Lecture Notes: 2304154 Physics and Electronics
Lecture 2 (2nd Half), Year: 2007
Physics Department, Faculty of Science, Chulalongkorn University
11/10/2007
Contents
1 Power in Ac Circuits
1
2 Signal Frequency and Impedances
2.1 First Order Filter Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Resonant and Second Order Filter Circuits . . . . . . . . . . . . . . . . . .
9
10
13
3 Problems
18
1
Power in Ac Circuits
Time Dependent Power
Given:
V (t) = Vm sin ωt
I(t) = Im sin(ωt + φ)
From: P (t) = V (t)I(t)
P (t) = Vm Im sin ωt sin(ωt + φ)
Trigonometry: 2 sin A sin B = cos(B − A) − cos(B + A)
1
P (t) = Vm Im [cos φ − cos(2ωt + φ)]
2
Power in ac circuit depends on time.
Power of Pure Resistive Load
V (t) = Vm sin ωt
I(t) = Im sin ωt
P (t) = Vm Im sin2 ωt
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V, I
0
V (t)
t
I(t)
P
t
0
Figure 1 Current, voltage, and power versus time for a purely resistive load.
V, I
V (t)
I(t)
t
0
P
t
0
Figure 2 Current, voltage, and power versus time for a purely inductive load.
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V, I
V (t)
I(t)
t
0
P
t
0
Figure 3 Current, voltage, and power versus time for a purely capacitive load.
Power of Pure Inductive Load
V (t) = Vm sin ωt
I(t) = Im sin(ωt − 90o )
1
P (t) = Vm Im [cos(−90o )
2
− cos(2ωt − 90o )]
1
= − Vm Im sin 2ωt
2
Power of Pure Capacitive Load
V (t) = Vm sin ωt
I(t) = Im sin(ωt + 90o )
1
P (t) = Vm Im [cos(90o )
2
− cos(2ωt + 90o )]
1
= Vm Im sin 2ωt
2
Active Power or Average Power
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Active power or average power (unit in Watt,W): the energy delivered to the electrical elements in one cycle divided by the period.
P = Pav
Because T =
2π
ω
and
R
cos(2θ + φ)dθ =
1
2
1
=
T
Z
T
P (t)dt
0
sin(2θ + φ)
Z 2π 1
1
P =
Vm Im cos φ − cos(2ωt + φ) d(ωt)
2π 0
2
1
= Vm Im cos φ = Vrms Irms cos φ
2
Power Factor and Power Angle
cos φ = power factor (P F )
φ
= power angle
Definition:
In case:
V (t) = Vm sin(ωt + φv )
I(t) = Im sin(ωt + φi )
φ = φv − φi
pure resistive load
pure inductive load
pure capacitive load
(φ = 0; P F = 1)
(φ = 90o ; P F = 0)
(φ = −90o ; P F = 0)
:
:
:
P
P
P
= 12 Vrms Irms
= 0
= 0
Power in Complex Number System
In general cases;
V = Vm φv = Vm cos φv + j Vm sin φv
I = Im φi = Im cos φi + j Im sin φi
V I ∗ = (Vm cos φv + j Vm sin φv )(Im cos φi − j Im sin φi )
= Vm Im cos(φv − φi ) + j Vm Im sin(φv − φi )
That is:
1
P = Re(V I ∗ )
2
1
Q = Im(V I ∗ )
2
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Im
|Z|
X
φ
Re
R
Figure 4 The load impedance in a complex plane
Reactive Power and Apparent Power
reactive power, Q (unit in Voltage–Amperes–Reactive, VAR): the energy flow back
and forth to the energy storage elements (inductances and/or capacitances).
1
Q = Im(V I ∗ ) = Vrms Irms sin φ
2
P 2 + Q2 = (Vrms Irms )2 cos2 φ + (Vrms Irms )2 sin2 φ
= (Vrms Irms )2 = AP 2
AP = Vrms Irms : apparent power (unit in Voltage–Amperes, VA)
The reactive power effects the power dissipation in the lines and transformers of a
power distribution system. Thus, the industrial factory has to pay for this kind of power.
That is ‘5–kW load’ means P = 5 kW, ‘5–kVA load’ means AP = 5 kVA, and ‘a load
absorbs 5–kVAR’ means Q = 5 kVAR.
Power Triangle
inductive load (I lags V )
AP
Q
(φ = positive)
φ
P
P
capacitive load (I leads V )
φ
Q
(φ = negative)
AP
Additional Power Relationship
Z = |Z|φ = R + j X
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XL
IC
I
V (t)
R
XC
IR
V (t) = 10− 900
I(t) = 0.1414− 1350
IR = 0.1− 1200
IC = 0.1− 900
XL = j 100 Ω
XC = −j 100 Ω
R = 100 Ω
Figure 5 Circuit for example 1
X=+
X=−
for inductance
for capacitance
R
|Z|
X
sin φ =
|Z|
cos φ =
P =
=
Q=
=
P =
Q=
Vm Im
Vm Im R
cos φ =
2
2 |Z|
2
Im
2
R = Irms
R
2
Vm Im
Vm Im X
sin φ =
2
2 |Z|
2
Im
2
X = Irms
X
2
2
VRrms
; VRrms : across resistance
R
2
VXrms
; VXrms : across reactance
X
Example 1.
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Find power and reactive power
φ = φv − φi
VSrms
Irms
= −90o − (−135o ) = 45o
√
√
= Vm / 2 = 10/ 2 = 7.071 V
√
√
= Im / 2 = 0.1414/ 2 = 0.1 A
P = VSrms Irms cos φ
= 7.071 × 0.1 cos(45o ) = 0.5 W
Q = VSrms Irms sin φ
= 7.071 × 0.1 sin(45o ) = 0.5 VAR
QR = 0
2
QL = Irms
XL = (0.1)2 (100) = 1.0 VAR
0.1
2
QC = ICrms
XC = ( √ )2 (−100)
2
= −0.5 VAR
QS = QL + QC
PL = PC = 0
0.1
2
PR = IRrms
R = ( √ )2 (100)
2
= 0.5 W
Example 2.
Find PS , Q, P F , and phasor current I
P FA = 0.5 leading: capacitive (XA = −, QA = −)
P FB = 0.7 lagging: inductive (XB = +, QB = +)
PA = Vrms Irms P FA = (104 )(0.5)
= 5 kW
q
QA = − APA2 − PA2
p
= − (104 )2 − (5000)2
= −8.67 kVAR
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I
V (t)
A
IA
IB
B
V (t) = 1414300
APA = 10 kVA
P FA = 0.5 leading
PB = 5 kW
P FB = 0.7 lagging
Figure 6 Circuit for example 2
APB = PB /P F = 5000/0.7 = 7.14 kVA
q
QB = APB2 − PB2
p
= (7142)2 − (5000)2
= 5.101 kVAR
PS = PA + PB = 10 kW
QS = QA + QB = −3.56 kVAR
QS
= −19.6o
φS = tan−1
PS
P FS = cos φS = −0.94 (94.2% leading)
p
APS = P 2 + Q2 = 10.6 kVA
1414
VSrms = √ = 1 kV
2
APS
Irms =
= 10.61 A
VSrms
√
Im = 2Irms = 15 A
φi = φv − φS = 30o − (−19.6o )
= 49.6o
I = 1549.6o A
Power Factor Correction
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AP
Q
φ
P
−Q
Figure 7 Power triangle for power factor correction
In heavy industry, Q causes higher currents in the power distribution system. The
energy rates charged to industry depend on the P F . To decrease P F , the capacitors is
placed in parallel with an inductive load to decrease the sum of Q.
Example 3. A 50 kW load operates from a 60–Hz 10–kV–rms line with a power factor of
60 % lagging. Compute the capacitance that must be placed in parallel with the load to
achieve a 90 % lagging power factor.
Load power angle (φL ) = cos−1 (0.6) = 53.13o
From power triangle concept;
QL = PL tan φL = 66.67 kVAR
After adding the capacitor, Pnew = PL = 50 kW and new power angle (φnew ):
φnew = cos−1 (0.9) = 25.84o
From power triangle concept;
Qnew = PL tan φnew = 24.22 kVAR
QC = Qnew − QL = −42.45 kVAR
XC = −
2
Vrms
= −2356 Ω
QC
Required capacitance;
C=
2
1
1
=
= 1.126 µF
2πf XC
2π(60)(2356)
Signal Frequency and Impedances
Frequency Dependent Impedance
Resistance:
R = Constant
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R, X
XL
R
XC
f
O
Figure 8 Graph shows the frequency effects the impedance value.
R
+
Vin
+
−
I
C
Vout
−
Figure 9 A first order lowpass filter.
Inductance:
XL = ωL = 2πf L
Capacitance:
XC =
2.1
1
1
=
ωC
2πf C
First Order Filter Circuits
First Order Lowpass Filter
From this circuit, first order differential equation can be written as:
RC
dVC (t)
+ VC (t) = Vs .
dt
Thus, it is called ‘first order circuit’.
Vin
Vin
=
ZT
R + 1/j 2πf C
1
= VC =
I
j 2πf C
1
Vin
=
×
j 2πf C
R + 1/j 2πf C
I=
Vout
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H(f )
|H(f )|
1
0o
0.707
0.5
−45o
0
0
fB 2fB 3fB 4fB
−90o
f
0
fB 2fB 3fB 4fB
f
Figure 10 Magnitude and phase of the first–order lowpass transfer function versus frequency.
Vout
1
=
Vin
1 + j 2πf RC
1
=
1 + j (f /fB )
H(f ) =
H(f ): transfer function
fB : half–power frequency
1
2πRC
fB =
In case, f = fB :
√
H(f ) = 1/ 2 = 0.707
P = Pmax /2
|H(f )| = p
1
1 + (f /fB )2
f
H(f ) = − arctan
fB
|H(f )|dB = 20 log |H(f )|
Pout
|
= 10 log|
Pin
Magnitude and Phase Plots of H(f )
f increases: H(f ) decreases
f increases: H(f ) decreases to -90o
The low frequency signal can be detected at Vout port. On another word, the low
frequency signal can be passed this circuit(Fig.10). (lowpass filter)
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L
+
Vin
+
−
R
Vout
−
Figure 11 Another first–order lowpass filter.
Example 4. Given an input signal Vin = 5 sin(20πt) + 5 sin(2000πt) is applied to the lowpass
RC filter. Let R = 1000/(2π) and C = 10µ F. Find an expression for the output signal.
fB =
1
1
=
= 100 Hz
2πRC
2π(1000/2π)(10 × 10−6 )
For the first component of Vin :
Vin1 (t) = 5 sin(20πt);
Vin1 = 50o
ω
20
=
= 10
2π
2π
1
1
H(10) =
=
= 0.995− 5.71o
1 + j (f1 /fB )
1 + j (10/100)
Vout1 = H(10)Vin1 = 4.975− 5.71o
f1 =
Vout1 (t) = 4.975 sin(20πt − 5.71o )
The second component of Vin :
Vin2 (t) = 5 sin(2000πt);
Vin2 = 50o
ω
2000
=
= 1000
2π
2π
1
1
=
= 0.0995− 84.29o
H(1000) =
1 + j (f2 /fB )
1 + j (1000/100)
Vout2 = H(1000)Vin2 = 0.4975− 84.29o
f2 =
Vout2 (t) = 0.4975 sin(2000πt − 84.29o )
Vout = 4.975 sin(20πt − 5.71o ) + 0.4975 sin(2000πt − 84.29o )
Another First Order Lowpass Filter
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C
+
Vin
+
−
R
Vout
−
Figure 12 A first order highpass filter.
First Order Highpass Filter
From this circuit:
Vin
Vin
=
ZT
R + 1/j 2πf C
Vout = VR = RI
Vin
=R×
R + 1/j 2πf C
j 2πf RC
=
1 + j 2πf RC
j(f /fB )
Vout
=
H(f ) =
Vin
1 + j (f /fB )
1
fB =
2πRC
I=
f /fB
|H(f )| = p
1 + (f /fB )2
f
o
H(f
)
= 90 − arctan
fB
Magnitude and Phase Plots of H(f )
f increases: H(f ) increases
f increases: H(f ) decreases to 0o
The high frequency signal can be detected at Vout port. On another word, the high
frequency signal can be passed this circuit(Fig.13). (highpass filter)
Another First Order Highpass Filter
2.2
Resonant and Second Order Filter Circuits
Series Resonance
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H(f )
|H(f )|
1
90o
0.707
0.5
45o
0
0
fB 2fB 3fB 4fB
0o
f
0
fB 2fB 3fB 4fB
f
Figure 13 Magnitude and phase of the first–order highpass transfer function versus frequency.
R
+
Vin
+
−
Vout
L
−
Figure 14 Another first–order highpass filter.
L
Vin
+
−
I
R
C
Figure 15 The series resonant circuit.
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From Kirchhoff’s Voltage Law:
d2 Q
dQ Q
+R
+
= Vm sin ωt
2
dt
dt
C
Q = Qm sin(ωt − φ)
Vm
Qm = p
L2 (ω 2 − ω02 )2 + R2 ω 2
1
1
1
=
− ωL
tan φ
R ωC
L
Compare with the equation of spring motion; natural frequency:
1
ω0 = √LC
If the frequency of Vin , ω = ω0 , Qm (or Im ) is maximum. It is called resonant
frequency.
On the other hands, resonant frequency:
f0 : the frequency at which the impedance is purely resistive (i.e.,the total reactance
is zero).
ZT = R + j (XL − XC )
resonant frequency can be found from:
XL − XC = 0
1
2πf0 L =
2πf0 C
1
f0 = √
2π LC
Define quality factor
2πf0 L
1
=
R
2πf0 CR
f
f0
ZT (f ) = R 1 + j Qs ( − )
f0
f
Qs =
current in this circuit:
I=
=
Vin
ZT (f )
Vin /R
1 + j Qs (f /f0 − f0 /f )
1
VR
=
Vin
1 + j Qs (f /f0 − f0 /f )
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|H(f )|
|H(f )|
Qs1 > Qs2 > Qs3 > Qs4
1
0.5
0
0.8
1
0.707
0.5
Qs4
Qs3 Qs2
Qs1
0.9
1.0
1.1
f
f0
1.2
f
0
fL f0 fH
Figure 16 Magnitude of the series resonant transfer function versus frequency.
+
Vin
+
−
R
C
L
Vout
−
Figure 17 The parallel resonant circuit.
Magnitude and Phase Plots of H(f )
The middle frequency signal can be detected at VR port(Fig.13). It can be called
second order bandpass filter.
Defind bandwidth B = fH − fL . At Qs 1:
f0
B∼
=
Qs
∼
fH = f0 + B/2
fL ∼
= f0 − B/2
Parallel Resonance
ZT =
1
1/R + j (2πf C − 1/2πf L)
f0 : the frequency at which the impedance is purely resistive (i.e.,the total reactance is
zero).
1
2πf0 L
1
f0 = √
2π LC
2πf0 C =
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|H(f )|
|H(f )|
Qp1 > Qp2 > Qp3 > Qp4
1
0.5
0
0.8
1
0.707
0.5
Qp4
Qp3 Qp2
Qp1
0.9
1.0
1.1
1.2
f
f0
f
0
fL f0 fH
Figure 18 Magnitude of the parallel resonant transfer function versus frequency.
|H(f )| (dB)
R
L
20
+
Vin
+
−
C
0
Vout
−20
−
−40
−60
f0 /10
first order filter
f0
second order filter
log f
10f0 100f0
Figure 19 Second–order lowpass filter and Transfer–function magnitudes
Define quality factor, Qp
R
= 2πf0 CR
2πf0 L
R
ZT =
1 + j Qp (f /f0 − f0 /f )
IR
Vout =
1 + j Qp (f /f0 − f0 /f )
Qp =
Magnitude and Phase Plots of H(f )
The middle frequency signal can be detected at VR port(Fig.18). It can be called
second order bandpass filter.
Defind bandwidth B = fH − fL . At Qp 1:
f0
B∼
=
Qp
Second Order Lowpass Filter
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|H(f )| (dB)
C
R
20
+
Vin
+
−
L
0
Vout
−20
−
−40
first order filter
second order filter
−60
f0 /10
f0
10f0 100f0
log f
Figure 20 Second–order highpass filter and Transfer–function magnitudes
R
|H(f )| (dB)
20
+
L
0
Vin
+
−
Vout
−20
−
−40
C
−60
f0 /10
f0
10f0
log f
Figure 21 Second–order band–reject (notch) filter and Transfer–function magnitudes
f0 =
1
√
;
2π LC
H(f ) =
2πf0 L
1
=
R
2πf0 CR
−j Qs (f0 /f )
=
1 + j Qs (f /f0 − f0 /f )
and Qs =
Vout
Vin
At Qp 1: |H(f )| reaches a high peak at f0 . Qs ∼
= 1 (maximally flat or Butterworth
function) is selected for filter design.
Second Order Highpass Filter
Second Order Band-Reject (Notch) Filter
3
Problems
1. (a) A voltage source V = 707.140o delivers 5 kW to a load with a power factor of
100 percent. Find the reactive power and the phasor current.
(b) Repeat if the power factor is 20 percent lagging.
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Figure 22 For Problem 4
Figure 23 For Problem 5
(c) For which power factor would the current ratings of the conductors connecting
the source to the load be higher? In which case could the wiring be a lower cost?
2. A load has an impedance
given by Z = 100 − j50Ω. The current flowing through
√
this load is I = 15 230o . Is the load inductive or capacitive? Determine the power
factor, power, and reactive power delivered to the load.
√
3. The phasor voltage across a certain load is 1000 230o , and the phasor current
√
through it is 15 260o . Determine the power factor, power, reactive power, and
impedance. Is the power factor leading or lagging?
4. Determine the power for each source shown in Figure 22. Also, state whether each
source is delivering or absorbing energy.
5. Find the power, reactive power, and apparent power delivered by the source in Figure
23. Find the power factor and state whether it is leading or lagging.
6. Derive and expression for the transfer function H(f ) = Vout /Vin of the filter shown in
Figure 11.
7. A first–order RC lowpass filter with a half–power frequency of 10 kHz is needed.
Determine the value of the capacitance if the resistance is 1 kΩ.
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8. A first–order lowpass filter has a break frequency of 500 Hz. The input is
Vin (t) = 5 + 3 sin(1000πt + 30o ) + 10 cos(20 × 103 πt)
Find an expression for the output voltage.
9. A first–order RC highpass filter is required to attenuate a 60–Hz input component by
40 dB. What value is required for the break frequency of the filter? By how many dB
is the 600–Hz component attenuated by this filter? If R = 1kΩ, what is the value of
C?
10. At the resonant frequency f0 = 1 MHz, a series resonant circuit with R = 50Ω has
VR = 2 V and VL = 20 V. Determine the values of L and C. What is the value of
VC ?
11. A series resonant circuit has B = 50 kHz, f0 = 400 kHz, and R = 20Ω. Determine
the values of L and C.
12. A parallel resonant circuit has f0 = 10 MHz and B = 200 kHz. The maximum value
of ZT is 10 kΩ. Determine the values of R, L, and C.
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