Read : Chapter 10 in Electric Circuits, 9 th Edition by Nilsson

1

Read: Chapter 10 in Electric Circuits, 10th Edition by Nilsson

Chapter 10 - Power Calculations in AC Circuits

Instantaneous Power: p(t) = v(t)i(t) = instantaneous power

T

-T

lim 1P p(t)dtT 2T

T

0

1P p(t)dtT

Average Power:• Instantaneous power is not commonly used.• Average power, P, is more useful. • P = PAVG = PDC = average or real power (in watts, W)

• P can be defined for any waveform as:

• For periodic waveforms with period T (including sinusoids), P can be expressed as:

1Chapter 10 EGR 272 – Circuit Theory II

Average power can be calculated:1) by inspection (for simple cases)2) by integration (for more complex cases)

Example: Find the average power absorbed to the resistor below.

10 v(t)

+

_

i(t)

t [s]

i(t) [A]

10

4

2 4 6 8 10 12 0


Example: Find the average power absorbed to the resistor below.

10 v(t)

+

_

i(t)

t [s]

v(t) [V]

20

2 4 6 8 10 12 00


RMS Voltage and CurrentVRMS = root-mean-square voltage (also sometimes called Veff = effective

voltage)VRMS = the square root of the average value of the function squared

T2

RMS0

1V v (t)dtT

T

2RMS

0

1I i (t)dtT

Note how the name RMS essentially gives the definition.


Calculating average power for a resistive load using RMS valuesA key reason that RMS values are used commonly in AC circuits is that they are used in power calculations that are very similar to those used in DC circuits. Recall that power in DC circuits can be calculated using:

2

2RMSRMS RMS RMS

VP V I I R

R

22VP V I I R

R

Similarly, instantaneous power to a resistor can be calculated using

Show that power can be also be calculated using RMS values as follows:

22v (t)p(t) v(t) i(t) i (t) R

R

Development:


Example: Find IRMS for the waveform below and use IRMS to calculate the average power absorbed to the resistor. Recall that P was calculated earlier using instantaneous power.

10 v(t)

+

_

i(t)

t [s]

i(t) [A]

10

4

2 4 6 8 10 12 0

RMS values can be calculated:1) by inspection (for simple cases)2) by integration (for more complex cases)


10 v(t)

+

_

i(t)

t [s]

v(t) [V]

20

2 4 6 8 10 12 00

Example: Find VRMS for the waveform below and use VRMS to calculate the average power absorbed to the resistor. Recall that P was calculated earlier using instantaneous power.


Development: Derive the expression for VRMS above.

RMS value of sinusoidal voltages and currentsUsing the definition of RMS voltage and current, it can be shown that for a sinusoidal voltage v(t) = Vpcos(wt) and a sinusoidal current i(t) = Ipcos(wt) that:

pRMS p

VV 0.707V

2 p

RMS p

II 0.707I

2

Example: Find the power absorbed by a 10 resistor with v(t) = 20cos(377t) V.


Superposition of PowerRecall that, in general, superposition applies to voltage and current, but not to power.

+ _

+ _

R

V 1 V 2

i Applying superposition to find the current i:i = i1 + i2 ori = current due to V1 + current due to V2 Show that this leads to: P P1 +

P2


T

0

2T

0

Rdt iT1 p(t)dt

T1 P

Development:

Power to a Complex LoadIf v(t) = Vm cos(wt) then i(t) = Im cos(wt - )

Useful identities:cos(wt - ) = cos()cos(wt) + sin()sin(wt)cos2 (wt) = ½ + ½ cos(2wt)sin(wt)cos(wt) = ½ sin(2wt)

Z v(t)

i(t) +

_

mm

V V 0 since I = I ZZ

2

2RMSm mRMS RMS RMS

VV IP cos ( ) V I cos ( ) cos ( ) I Z cos ( ) 2 Z

Show that:


T

0

T

0

v(t)i(t)dtT1 p(t)dt

T1 P

Development:

Key definitions:Recall that real or average power, P, was just defined as:

Since cos is an even function (i.e., cos() = cos(-)), calculating cos() does not reveal the sign of the angle, so the terms leading and lagging are used with power factor.• The power factor is leading if < 0, (i.e., I leads V or the load is capacitive).• The power factor is lagging if > 0, (i.e., I lags V or the load is inductive).

Apparent power = (VRMS)(IRMS) measured in volt-amperes, VA

preal powerpower factor p.f. F cos( )

apparent power

2

2RMSm mRMS RMS RMS

VV IP cos ( ) V I cos ( ) cos ( ) I Z cos ( ) 2 Z

Other important terms are apparent power and power factor, as defined below:


Examples: Determine P and the p.f. for each case below:

Z 10V RMS

+

_

a) Z 4 b) Z j3 c) Z -j3

d) Z 4 j3 e) Z 4 - j3


Complex Power:

The use of phasors with RMS values is very convenient for power calculations.

m m

RMS RMS

If v(t) V cos (wt), then V V 0 (phasor using the peak value of v(t))

but we can also define V V 0 (phasor using the RMS value of v(t))

Z VRMS

+

_

IRMS

RMSRMS RMS RMS V V 0 implies that I I

RMS RMSRMS RMS

V V 0since I = IZZ

*RMSRMSS complex power V I (in volt-amperes, VA) Define


*RMSRMSS complex power V I (in volt-amperes, VA)

Show that

Since

RMS RMSS S V I P jQ

RMS RMSP real or average power V I cos( ) (in watts, W)

RMS RMSQ reactive power V I sin( ) (in volt-amperes reactive, VAR)

2 2RMS RMSS apparent power V I P Q (in volt-amperes reactive, VAR)

Pp.f. power factor cos( ) (expressed as leading if 0 or lagging if 0)S


where

Development:

Complex Power DiagramA complex power diagram is sometimes used to illustrate the relationship between P, Q, |S|, and S.

Re

Im

P

QS

|S|

P

Q|S|

or

Example:If VRMS = 1000 V and IRMS = 5 -30 A, find P, Q, |S|, and S and illustrate them on a complex power diagram. Also find the power factor.


2

2RMSR RMS L C

VP I R and P P 0

R

2

2RMSL C RMS R L C

V -1Q (or Q ) I X and Q 0 where X wL and X X wC

Development:

Other useful relationshipsSubstitute VRMS = ZIRMS and Z = R + jX into S = VRMSIRMS

* to show that:


Two approaches for performing power calculations:

1.

2. Find the voltage or current for each component (magnitude only). Then calculate P or Q for each component and add the results for total P and Q.

RMSTotal RMS

Total

VFind Z and I Z

*RMSRMSThen find S V I P jQ

TT component component T TTThen use P P , Q Q , and S P jQ


Example: A) By first finding the total circuit impedance

Find S, S , P, Q, and p.f.

+

_

100V RMS

10

132.6 uF 60 Hz

15

26.5 mH


Example: (continued)B) By finding P or Q for each component (use mesh equations first to find the

currents for each component).

Find S, S , P, Q, and p.f.

+

_100V RMS

10

132.6 uF 60 Hz

15

26.5 mH


A 61.2-7.071 I - I I

A 9.812.828 I

A 41.8-5.099 I

0100

II

j10-15j10-j10-j1001

0 Ij20- I15 )I - Ij10( :2mesh KVL,

0 )I - Ij10( I100100- :1mesh KVL,

21L

2

1

2

1

2212

211

Power Factor CorrectionPower companies charge their customers based on the real power (P) used. If the customer has a significant amount of reactive power (or a low power factor) this has no effect on P, but results in higher current levels. Power companies “encourage” their large customers to “correct” their power factor. If their power factor drops too low (perhaps less than 0.9), the power company may charge higher rates.

Large companies often have lagging power factors due to large amounts of machinery (inductive loads) and they can correct their power factor by adding in large parallel capacitors (or use synchronous motors). This generates negative reactive power that cancels the positive reactive power due to the inductive loads resulting in a power factor which is near or close to unity. The corrected power factor results in lower current levels and thus lower line losses (I2 R) as the power company delivers the power to the customer. Use user also benefits from lower current levels since maximum current ratings on wire and equipment can be reduced.


Illustration - Power Factor Correction

Circuit

+

_

I1

Original Circuit

Circuit

+

_

I2

New Circuit (Note that |I2| < |I1| )

I1

IC

C

Original Circuit

P

Q

Re

Im

S = P + jQ

Added Capacitor

QC = -Q

Re

Im

S = P + jQ

New Circuit

PRe

Im

S = P + jQ - jQ = P

+ =


Example: Add a capacitor in parallel with the source in order to: A) correct the power factor to unityAlso calculate the total source current and compare it to the original current.Note that this is the same circuit analyzed previously.

+

_100V RMS

10

132.6 uF 60 Hz

15

26.5 mH


Example: (continued) Add a capacitor in parallel with the source in order to: B) correct the power factor to 0.95, laggingAlso calculate the total source current and compare it to the original current.

Circuit p.f. C (added) current, |I| Reduction in |I| Original - - - - - - Part A 1.0 (unity) Part B 0.95, lagging

Summary: (Parts A and B)


Power Factor in SystemsJust as total power can be determined by adding the power dissipated by each component, so can the total power for a system be determined by adding the power dissipated by each subsystem.

TTotal subsystem subsystem T TTotalThen use P P , Q Q , and S P jQ

Example: Find the power factor of the entire system below that is comprised of three subsystems as shown below.

Subsystem 1:|S| = 2000 VAp.f. = 0.8, lagging

Subsystem 2:P = 2000 Wp.f. = 0.9, leading

Subsystem 3:|S| = 2000 VAQ = 1500 VAR

+

_


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Read : Chapter 10 in Electric Circuits, 9 th Edition by Nilsson