chapter 11: ac power analysis -...

Chapter 11: AC Power Analysis

11.1 Instantaneous and Average Power11.2 Maximum Average Power Transfer11.3 Effective or RMS Value11.4 Apparent Power and Power Factor11.5 Complex Power 11.6 Conservation of AC Power11.7 Power Factor Correction11.8 Power Measurement11.9 Summary

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The choice of ac over dc allowed high-voltage power transmission from the power generating plant to the consumer.

11.1 Instantaneous and Average Power (1)

• The instantaneously power (in watts), p(t)

2

) 2( cos 21 ) ( cos

21

) ( cos ) ( cos )( )( )(

ivmmivmm

ivmm

tIVIV

ttIVtitvtp

θθωθθ

θωθω

+++−=

++==

p(t) > 0: power is absorbed by thecircuit; p(t) < 0: power is absorbedby the source.

sinusoidal power at 2ωtconstant power

• The average power, P, is the average of p(t) over one period.1. P is not time dependent. 2. When θv = θi , it is a purely resistive load case. 3. When θv– θi = ± 90o, it is a purely reactive load case. 4. P = 0 means that the circuit absorbs no average power.

) ( cos 21 )(1

0 ivmm

TIVdttp

TP θθ −== ∫

( ) cos( )( ) cos( )

m v

m i

v t V ti t I t

ω θω θ

= += +

Example: A current flows through an impedance . Find the average power delivered to the impedance.

20 30= ∠ °I

3Answer: 7.417 kW

40 22= ∠− °ΩZ


1 cos( )2 m m v iP V I θ θ= −

R: L or C:

1 Re2

VIP ∗ =

221 1 12 2 2

Im m mP V I I R R= = =1 cos90 02 m mP V I= =

Example: Determine the average power generated by each source and the average power absorbed by each passive element in Fig.

4


by mesh analysis

1.5 Power and Energy (1)

• Power is the time rate of expending or absorbing energy, measured in watts (W).

• Mathematical expression:

5

vidtdq

dqdw

dtdwp =⋅==

Passive sign conventionP = + vi p = – vi

absorbing power supplying power

11.2 Maximum Average Power Transfer (1)

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Th Th ThR jX= +Z

• The maximum average power can be transferred to the load if XL = –XTh and RL = RTh, i.e.,

2Th

maxTh8

PR

∴ =V

• If the load is purely real, i.e. XL = 0, then 2 2Th Th Th LR R X= + = Z

L L LR jX= +Z

( ) ( )Th Th

Th Th Th

V VIZ ZL L LR jX R jX

= =+ + + +

( ) ( )

22 Th

2 2Th Th

/ 212

VI L

LL L

RP R

R R X X= =

+ + +

Th Th ThZL L LR jX R jX Z ∗= + = − =

• The load impedance must be equal to the complex conjugate of the Thevenin impedance.

2Th

Th max 4L

L

VR R PR

= ⇒ =

22 Th

Th

L LL

VP i R RR R

= = +

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- There are applications where it is desirable to maximize the power delivered to a load. Also, power utility systems are designed to transport the power to the load with the greatest efficiency by reducing the losses on the power lines.

- If the entire circuit is replaced by its Theveninequivalent except for the load, the power delivered to the load is:

The power transfer profile with different RL

- Maximum power is transferred to the load resistance equals the Thevenin resistance as seen from the load.

4.7 Maximum Power Transfer (1)

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Example: For the circuit shown below, find the load impedance ZLthat absorbs the maximum average power. Calculate Pmax.

11.2 Maximum Average Power Transfer (2)

11.3 Effective or RMS Value (1)

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2 2 2

0 0

1 T T

effRP i Rdt i dt I R

T T= = =∫ ∫

Hence, Ieff is equal to: rms

T

eff IdtiT

I 1

0

2 == ∫Note: The root mean square (rms) value is a constant itself which depending on the shape of the function i(t). The effective of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.

• The total power dissipated by R is given by:

rms 2mII =

• The average power can be written in terms of the rms values:

rms rms1 cos( - ) cos( )2 m m v i v iP V I V Iθ θ θ θ= = −

• The rms value of a i(t) = Imcos(ωt) is given by:

Note: If you express amplitude of a phasor source(s) in rms, then all the answer as a result of this phasor source(s) must also be in rms value.

11.3 Effective or RMS Value (2)

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Example: Find the rms value of the current waveform. If the current flows through a 9-Ω resistor, calculate the average power absorbed by the resistor.

Example: Find the rms value of the full-wave rectified sine wave. Calculate the average power dissipated in a 6-Ω resistor.

11.4 Apparent Power and Power Factor (1)

• Apparent Power (in VA), S, is the product of the rms values of voltage and current.

• It is measured in volt-amperes or VA to distinguish it from the average or real power which is measured in watts.

• Power factor is the cosine of the phase difference between the voltage and current. It is also the cosine of the angle of the load impedance.

rms rms cos( ) cos( )v i v iP V I Sθ θ θ θ= − = −

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Apparent Power, S Power Factor, pf

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Purely resistiveload (R)

θv– θi = 0, pf = 1

P/S = 1, all power are consumed

Purely reactiveload (L or C)

θv– θi = ±90o, pf = 0

P = 0, no real power consumption

Resistive & reactive load (R & L/C)

θv– θi > 0θv– θi < 0

• Lagging - inductive load• Leading - capacitive load

11.4 Apparent Power and Power Factor (2)

Example: Determine the power factor of the entire circuit by the source. Calculate the average power delivered by the source.

11.5 Complex Power (1)

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• Complex power (in VA) S is the product of rms voltage phasorand complex conjugate of rms current phasor:

rms rms

rms rms

2

2

m v v

m i i

V V

I I

θ θ

θ θ

= ∠ → = = ∠

= ∠ → = = ∠

VV V

II I

rms rms rms rms12

S VI V I∗ ∗= = = ∠ −v iV I θ θ

where

rms rms rms rmscos( ) sin( )v i v iV I jV Iθ θ θ θ= − + −S

S = P + j QP: is the average power in watts delivered to a load and it is the only useful power.Q: is the reactive power exchange between the source and the reactive part of the

load. It is measured in VAR. Q = 0 for resistive loads (unity pf). Q < 0 for capacitive loads (leading pf). Q > 0 for inductive loads (lagging pf).

(l) The impedance triangle where Z = R + jX. (r) The complex power triangle where S = P + jQ.


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Apparent Power: S = |S| = Vrms × Irms = Real power: P = Re(S) = I2

rms R = S cos(θv – θi)Reactive Power: Q = Im(S) = I2

rms X = S sin(θv – θi) Power factor: pf = P/S = cos(θv – θi)

2 2P Q+


rms rms rms rmscos( ) sin( )v i v iV I jV Iθ θ θ θ= − + −S

S = P + j Q

Impedance Triangle

Power Factor

Power Triangle

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Example: For a load, . Determine: (a) the complex and apparent powers, (b) the real and reactive powers, and (c) the power factor and the load impedance.

o orms rms110 85 V, 0.4 15 A= ∠ = ∠V I

11.6 Conservation of AC Power

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The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads. For parallel connection:

*1 2 1 2 1 2( ) ∗ ∗ ∗= = + = + = +S VI V V I V I V I S S

The same results can be obtained for a seriesconnection.

Example: In the circuit, the 60-Ω resistors absorbs an average power of 240 W. Find V and the complex power of each branch of the circuit. What is the overall complex power of the circuit? (Assume the current through the 60-Ω resistor has no phase shift.)

( )*1 2 1 2 1 2

∗ ∗ ∗= = = + = +S VI V I + I VI VI S S

11.7 Power Factor Correction (1)

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• Most domestic and industrial loads, such as washing machines, air conditioners, and induction motors are inductive. They have a low, lagging power factor. The load cannot be changed, but the power factor can be increased without altering the voltage or current to the original load.

• Power factor correction: the process of increasing the power factor without altering the voltage or current to the original load.

• To mitigate the inductive aspect of the load, a capacitor is added in parallel with the load. The power factor has improved.

11.7 Power Factor Correction (2)

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QC = Q1 – Q2 = P (tan θ1 - tan θ2) = ωCV 2rms

Q1 = S1 sin θ1 = P tan θ1

Q2 = P tan θ2P = S1 cos θ1

1 22 2

(tan tan )

c

rms rms

Q PCV V

θ θω ω

−= =

Power factor correction is necessary for economic reason.

• With the same supplied voltage, the current draw is less by adding the capacitor. Since power companies charge more for larger currents because it leads to larger power losses. Overall, the power factor correction benefits the power company and the consumer. By choosing a suitable size for the capacitor, the power factor can be made to be unity.

• The real power dissipated in the load is not affected by the shunt capacitor.• Although it is not as common, if a load is capacitive in nature, the same

treatment with an inductor can be used.

11.8 Power Measurement (1)

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• The wattmeter is the instrument for measuring the average power.• The meter consists of two coils; the current and voltage coils. − The current coil is designed with low impedance and is

connected in series with the load.− The voltage coil is designed with very large impedance and is

connected in parallel with the load.

Basic structure Equivalent Circuit with load

11.8 Power Measurement (2)

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If )cos()( vm tVtv θω += and )cos()( im tIti θω +=

1rms rms 2cos( ) cos ( )v i m m v iP V Iθ θ θ θ= − = −V I

• The induced magnetic field from both causes a deflection in the current coil.

• Ideally, the configuration does not alter the load and affect the power measured.

• The physical inertia of the moving coil results in the output being equal to the average power.

11.9 Summary (1)

11.9 Summary (2)

chapter 11: ac power analysis -...

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