KCL @ node n: b abn Ib = 0 ⇒ b abn Ib = 0


n

AC Power: 14 – 7 / 11





abn



P

abn xn (e.g. V4 = x3 − x2 ) P P ∗ KCL @ node n: b abn Ib = 0 ⇒ b abn Ib = 0 P P P ∗ ∗ Tellegen: b V b Ib = b n abn xn Ib


n

AC Power: 14 – 7 / 11





abn



P

abn xn (e.g. V4 = x3 − x2 ) P P ∗ KCL @ node n: b abn Ib = 0 ⇒ b abn Ib = 0 P P P ∗ ∗ Tellegen: b V b Ib = b n abn xn Ib P P = n b abn Ib∗ xn E1.1 Analysis of Circuits (2017-10213)

n

AC Power: 14 – 7 / 11





abn



P

abn xn (e.g. V4 = x3 − x2 ) P P ∗ KCL @ node n: b abn Ib = 0 ⇒ b abn Ib = 0 P P P ∗ ∗ Tellegen: b V b Ib = b n abn xn Ib P P P P ∗ = n b abn Ib xn = n xn b abn Ib∗ E1.1 Analysis of Circuits (2017-10213)

n

AC Power: 14 – 7 / 11





abn



P

abn xn (e.g. V4 = x3 − x2 ) P P ∗ KCL @ node n: b abn Ib = 0 ⇒ b abn Ib = 0 P P P ∗ ∗ Tellegen: b V b Ib = b n abn xn Ib P P P P P ∗ ∗ = n b abn Ib xn = n xn b abn Ib = n xn × 0 E1.1 Analysis of Circuits (2017-10213)

n

AC Power: 14 – 7 / 11





abn



P

abn xn (e.g. V4 = x3 − x2 ) P P ∗ KCL @ node n: b abn Ib = 0 ⇒ b abn Ib = 0 P P P ∗ ∗ Tellegen: b V b Ib = b n abn xn Ib P P P P P ∗ ∗ = n b abn Ib xn = n xn b abn Ib = n xn × 0 P P P Note: and also b Sb = 0 ⇒ b Pb = 0 b Qb = 0. E1.1 Analysis of Circuits (2017-10213)

n

AC Power: 14 – 7 / 11

Power Factor Correction 14: Power in AC Circuits


Ve = 230. Motor modelled as 5||7j Ω. e e Ie = VR + ZVL


AC Power: 14 – 8 / 11



Ve = 230. Motor modelled as 5||7j Ω. e e Ie = VR + ZVL = 46 − j32.9 A


AC Power: 14 – 8 / 11



Ve = 230. Motor modelled as 5||7j Ω. e e Ie = VR + ZVL = 46 − j32.9 A = 56.5∠ − 36◦


AC Power: 14 – 8 / 11



Ve = 230. Motor modelled as 5||7j Ω. e e Ie = VR + ZVL = 46 − j32.9 A = 56.5∠ − 36◦ S = Ve Ie∗ = 10.6 + j7.6 kVA


AC Power: 14 – 8 / 11



Ve = 230. Motor modelled as 5||7j Ω. e e Ie = VR + ZVL = 46 − j32.9 A = 56.5∠ − 36◦ S = Ve Ie∗ = 10.6 + j7.6 kVA = 13∠36◦ kVA


AC Power: 14 – 8 / 11



Ve = 230. Motor modelled as 5||7j Ω. e e Ie = VR + ZVL = 46 − j32.9 A = 56.5∠ − 36◦ S = Ve Ie∗ = 10.6 + j7.6 kVA = 13∠36◦ kVA P = cos 36◦ = 0.81 cos φ = |S|


AC Power: 14 – 8 / 11



Ve = 230. Motor modelled as 5||7j Ω. e e Ie = VR + ZVL = 46 − j32.9 A = 56.5∠ − 36◦ S = Ve Ie∗ = 10.6 + j7.6 kVA = 13∠36◦ kVA P = cos 36◦ = 0.81 cos φ = |S| Add parallel capacitor of 300 µF:


AC Power: 14 – 8 / 11




ZC =


1 jωC

= −10.6j Ω

AC Power: 14 – 8 / 11




ZC =


1 jωC

= −10.6j Ω ⇒ IeC = 21.7j A

AC Power: 14 – 8 / 11




1 = −10.6j Ω ⇒ IeC = 21.7j A ZC = jωC Ie = 46 − j11.2 A = 47∠ − 14◦ A


AC Power: 14 – 8 / 11





SC = Ve IeC∗ = −j5 kVA


AC Power: 14 – 8 / 11





SC = Ve IeC∗ = −j5 kVA S = Ve Ie∗ = 10.6 + j2.6 kVA


AC Power: 14 – 8 / 11





SC = Ve IeC∗ = −j5 kVA S = Ve Ie∗ = 10.6 + j2.6 kVA = 10.9∠14◦ kVA


AC Power: 14 – 8 / 11





SC = Ve IeC∗ = −j5 kVA S = Ve Ie∗ = 10.6 + j2.6 kVA = 10.9∠14◦ kVA P cos φ = |S| = cos 14◦ = 0.97


AC Power: 14 – 8 / 11





SC = Ve IeC∗ = −j5 kVA S = Ve Ie∗ = 10.6 + j2.6 kVA = 10.9∠14◦ kVA P cos φ = |S| = cos 14◦ = 0.97 Average power to motor, P , is 10.6 kW in both cases.


AC Power: 14 – 8 / 11





SC = Ve IeC∗ = −j5 kVA S = Ve Ie∗ = 10.6 + j2.6 kVA = 10.9∠14◦ kVA P cos φ = |S| = cos 14◦ = 0.97 power to motor, P , is 10.6 kW in both cases. Average e I , reduced from 56.5 ց 47 A (−16%) ⇒ lower losses. E1.1 Analysis of Circuits (2017-10213)

AC Power: 14 – 8 / 11





SC = Ve IeC∗ = −j5 kVA S = Ve Ie∗ = 10.6 + j2.6 kVA = 10.9∠14◦ kVA P cos φ = |S| = cos 14◦ = 0.97 power to motor, P , is 10.6 kW in both cases. Average e I , reduced from 56.5 ց 47 A (−16%) ⇒ lower losses. Effect of C : VARs = 7.6 ց 2.6 kVAR E1.1 Analysis of Circuits (2017-10213)

AC Power: 14 – 8 / 11





SC = Ve IeC∗ = −j5 kVA S = Ve Ie∗ = 10.6 + j2.6 kVA = 10.9∠14◦ kVA P cos φ = |S| = cos 14◦ = 0.97 power to motor, P , is 10.6 kW in both cases. Average e I , reduced from 56.5 ց 47 A (−16%) ⇒ lower losses. Effect of C : VARs = 7.6 ց 2.6 kVAR , power factor = 0.81 ր 0.97. E1.1 Analysis of Circuits (2017-10213)

AC Power: 14 – 8 / 11

Ideal Transformer 14: Power in AC Circuits


A transformer has ≥ 2 windings on the same magnetic core.


AC Power: 14 – 9 / 11



A transformer has ≥ 2 windings on the same magnetic core. Ampère’s law:


P

Nr Ir =

lΦ µA ;

Vr dΦ = Faraday’s law: N dt . r

AC Power: 14 – 9 / 11




P

Vr dΦ = Faraday’s law: N dt . r N1 : N2 + N3 shows the turns ratio between the windings. The • indicates the voltage polarity of each winding.

Ampère’s law:


Nr Ir =

lΦ µA ;

AC Power: 14 – 9 / 11




P


Ampère’s law:

Nr Ir =

lΦ µA ;

V1 V2 V3 Since Φ is the same for all windings, N = = N2 N3 . 1


AC Power: 14 – 9 / 11




P


Ampère’s law:

Nr Ir =

lΦ µA ;


Assume µ → ∞ ⇒ N1 I1 + N2 I2 + N3 I3 = 0


AC Power: 14 – 9 / 11




P


Ampère’s law:

Nr Ir =

lΦ µA ;


Assume µ → ∞ ⇒ N1 I1 + N2 I2 + N3 I3 = 0 These two equations allow you to solve circuits and also P imply that Si = 0.


AC Power: 14 – 9 / 11




P


Ampère’s law:

Nr Ir =

lΦ µA ;


Assume µ → ∞ ⇒ N1 I1 + N2 I2 + N3 I3 = 0 These two equations allow you to solve circuits and also P imply that Si = 0. Special Case:


AC Power: 14 – 9 / 11




P


Ampère’s law:

Nr Ir =

lΦ µA ;


Assume µ → ∞ ⇒ N1 I1 + N2 I2 + N3 I3 = 0 These two equations allow you to solve circuits and also P imply that Si = 0. Special Case: For a 2-winding transformer this simplifies to N2 N1 V2 = N V1 and IL = −I2 = N I1 1


2

AC Power: 14 – 9 / 11




P


Ampère’s law:

Nr Ir =

lΦ µA ;



Hence VI 1 = 1


2

N1 N2

2

V2 IL

=

N1 N2

2

Z

AC Power: 14 – 9 / 11




P


Ampère’s law:

Nr Ir =

lΦ µA ;



Hence VI 1 = 1

2

N1 N2

2

V2 IL

=

N1 N2

2

Z

Equivalent to a reflected impedance of E1.1 Analysis of Circuits (2017-10213)

N1 N2

2

Z AC Power: 14 – 9 / 11

Transformer Applications 14: Power in AC Circuits



AC Power: 14 – 10 / 11



Power Transmission Suppose a power transmission cable has 1 Ω resistance. 100 kVA@ 1 kV = 100 A ⇒ Ie2 R = 10 kW losses.


AC Power: 14 – 10 / 11



Power Transmission Suppose a power transmission cable has 1 Ω resistance. 100 kVA@ 1 kV = 100 A ⇒ Ie2 R = 10 kW losses. 100 kVA@ 100 kV = 1 A ⇒ Ie2 R = 1 W losses.


AC Power: 14 – 10 / 11




Voltage Conversion

Electronic equipment requires ≤ 20 V but mains voltage is 240 V ∼.


AC Power: 14 – 10 / 11




Voltage Conversion

Electronic equipment requires ≤ 20 V but mains voltage is 240 V ∼. Interference protection Microphone on long cable is susceptible to interference from nearby mains cables. An N : 1 transformer reduces the microphone voltage by N but reduces interference by N 2 .


AC Power: 14 – 10 / 11




Voltage Conversion

Electronic equipment requires ≤ 20 V but mains voltage is 240 V ∼. Interference protection Microphone on long cable is susceptible to interference from nearby mains cables. An N : 1 transformer reduces the microphone voltage by N but reduces interference by N 2 . Isolation There is no electrical connection between the windings of a transformer so circuitry (or people) on one side will not be endangered by a failure that results in high voltages on the other side.


AC Power: 14 – 10 / 11

Summary 14: Power in AC Circuits


• Complex Power: S = Ve Ie∗ = P + jQ where Ve = Vrms =


√1 V 2

.

AC Power: 14 – 11 / 11



• Complex Power: S = Ve Ie∗ = P + jQ where Ve = Vrms = 2 2 |Ve | e ◦ For an impedance Z : S = I Z = Z ∗


√1 V 2

.

AC Power: 14 – 11 / 11



• Complex Power: S = Ve Ie∗ = P + jQ where Ve = Vrms = √12 V . 2 2 |Ve | e ◦ For an impedance Z : S = I Z = Z ∗ ◦ Apparent Power: |S| = Ve Ie used for machine ratings.


AC Power: 14 – 11 / 11



• Complex Power: S = Ve Ie∗ = P + jQ where Ve = Vrms = √12 V . 2 2 |Ve | e ◦ For an impedance Z : S = I Z = Z ∗ ◦ Apparent Power: |S| = Ve Ie used for machine ratings. ◦ Average Power: P = ℜ (S) = Ve Ie cos φ (in Watts) ◦ Reactive Power: Q = ℑ (S) = Ve Ie sin φ (in VARs)


AC Power: 14 – 11 / 11



• Complex Power: S = Ve Ie∗ = P + jQ where Ve = Vrms = √12 V . 2 2 |Ve | e ◦ For an impedance Z : S = I Z = Z ∗ ◦ Apparent Power: |S| = Ve Ie used for machine ratings. ◦ Average Power: P = ℜ (S) = Ve Ie cos φ (in Watts) ◦ Reactive Power: Q = ℑ (S) = Ve Ie sin φ (in VARs) ◦ Power engineers always use Ve and Ie and omit the ~.


AC Power: 14 – 11 / 11



• Complex Power: S = Ve Ie∗ = P + jQ where Ve = Vrms = √12 V . 2 2 |Ve | e ◦ For an impedance Z : S = I Z = Z ∗ ◦ Apparent Power: |S| = Ve Ie used for machine ratings. ◦ Average Power: P = ℜ (S) = Ve Ie cos φ (in Watts) ◦ Reactive Power: Q = ℑ (S) = Ve Ie sin φ (in VARs) ◦ Power engineers always use Ve and Ie and omit the ~. • Tellegen: In any circuit


P

b

Sb = 0 ⇒

P

b

Pb =

P

b

Qb = 0

AC Power: 14 – 11 / 11




P

b

Sb = 0 ⇒

P

b

Pb =

P

b

Qb = 0

• Power Factor Correction: add parallel C to generate extra VARs


AC Power: 14 – 11 / 11




P

b

Sb = 0 ⇒

P

b

Pb =

P

b

Qb = 0

• Power Factor Correction: add parallel C to generate extra VARs • Ideal Transformer: Vi ∝ Ni and


P

Ni Ii = 0 (implies

P

Si = 0)

AC Power: 14 – 11 / 11




P

b

Sb = 0 ⇒

P

b

Pb =

P

b

Qb = 0

• Power Factor Correction: add parallel C to generate extra VARs • Ideal Transformer: Vi ∝ Ni and

P

Ni Ii = 0 (implies

P

Si = 0)

For further details see Hayt Ch 11 or Irwin Ch 9.


AC Power: 14 – 11 / 11