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16.202: Effective or RMS Values
• Alternating Current (AC) waveforms may be characterized
with:
– Peak value ; Peak-to-Peak value ; Average value
– Effective or Root Mean Square (RMS) values
• RMS value of i(t) = Imcos(ωt + φi)
Irms =



=




1
T
T
0

2 cos2 (ωt + φ )dt
Im
i

2 1
Im

T

(1
+
cos(2ωt
+
2φ
))
i  dt
0
T
2
Irms =
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Prof.
K. Chandra, February 17, 2006
Im
√
2
16.202: Circuit Theory II; ECE, UMASS Lowell
2
• The effective value provides a comparison between AC and DC
circuits
• Average power absorbed by a resistance R Ohms :
P =
2
Im
2 R
=
2
I
√m R
2
2
= Irms
R Watts
• A DC source of Irms Amperes will supply same average power
as an AC source of peak amplitude Im and effective value
Im
Irms = √
2
Note:
Im
Irms = √
2
Vm
Vrms = √
2
VmIm
Pavg =
cos(θv − θi)
2
= Vrms Irmscos(θv − θi)
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
3
• Oscilloscope may be used to measure the peak-to-peak or peak
values.
• The multimeter measures the effective or rms value of the signal.
• Power company specification of 110/220 Volts (available from
outlet) is an RMS value.
Examples
1. Find effective value of current shown:
Im
-T
0
T
2T
Figure 1: Sawtooth function
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
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1 T 2
=
i (t)dt
T 0
2 2
t
1 T Im
dt
=
0
2
T
T
2
I
= m
3
Im
√
Irms =
3
2. Find effective value of i(t) = 4cos(100t) + 5cos(100t + 150o)
I = 4 0 + 5 150o = 4 + 5cos(150) + j 5 sin(150)
2
Irms
Irms
= −0.33√+ j2.5 = 2.52 −83.157o
= 2.52/ 2 = 1.782
Amps.
(2)
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
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Apparent Power and Power Factor
• Average Power : P =
1
2 Vm Im cos(θv
− θi )
• P = VrmsIrmscos(θv − θi) = S cos(θv − θi)
• S = VrmsIrms is termed Apparent Power (volt-amps: VA)
• The term cos(θv − θi) is the power factor (pf )
• P = S pf
pf = P/S = cos(θv − θi)
• Angle θv − θi is the power factor angle
Vm Vrms • Note: Impedance Z = V
=
θ
−
θ
=
v
i
I
Irms θv − θi
m
I
• Power factor pf : Cosine of angle of load impedance 0 < pf < 1
• Load Resistive : Z = R : pf=1; Reactive:Z = jX:pf=0
• In general Z = R + jX : pf is leading or lagging:
– Leading power factor: Current Leads Voltage (Capacitive)
– Lagging power factor: Current Lags Voltage (Inductive)
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
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Power Factor Impact
• Power Factor has an economic impact on consumers of large
power (industrial loads)
• A load with a low power factor that consumes P watts of power
draws higher current from a constant voltage source.
• Higher currents increase line losses and increase the amount of
supplied power
• Such loads can be charged at a higher rate by power companies
• Ideally a pf of 1.0 is desired
• Most loads that consume a large amount of power are inductive
• Inductive load can be changed by adding capacitors to increase
the pf towards unity value, thus optimizing cost.
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
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Example
1
5
120 0 rms
60 Hz
10
10mH
Figure 2:
(a) Find apparent and real power supplied by the source.
Real Power Supplied: Vrms Irms pf
Apparent Power: S = Vrms Irms
0
Source Current Is = 120
Z
in
Zin
10 (5 + j(2 ∗ π ∗ 60 ∗ 10 ∗ 10−3))
= 1+
15 + j(2 ∗ π ∗ 60 ∗ 10 × 10−3)
= 1 + 4.05 22.89 = 4.73 + j1.575 = 4.985 18.42
Therefore :
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
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• Is(rms) =
120
4.985 18.42
= 24.07 −18.42
• Apparent Power : S = 120 × 24.07 = 2888.6 V A (2.888
kVA )
• Power Factor : pf = cos(0 + 18.47) = 0.9485
Lagging
(b) If a pf of 0.92 lagging is desired, to what value should
the inductance be changed to ?
Zin
Zin
cos−1 (pf )
tan[cos−1(pf )]
=
=
=
=
A θv − θi = A cos−1(pf )
R + jX
tan−1(X/R)
X/R
In circuit above, replace the inductor by unknown X,
Zin
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Prof.
K. Chandra, February 17, 2006
10(5 + jX)
= 1 +
15 + jX
16.202: Circuit Theory II; ECE, UMASS Lowell
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975 + 11X 2
j100X
=
+
226
226
100X
975 + 11X 2
tan[cos−1(0.92)] = 0.426
100X
0.426 =
975 + 11X 2
415.35 + 4.686X 2 − 100X = 0
tan[cos−1(pf )] =
Solving for X: X = 5.649 or X = 15.691
Using ωL = X and ω = 120π
L = 14.9mH or L = 41.62 mH
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
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Complex Power
• Average (P ) and Apparent power (S) are both real quantities
• Consider power represented as a complex number
∗
VI
S = 2 = Vm2Im θv − θi = VrmsI∗rms
• Magnitude: |S| =
•S=
Vm Im
2 cos(θv
Vm Im
2
: Apparent Power
− θi) + j Vm2Im sin(θv − θi)
• Real part of S is
Vm Im
2 cos(θv
− θi) is Average Power P
• S = P + jQ
• Where Q =
Vm Im
2 sin(θv
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Prof.
K. Chandra, February 17, 2006
− θi) is Reactive Power
16.202: Circuit Theory II; ECE, UMASS Lowell
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• Units
– Average Power P : Watts
– Complex Power S : volt-amps (VA)
– Reactive power Q : volt-amps reactive (VAR)
• In terms of load impedance Z
V
I
Vrms
=
Irms
Z =
S = VrmsI∗rms
= Z IrmsI∗rms
2
2
= Irms
Z = Vrms
/Z
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
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Since Z = R + jX
2
S = Irms
Z
2
= Irms
[R + jX] = P + jQ
Matching Real and Imaginary Parts
2
2
P = Re[S] = Irms
R
Q = Im[S] = Irms
X
• Complex power incorporates all power information of a load
• Real power P is the average power in watts delivered to the
load
• Reactive power Q represents energy exchange between source
and reactance of load
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell
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Summary
• RMS/Effective values : Vrms =
Vm
√
2
Irms =
Im
√
2
• Average Power P = VrmsIrmscos(θv − θi) (Watts)
• Apparent Power S = VrmsIrms (VA)
• Power Factor (pf) : cos(θv − θi)
∗
• Complex Power S = V2 I = P + jQ
• Reactive Power Q =
Vm Im
2 sin(θv
− θi )
• Current Leads Voltage: Leading PF
• Current Lags Voltage: Lagging PF
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Prof.
K. Chandra, February 17, 2006
16.202: Circuit Theory II; ECE, UMASS Lowell