EE 204
Lecture 26
The Root Mean Square (RMS) Value
The RMS Value of a Periodic Function of Time:
The RMS [Root Mean Square] value of the periodic function f (t ) is defined as:
to +T
f rms =
∫
t
f 2 (t )dt
o
T
= [ f 2 (t )]avg
The RMS value is also called the effective value ⇒ f eff = f rms
Example 1:
Find the average and effective (rms) values of the sawtooth waveform.
Solution:
By inspection Yav = 25
2
For the interval 0 < t < 2, y = 25t ; then Yrms
=
Yrms = 28.9
1 T 2
1 2
y dt = ∫ 625t 2 dt = 834, and
∫
T 0
2 0
Example 2:
Find the average and effective (rms) values of the waveform shown.
y
10
t[s]
0
0.01
0.02
0.03
0.04
Solution:
For 0 < t < 0.01, y = 10;
For 0.01 < t < 0.03, y = 0; the period is 0.03s
1 0.01
10(0.01)
10dt =
Yav =
= 3.33
∫
0.03 0
0.03
1 0.01 2
2
and Yrms
10 dt = 33.3 therefore Yrms = 5.77
=
0.03 ∫0
The RMS Value of the General Sinusoidal Function:
Let us find the RMS value of the general sinusoidal function f (t ) = A cos(ωt + θ )
f rms = [ f 2 (t )]avg = [ A2 cos 2 (ωt + θ )]avg = [ A2
1 + cos(2ωt + 2θ )
]avg
2
f rms =
A
A
[1 + cos(2ωt + 2θ )]avg =
[1]avg + [cos(2ωt + 2θ )]avg
2
2
f rms =
A
1+ 0
2
∴ f rms =
[because the average value of a general sinusoid equals zero]
A
2
The R.M.S. value is commonly used in sinusoidal circuits.
For instance, the sinusoidal voltage v(t ) = V cos(ωt + θ v )
Has the R.M.S. value Vrms =
V
2
The value V represents the peak voltage
And the value Vrms =
V
represents the R.M.S. voltage
2
Similarly, the sinusoidal current i(t ) = I cos(ωt + θi )
Has a peak current I and an R.M.S. current I rms =
I
2
Household voltage varies sinusoidally in time. The commonly used 120 V household
voltage is actually an R.M.S measure, it is not a measure of the peak voltage.
This means that if the outlets at a certain household provide 120 V (actually
120 V (rms ) ), then:
120 V (rms)
⇒
V = 120 × 2 = 169.71 170 V (peak voltage)
Figure 1
We shall use the units V (rms) & I (rms ) when the R.M.S. measure is used.
Example 1:
The outlets at a certain household provide 110 V (rms ) at 60 Hz. Write an expression
for the voltage v(t ) supplied by those outlets. Assume the phase angle to be zero.
Solution:
v(t ) = V cos(ωt + θ v ) = V cos(ωt + 0) = V cos ωt
V = Vrms × 2 = 110 V (rms ) × 2 = 155.56 V
ω = 2π f = 2π × 60 = 120π = 376.991 377 [rad / s ]
v(t ) = 155.56 cos 377t V
(supply voltage as a function of time)
Power Expressions in terms of the R.M.S. Voltage and Current:
The power expressions must be revised when R.M.S. values are used.
For instance, the average power expression P =
Can be rewritten as P = (
1
IV cos θ
2
I
V
)( ) cos θ = I rmsVrms cos θ
2
2
The expression P = I rmsVrms cos θ , which assumes R.M.S. values for the voltage and
1
current does not have the factor of .
2
Similarly, when R.M.S. values of the voltage and current are used, we must drop the
1
factor from the expressions for the reactive power, apparent power and the
2
complex power. Thus:
P=
1
IV cos θ
2
⇒
P = I rmsVrms cos θ
(average power)
Q=
1
IV sin θ
2
⇒
Q = I rmsVrms sin θ
(reactive power)
S=
1
IV
2
⇒
S = I rmsVrms
(apparent power)
S=
1 ∗
I V
2
⇒
∗
S = I rms
Vrms
(complex power)
It is important to keep in mind that the values of the above powers remain the same,
whether we use peak values or R.M.S. values.
There is no such thing as R.M.S. power. This measure applies only to voltages and
currents.
Example 2:
For the load impedance Z , calculate:
a) the load current I
b) the complex power
c) the average and the reactive powers
d) the current i (t ) as a function of time if the source frequency is 50 Hz.
I
VS =
220 0o
Z = 15 60o Ω
Figure 2
Solution:
a) The supply voltage Vs = 220 0o V (rms ) is clearly given in the R.M.S. measure.
I =
Vs 220 0o V (rms )
=
= 14.667 −60o A(rms )
Z
15 60o Ω
[Since Vs is given in R.M.S. units ⇒ I is also in R.M.S. units]
b) The voltage and current associated with the load Z
Vs = 220 0o V (rms ) & I = 14.667 −60o A(rms ) (both in R.M.S. units).
Therefore, to we drop the factor
1
in all power expressions:
2
∗
S = I rms
Vrms
⇒
S = (14.667 −60o )∗ × (220 0o ) = (14.667 +60o ) × (220 0o )
S = 3226.74 +60o VA (the load complex power)
c) The load average and reactive powers can easily be obtained from S :
P = Re( S ) = Re(3226.74 +60o ) = Re(1613.37 + j 2794.44) = 1613.37 W
Q = Im( S ) = Im(1613.37 + j 2794.44) = 2794.44 VAR
f = 50 Hz , then:
d) Since I = 14.667 −60o A(rms ) &
i(t ) = 14.667 × 2 cos(2π × 50t − 60o ) = 20.742 cos(314.16t − 60o ) A
I
VS =
220 0o
Z = 15 60o Ω
Figure 2
Power Factor Correction:
The addition of a reactive element (capacitance or inductance) in parallel with the
load in order to increase the power factor is known as “power factor correction”
Figure 3
When the load has a lagging pf, power factor correction is done by connecting a
capacitor in parallel with the load.
Figure 4
When pf of the load is leading, power factor correction is done by connecting an
inductor in parallel with the load.
Figure 5
There are a number of methods to carryout the power correction procedure. The most
efficient approach utilizes the complex power.
Example 3:
The load has a power factor of 0.6 (lagging) and it absorbs 8 KW of average power
when connected to the 220 0o V (rms ) , 60 Hz source. Calculate the capacitance C to
be connected in parallel with this load so that the “combined load” has a pf of 0.90
(lagging).
IL
VS =
220 0o
VL
Figure 6
Solution:
Because the capacitor is connected to the load in parallel, the load voltage ( VL ) and
current ( I L ) do not change. Therefore, the power absorbed by the load also does not
change.
The load average power PL = 8000 W
The pf angle of the load θ L = + cos −1 (0.6) = 53.130o ( θ L > 0 due to the lagging pf)
IL
VS =
220 0o
VL
1
j ωC
Figure 7
Using the load power triangle:
tan θ L =
QL
PL
⇒
tan 53.130o =
QL
8000
⇒
QL = 10666.67 VAR
Figure 8
Therefore, load the complex power is:
S L = PL + jQL = 8000 + j10666.67 VA
The capacitor does not absorb average power, therefore:
The combined load average power Pcom = PL = 8000 W
The pf angle of the combined load θ com = + cos −1 (0.9) = 25.842o
IL
VS =
220 0o
VL
1
j ωC
Figure 7
Using the combined load power triangle:
tan θ com =
Qcom
Pcom
⇒
tan 25.842o =
Qcom
8000
⇒
Qcom = 3874.58 VAR
Figure 9
Therefore, the combined load complex power:
Scom = Pcom + jQcom = 8000 + j 3874.58 VA
The combined load complex power ( Scom ) equals the sum of the load complex power
( S L ) and the capacitor complex power ( Sc ).
Scom = S L + Sc
⇒
(8000 + j 3874.58) = (8000 + j10666.67) + Sc
Giving:
Sc = (8000 + j 3874.58) − (8000 + j10666.67)
Sc = − j 6792.09 = 6792.09 −90o VA
This result is to be expected, since the real part of capacitor complex power must be
zero.
IL
VS =
220 0o
VL
1
j ωC
Figure 7
Sc = I c∗Vc = 6792.09 −90o
I c∗ =
6792.09 −90o
= 30.873 −90o A(rms )
220 0o
V
1
= c
jωC I c
C=
& Vc = Vs = 220 0o V (rms )
⇒
1
j 2π fC
=
Vc
Ic
⇒
⇒ I c = 30.873 +90o = j 30.873 A(rms )
1
220 0o
=
j 2π × 60 × C j 30.873
30.873
= 3.722 × 10−4 F = 372.2 µ F
2π × 60 × 220
IC
VS =
220 0o
VC
1
j ωC
Figure 10
Example 4:
The load has a power factor of 0.7 (lagging). It absorbs 5 KW of average power when
connected to the120 0o V (rms ) , 60 Hz source.
a) Calculate the capacitance C to be connected in parallel with this load so that the
combined load has a unity pf.
b) Calculate the magnitude of the current supplied by the voltage source before and
after power factor correction.
I
VS =
120 0o
Figure 11
Solution:
a) The same procedure will be used to calculate the value of C.
θ L = + cos −1 (0.7) = 45.557o
IC
VS =
120 0o
VC
1
j ωC
Figure 12
QL = 5000 tan 45.557o = 5098.17 VAR
Figure 13
S L = PL + jQL = 5000 + j 5098.17 VA
The combined load has pf = 1 (unity pf) ⇒ θ com = 0
Therefore, the combined load is purely resistive. We know that a purely resistive
impedance has zero reactive power. Thus:
Scom = Pcom + jQcom = Pcom + j 0 = Pcom
Since the combined load average power equals the load average power, then:
Scom = Pcom = PL = 5000 VA
Scom = S L + Sc
⇒
5000 = (5000 + j 5098.17) + Sc
Sc = − j 5098.17 = 5098.17 −90o VA
Sc = I c∗Vc = 5098.17 −90o
I c∗ =
5098.17 −90o
= 42.485 −90o A(rms )
120 0o
V
1
= c
jωC I c
C=
& Vc = Vs = 120 0o V (rms )
⇒
⇒ I c = 42.485 +90o = j 42.485 A(rms )
1
120 0o
=
j 2π × 60 × C j 42.485
42.485
= 939.12 µ F
2π × 60 ×120
IC
VS =
120 0o
[V] rms
f=60Hz
Load
5 kW
p.f.=0.7
lagging
+
VC
1
j ωC
Figure 12
b) Before pf correction, the supply current equals the load current:
Is = IL
Using PL = I LVL cos θ L
⇒ 5000 = I L120 × 0.7 ⇒ I L = I s = 59.514 A(rms )
IS
IL
VS =
120 0o
Figure 14
After pf correction, the supply current does not equal the load current, because:
I s = I L + Ic
Using
Pcom = I comVcom cos θ com = I sVs cos θ com
5000 = I s 120 × 1 ⇒
I s = 41.667 A(rms )
Thus:
I s = 59.514 A(rms )
(before pf correction)
I s = 41.667 A(rms ) (after pf correction)
We can see that when the pf is raised from 0.7 to unity, the supply current drops
In practice, pf correction is done in order to reduce the supply current I s
IS
IL
IC
VS =
120 0o
1
j ωC
Figure 15
Household and industrial voltages are supplied by electric power companies. The
generators at a power company are connected to the users using transmission lines.
In this simplified diagram, Vs represents the generator voltage.
A practical transmission line has a nonzero resistance R
This nonzero resistance results in electric power loss in the transmission line, because
PR = I R2 R = I s2 R
By increasing the pf of the load, I s is reduced resulting in reduced PR . The power
company will save money in this case.
IS
VS
VL
Figure 16
When the load has a leading pf, a similar procedure can be used for pf correction.
However, an inductive reactance must be connected in parallel with the load.