ECE 1100: Introduction toECE 1100: Introduction toElectrical and Computer EngineeringElectrical and Computer Engineering
Notes 20
Power in AC Circuits and RMS
1
Spring 2011
Wanda WosikAssociate Professor, ECE Dept.
Notes prepared by Dr. Jackson
AC PowerAC Power
22 cosp
abs
V tv tP t
R R
R []+-v (t)
Goal: Find the average power absorbed by resistor:
cos
cos 2 [V]
p
p
v t V t
V f t
f = frequency [Hz]
Vp = peak voltage
Note: The phase of the voltage wave is assumed to be zero here for convenience.
AVEabsP
2
AC Power (cont.) AC Power (cont.)
2
2cospabs
VP t t
R
T = 1/f [s]c os
(t )
Tp = T / 2 = 0.5 / f [s]
c os2
(
t)
3
AC Power (cont.)AC Power (cont.)
0
22
0
22
0
22
0
1
1cos
1cos
1cos
p
p
p
T
AVEabs abs
p
T
p
p
T
p
p
Tp
P P t dtT
Vt dt
T R
Vt dt
T R
Vt dt
T R
Note: We obtain the same result if we integrate over Tp or T.
4
AC Power (cont.)AC Power (cont.)
2
2
0
1cos
TpAVE
abs
VP t dt
T R
Consider the integral that needs to be evaluated:
2
0
cosT
cI t dt
5
AC Power (cont.)AC Power (cont.)
2 2
0 0
2
0
0
0
cos cos 2
cos 2 /
1 cos 2*2 /
2
sin 2*2 /
2 2(2*2 / )
2
T T
c
T
T
T
I t dt f t dt
t T dt
t Tdt
t Tt
T
T
“The average value of cos2 is 1/2.”6
AC Power (cont.)AC Power (cont.)
21
2pAVE
abs
V TP
T R
Hence 21
2pAVE
abs
VP
R
Ic = T/2
2
2
0
1cos
TpAVE
abs
VP t dt
T R
so
7
R []+-v (t)
cospv t V t
21
2pAVE
abs
VP
R
SummarySummary
8
Effective Voltage Effective Voltage VVeffeff
21
2pAVE
abs
VP
R
2p
eff
VV
2effAVE
abs
VP
R
Define:
Then we have:Note: Veff is used the
same way we use V in a DC power calculation.
9
Effective Voltage Effective Voltage VVeffeff
2effAVE
abs
VP
R
R []+-v (t)R []V
2AVE
abs abs
VP P
R
DC AC
same formula
10
ExampleExampleIn the U. S., 60 Hz line voltage has an effective voltage of 120 [V]. Describe the voltage waveform mathematically.
Veff = 120 [V]
2 120 2 169 71 [V]p effV V .
cos cos 2p pv t V t V ft
169 71cos 2 60v t . t so
11
ExampleExample60 Hz line voltage is connected to a 144 [] resistor.Determine the average power being absorbed.
R = 144 []+-
120 [V] (eff)
22 120100
144effAVE
abs
VP
R
100 [W]AVEabsP
12
RMS (Root Mean Square)RMS (Root Mean Square)
This is a general way to calculate the effective voltage for any periodic waveform (not necessarily sinusoidal).
t
v(t)
T digital pulse waveform
tp
Duty cycle: D = tp / T13
RMS (cont.)RMS (cont.)
2effAVE
abs
VP
R
2 2
0
1 TeffV v t
dtR T R
Hence,
2
0
1 TAVE
abs
v tP dt
T R Also,
By definition,
14
RMS (cont.)RMS (cont.)
2 2
0
1 TeffV v t
dtR T R
Hence
2
0
1 T
effV v t dtT
15
RMS (cont.)RMS (cont.)
Define
VRMS is the root (square root) of the mean (average) of the square of the voltage waveform
2
0
1 T
RMSV v t dtT
Veff = VRMS
Comparing with the formula for Veff , we see that
16
RMS (cont.)RMS (cont.)
For sinusoidal (AC) signals, 2
pRMS
VV
For other periodic signals, there will be a different relationship between VRMS and Vp.
(See the example at the end of these notes.)
17
RMS CurrentRMS Current
R []+-
i (t)
v (t)
The concept of effective (RMS) current works the same as for voltage.
cospi t I t
2eff RMS pI I I / 2AVEabs RMSP R I
2 2 2cosabs pP t i t R I R t
21
2AVE
abs pP I R
Define:
18
RMS Current (cont.)RMS Current (cont.)RMS current can be easily related to RMS voltage.
cos
cos
p
p
Vv ti t t
R RI t
cospv t V t
2
2p pRMS
RMS pp
V / VVR
I II /
R []+-
i (t)
v (t)
pp
VI
R
where
19
ExampleExample60 Hz line voltage is connected to a 144 [] resistor. Determine the RMS current and the average power absorbed (using the current formula).
1200 83333
144RMS
RMS
VI .
R
0 83333 [A]RMSI .
120 [V] (RMS)
R = 144 []+-
IRMS 22 144 0 83333 100AVEabs RMSP R I .
100 [W]AVEabsP
20
RMS Voltage and CurrentRMS Voltage and Current
Power can also be expressed in terms of both RMS voltage and current.
2AVE RMS RMS
abs RMS RMS RMS
V VP V V I
R R
AVEabs RMS RMSP V I
R []+-
IRMS
VRMS
-
+
21
ExampleExample60 Hz line voltage is connected to a 144 [] resistor. Determine the average power (using the voltage-current formula).
0 83333 [A]RMSI .
R = 144 []+-
120 [V] (RMS)
IRMS
100 [W]AVEabsP
120 [V]RMSV
120 0 83333AVEabs RMS RMSP V I .
22
Summary of AC PowerSummary of AC Power
2AVE RMS
abs
VP
R
AVEabs RMS RMSP V I
2AVEabs RMSP R I
RMSRMS
VI
R
2p
RMS
VV
2p
RMS
II
R []+- VRMS
IRMS
-
+
23
Example (non-sinusoidal)Example (non-sinusoidal)Find the RMS voltage of a sawtooth waveform:
for 0pv t V t / T t T
t
v (t)
T
Vp
2
2
0 0
1 1T T
RMS p
tV v t dt V dt
T T T
24
Example (cont.)Example (cont.)
2
0
2 22 3
2 200
23 2
2
1
1 1 1
3
1 1 1
3 3
T
RMS p
TTp p
pp
tV V dt
T T
V Vt dt t
T T T T
VT V
T T
Hence 3RMS pV V /25
Example (sawtooth wave)Example (sawtooth wave)
t
v (t)
T
Vp
R []+-
v (t)
Given: Vp = 10 [V]
R = 100 []
Find the average power absorbed by the resistor.
26
Example (cont.)Example (cont.)
t
v (t)
T
Vp = 10 [V]
100 []+-
v (t)
3RMS pV V /2
AVE RMSabs
VP
R
25 7735
0 33333100
AVEabs
.P .
10 3 5 7735 [V]RMSV / .
0 33333 [W]AVEabsP .
(for sawtooth)
27