circuit analysis

1. 1
Topic 2b
AC Circuits Analysis

2. 2
Example
An ac generator with rms voltage of
110 V is connected in series with a
35-Ω resistor and 1-µF capacitor. At
what frequency must the generator
operate if it is to maintain a current
of 1.2 Amps.

3. 3
Example
Z = V/I = 110/1.2 = 91.7 Ω
Z2
= R2
+ (1/ ωC)2
; f = ω/2π = 1.9 kHz

4. 4
Example
An ac generator emf is ε = ε0∙sinωt,
with ε0∙ = 22.8 V and ω = 353 rad/s. It
is connected to a 17.3 H inductor.
When the current is a maximum, what
is the emf of the generator? When
the emf of the generator is -11.4 V
and increasing in magnitude, what is
the current?

5. 5
Example
mAXVI L 73.3)3.17353/(8.22/maxmax =⋅==
)30180cos(
3sin
3018030360
2/18.22/4.11sin
max


+⋅=
⇒
+−=⇒
−=−=
II
quadrantrdmagnitudegincrea
or
t
ϕ
ω
ϕ

6. 6
Example
An ac generator emf is ε = ε0∙sin ωt,
with ε0∙ = 28.6 V and ω = 333 rad/s. It
is connected to a 4.15 µF capacitor.
When the current is a maximum, what
is the emf of the generator? When
the emf of the generator is -14.3 V
and increasing in magnitude, what is
the current?

7. 7
Example
ϕ
mAXVI C 5.39)1015.4333/1/(6.28/ 6
maxmax =×⋅== −
)30180cos(
3sin
3018030360
2/16.28/3.14sin
max


+⋅=
⇒
+−=⇒
−=−=
II
quadrantrdmagnitudegincrea
or
t
ϕ
ω

8. 8
Example
A 0.4-H inductor and a 220-Ω
resistor are connected in series to an
ac generator with an rms voltage of
30 V and a frequency of 60 Hz. Find
the rms values of the current, voltage
across the resistor, and voltage
across the inductor.

9. 9
Example
Irms = Vrms/Z = 0.112 A
(VL)rms = IrmsXL= 17 V
(VR)rms = IrmsR = 24.6 V

10. 10
Example
An ac generator has a frequency of 1200 Hz
and a constant rms voltage. When a 470-Ω
resistor is connected between the terminals
of the generator, an average power of 0.25
W is dissipated in the resistor. Then, a
0.08-H inductor is connected in series with
the resistor, and the combination is
connected between the generator terminals.
What is the average power dissipated in the
series combination?

11. 11
Example

12. 12
Example
A person is working near the secondary of
a transformer, as shown. The primary
voltage is 120 V at 60.0 Hz. The
capacitance Cs
, which is the stray
capacitance between the hand and the
secondary winding, is 20.0 pF. Assuming the
person has a body resistance to ground Rb
=
50.0 kΩ, determine the rms voltage across
the body.

13. 13
Example

14. 14
Example
A typical "light dimmer" used to dim the
stage lights in a theater consists of a
variable inductor L (whose inductance is
adjustable between zero and Lmax) connected
in series with a lightbulb. The electrical
supply is 110 V (rms) at 62.5 Hz; the
lightbulb is rated as "110 V, 900 W.“ What
Lmax is required if the rate of energy
dissipation in the lightbulb is to be varied
by a factor of 5 from its upper limit of 900
W? Assume that the resistance of the
lightbulb is independent of its
temperature. Could a variable resistor be
used instead?

15. 15
Example
( )
( )
mH
R
L
R
LR
LR
LR
ZV
ZV
RI
RI
P
P
rms
rms
rms
rms
3.71
)60(2
900/)110(22
)(
)(
)(
5
5
/
/
2
max
2
2
max
2
2
min
2
2
max
2
2
max
2
2
min
2
2
min,
2
max,
min
max
===⇒
+
=
+
+
=
==
⋅
⋅
=
πω
ω
ω
ω

16. 16
Modern dimmer

17. 17
Phasors for a series RLC circuit
Max or rms: V0
2
= VR
2
+ (VL
- VC
)2
Vrms
= Irms
Z
VR
= Irms
R, VC
= Irms
XC
, and VL
= Irms
XL

18. 18
Power factor for a series RLC circuit

19. 19
Example
A series RLC circuit contains a 148-Ω
resistor, a 1.50-μF capacitor, and a
35.7-mH inductor. The generator has
a frequency of 512 Hz and an rms
voltage of 35 V. What is (a) the rms
voltage across each circuit element
and (b) the average electric power
consumed by the circuit.

20. 20
Example

21. 21
rms current in RL and RC circuits depends on
frequency

22. 22
When the phase difference between current and
voltage is zero (Xc=XL), the circuit is in resonance

23. 23
How does the variable air capacitor work?

24. 24
Phasors
• Sinusoids are easily expressed in terms of
phasors, which are more convenient to work
with than sine and cosine functions.
• A phasors is a complex number that
represents the amplitude and phase of a
sinusoid.

25. 25
Complex number
• A complex number z can be written in rectangular
form as z = x + jy where : x is the real part
and y is the imaginary part of z.
• The complex number z can also be written in polar
form or exponential form as
where r is the magnitude of z and φ is the the phasor
of z.
1−=j
φ
φ j
rerz =∠=
φ
φ j
rerjyxz =∠=+=

26. 26
Complex number
• Rectangular form
• Polar form
• Exponential form
• Given x, y can get r and φ
• Know r and φ, can obtain x and y
jyxz +=
φ∠= rz
φj
rez =
.tan, 122
x
y
yxr −
=+= φ
.sin,cos φφ ryrx ==

27. 27
Important operation
• Addition:
• Subtraction:
• Multiplication:
• Division:
• Reciprocal:
• Square root:
)()( 212121 yyjxxzz −+−=−
)()( 212121 yyjxxzz +++=+
)( 212121 φφ +∠= rrzz
)( 21
2
1
2
1
φφ −∠=
r
r
z
z
( )φ−∠=
rz
11
( )2/φ∠= rz

28. 28
Complex conjugate
• Complex conjugate of z is
Real part of e jφ
.
Imaginary part of ejφ
.
( ) φ
φ j
rerjyxz −∗
=−∠=−=
j
j
−=
1
( )
( )φ
φ
φ
φ
φ
φφ
j
j
j
e
e
je
Imsin
Recos
sincos
=
=
±=±

29. 29
Real and imaginary part
• can be express as
• V is the phasor representation of the sinusoid v(t).
)cos()( φω += tVtv m
( )
( )
( ) φ
φω
φω
ωφ
φω
∠===
=
=+= +
m
j
m
tj
tjj
m
tj
mm
VeVetv
eeVtv
eVtVtv
VV whereRe)(
Re)(
Re)cos()( )(

30. 30
Sinusoidal-phasor transformation
Time –domian
representation
Phasordomian
representation
)cos( φω +tVm
)cos( φω +tIm
)sin( φω +tVm
)sin( φω +tIm
φ∠mV
)90( o
mV −∠ φ
φ∠mI
)90( o
mI −∠ φ
( )
tion)representadomain-(Phasortion)representadomain-Time(
cos)( φφω ∠=⇔+= mm VtVtv V

31. 31
Derivative and Integral
• The derivative of v(t) is transformed to the phasor
as jωV.
• The integral of v(t) is transformed to the phasor as
V/ jω.
domain)(Phasordomian)(Time
j Vω⇔
dt
dv
domain)(Phasordomian)(Time
jω
V
⇔∫vdt

32. 32
Differences between v(t) and V
1. v(t) is the instantaneous or time domain
representation, while V is the frequency or
phasor-domain representation.
2. v(t) is time dependent, while V is not
3. v(t) is always real with no complex term,
while V is generally complex.

33. 33
Example
Transform these sinusoids to phasors
(a)
(b)
Solution
(a)
(b)
( )o
ti 4050cos6 −=
( )o
tv 5030sin4 +−=
( )o
I 406 −∠=
( )
( )
( )
o
o
oo
o
t
t
tv
1404
14030cos4
905030cos4
5030sin4
∠=
+=
++=
+−=
V
sin (ωt±180o
) = −sin ωt
Cos (ωt±180o
) = −cos ωt
sin (ωt±90o
) = ±cos ωt
cos (ωt±90o
) = sin ωt

34. 34
Example
Find the sinusoids represented by these phasors:
(a) I = -3 +j4
(b)
Solution
(a) I = -3 +j4 = 5∠(126.87o
)
(b)
o
j
ej 20
8 −
=V
( )o
tti 87.126cos5)( += ω
( )( )
( )o
o
ooj
ttv
ej
o
70cos8)(
708
)20(89018 20
+=
∠=
−∠∠== −
ω
V
o
j 901∠=

35. 35
Phasor relationships for circuit
elements
• For resistor R , V = RI I = V/R
• For inductor L, V = jωLI
• For capacitor C, I = jωCV
Cjω
I
V =
Ljω
V
I =

36. 36
Example
The voltage is applied
to a 0.1-H inductor. Find the steady-state current
through the inductor.
Solution
ω = 60, V = 12∠45o
( )o
tv 4560cos12 +=
( )o
o
oo
jLj
452
906
4512
)1.0)(60(
4512
−∠=
∠
∠
=
∠
==
ω
V
I
( )A4560cos2)( o
tti −=

37. 37
Impedance and Admittance of passive elements
• Impedance Z of a circuit is the ratio of the phasor
voltage V to the phase current I, measured in ohms.
• The admittace Y is the reciprocal of impedance,
measured in siemens.
Element Impedance Admittance
R Z = R
L Z = jωL
C
Cjω
1
=Z
R
1
=Y
Cjω=Y
Ljω
1
=Y

38. 38
Impedance Z
• The complex quantity Z may be represented by
rectangular form as Z = R +jX.
where R = ReZ is the resistance and
X = Im Z is the reactance.
θθ
θ
θ
sinX,cos
tan, 1-22
ZZ
Z
ZZ
==
=+=
∠=+=
R
R
X
XR
jXR

39. 39
Example
Find v(t) and i(t) in the circuit shown.
+_
i 5Ω
0.1F v
+
_
vs=10cos4t

40. 40
Solution
∀ ω = 4. Vs = 10∠0o
.
• Voltage across the capacitor
( )
o
o
oo
s
j
j
j
57.26789.1
57.2659.5
010
5.25
010
5.25
)1.0)(4(
1
5
∠=
−∠
∠
=
−
∠
==
−=+=
Z
V
I
Z
)43.634cos(47.4)(
)57.264cos(769.1)(
)43.63(47.4
904.0
57.26789.1
)1.0)(4(
57.26789.1
o
o
o
o
oo
c
ttv
tti
jCj
−=
+=
−∠=
∠
∠
=
∠
===
ω
I
IZV

41. 41
Impedance combination
• Consider N series-connected impedance,
V = V1 + V2 +… +VN = I(Z1+Z2+…+ZN)
• Consider N parallel-connected impedance,
N21eq Z...ZZ
I
V
Z +++==
N21eq
N21eq
N21
N21
Y...YYY
Z
1
...
Z
1
Z
1
V
I
Z
1
Z
1
...
Z
1
Z
1
VI...III
+++=
+++==






+++=+++=

42. 42
Example
Find the input impedance in the circuit shown.
Assume that the circuit operates at ω = 50 rad/s.
Zin
2mF
0.2H
8Ω
3Ω
10mF

43. 43
Solution
• Impedance of 2mF capacitor
• Impedance of the 3-Ω resistor in series with 10
mF capacitor
• Impedance of the 8-Ω resistor in series with 0.2 H
inductor
Ω−=== 10
)002.0)(50(
11
1 j
jCjω
Z
Ω−=+=+= )23(
)01.0)(50(
1
3
1
32 j
jCjω
Z
Ω+=+=+= )108()2.0)(50(882 jjLjωZ

44. 44
Solution
• The input impedance is
07.1122.3
07.122.310
811
)811)(1444(
10
811
)108)(23(
10
)108()23(
)108)(23(
10
22
j
jj
jj
j
j
jj
j
jj
jj
j
−=
−+−=
+
−+
+−=
+
+−
+−=
++−
+−
+−=+= 321in ZZZZ

45. 45
Example
• Determine vo(t) in the circuit.
+_
+
_vo
5H10mF
60Ω
20sin(4t-15o
)

46. 46
Solution
• Vs = 20∠(-105o
)
• Impedance of 10 mF capacitor
• Impedance of 5-H inductor ZL= j(4)(5)=j20Ω
• Impedance of the parallel combination of 10-mF
capacitor and 5-H inductor
Ω−== 25
)01.0)(4(
1
j
j
cZ
V)04.744cos(15.17)(
04.7415.17)105(20)(96.308575.0(
)105(20(
10060
100
60
100
5
500
)20()25(
)20)(25(
2025
o
o
ooo
o
so
ttv
j
j
j
jjj
jj
jj
−=
−∠=−∠∠=
−∠
+
=
+
=
Ω=
−
=
+−
−
=−=
V
Z
Z
V
Z

47. 47
Effective or RMS Value
• The idea of effective value arises from the
need to measure the effectiveness of a
voltage or current source in delivering
power to a resistive load.
• The effective value of a period current is the
dc current that deliver the same average
power to a resistor as the periodic current.

48. 48
Root Mean Square Value
• Effective value for current
• Effective value for voltage
• The effective value of a periodic signal is them root
mean square (rms) value
∫=
T
eff dti
T
i
0
21
∫=
T
eff dtv
T
v
0
21
∫=
T
rms dtx
T
X
0
21

49. 49
RMS value for Sinusoidal function
• The rms value of a constant is the constant itself.
• For sinusoidal i(t) = Im cos ωt, the effective value
is
• For v(t) = Vm cos ωt, the effective value is
.
2
)2cos1(
2
1
cos
1 2
22 m
T
o
m
T
o
mrms
I
dtt
T
I
tdtI
T
I =+== ∫∫ ωω
.
2
m
rms
V
V =

50. 50
Average power
• The average power absorbed by a resistor R can
be written as
• For instance, the 230 V available at every
household is the rms value.
• It is convenient in power analysis to express
voltage and current in their rms value.
R
V
RIP rms
rms
2
2
==

51. 51
Example
Determine the rms value of the current waveform
as shown. If the current is passed through a 2-Ω
resistor, find the average power absorbed by the
resistor.
0
10
-10
i(t)
t2 4 6 8 10

52. 52
Solution
• The period of the waveform T = 4.
• Can write the current waveform as
• The power absorbed by a 2-Ω resistor is
At
t
dtdttdti
T
I
t
tt
ti
T
rms
165.8200
3
200
4
1
2
4
100
0
2
3
25
4
1
)10()5(
4
11
42,10
20,5
)(
3
4
2
2
2
0
2
0
2
=



+=





+×=






−+==



<<−
<<
=
∫∫∫
.3.133)2()165.8( 22
WRIP rms ===

53. 53
Example
The waveform shown is a half-wave
rectified sine wave. Find the rms value and
the amount of average power dissipated in a
10-Ω resistor.
10
0
v(t)
tπ 2π 3π

54. 54
Solution
• The period of the voltage is T = 2π, and
( )
W
R
v
P
v
t
t
dtttdtv
t
tt
tv
rms
rms
rms
5.2
10
5
isabsorbedpowerAverage
5
250
2
2sin25
02
2sin
4
100
2cos1
2
100
2
1
sin100
2
1
2,0
0,sin10
)(
22
00
22
===
=
=





−−=





−=
−==



<<
<<
=
∫∫
π
π
π
π
π
ππ
ππ
π
ππ

55. 55
Maximum Power Transfer
• Why use Thevenin and Norton equivalents?
– Very easy to calculate load related quantities
– E.g. Maximum power transfer to the load
• It is often important to design circuits that
transfer power from a source to a load. There
are two basic types of power transfer
– Efficient power transfer: least power loss (e.g.
power utility)
– Maximum power transfer (e.g.
communications circuits)
• Want to transfer an electrical signal (data,
information etc.) from the source to a
destination with the most power reaching
the destination. Power is small so
efficiency is not a concern.

56. 56
Maximum Power Transfer Theorem
Given a Thévenin equivalent circuit of a
network, the maximum power that can be
transferred to an external load resistor is
obtained if the load resistor equals the
Thévenin resistance of the network. Any
load resistor smaller or larger than the
Thévenin resistance will receive less power.

57. 57
Maximum Power Demonstrated (1/2)
PL is max if
and
Th
Th L
V
I
R R
=
+
( )
2
2
L
L Th
Th L
R
P V
R R
=
+0L
L
dP
dR
=

58. 58
Maximum Power Demonstrated (2/2)
• Maximum Power Transfer Theorem applied
when matching loads to output resistances
of amplifiers
• Efficiency is 50% at maximum power
transfer
⇒
and
0L
L
dP
dR
= L ThR R=