circuit analysis

1. 1
Topic 2a :AC Circuits Analysis
• Capacitors network
• Inductors network
• AC analysis – average power, rms power
• Frequency response of RC and RL circuits.

2. 2
Capacitors
• A capacitor is a passive element designed to store
energy in its electric field.
• A capacitor consists of two conducting plates
separated by an insulator (or dielectric).
• When a voltage source v is connected to the
capacitor, the source deposits a charge q on one
plate and a negative charge –q on the other.
• q = Cv, where C is the capacitance of the
capacitor.

3. 3
Capacitance
• Capacitance is the ratio of the charge on one plate of a
capacitor to the voltage difference between the two plates,
measured in Farad (F).
• For a parallel-plate capacitor. The capacitance is given by
where A is the surface area of each plate,
d is the distance between the two plates, and
 is the permitivity of the dielectric material.
d
A
C



4. 4
Symbols for capacitors
Variable capacitor
Fixed capacitor
Fixed capacitor

5. 5
Current- voltage relationship of the
capacitor
• Since different q = Cv gives
• Voltage current relationship
where v(to) is the voltage across the capacitor at time
to.
,
dt
dq
i  .
dt
dv
Ci 
)(
1
or
1
o
t
t
t
tvidt
C
vidt
C
v
o
 

6. 6
Power delivered to capacitor
• The instantaneous power delivered to a capacitor
is
• Energy stored in the capacitor is
• v(-) = 0 then
.
dt
dv
Cvvip 

  
t
t
CvvdvCdt
dt
dv
vCpdtw
tt
2
2
1
.
2
or
2
1 2
2
C
q
wCvw 

7. 7
Important properties of capacitor
1. When the voltage across the capacitor is not
changing with time, the current through the
capacitor is zero.
A capacitor is an open circuit to dc.
2. The voltage on the capacitor must be
continuous.
The voltage on a capacitor cannot change
abruptly (suddenly).
2. The ideal capacitor does not dissipate energy.
3. A real capacitor has a parallel leakage resistance.
The leakage resistance may be as high as 100
M.

8. 8
Example
(a) Calculate the charge stored on a 3-pF capacitor
with 20V across it.
(b) Find the energy stored in the capacitor.
Solution
(a)
(b) Energy stored
.6020103 12
pCCvq  
.60020103
2
1
2
1 2122
pJCvw  

9. 9
Example
The voltage across a 5-F capacitor is
Calculate the current through it.
Solution
The current is
V.6000cos10)( ttv 
A.6000sin3.0
6000sin106000105
6000cos10
105)(
6
6
t
t
dt
t)d(
dt
dv
Cti






10. 10
Example
If a 10-F capacitor is connected to a voltage
source with
determine the current through the capacitor.
Solution
The current is
V.2000sin50)( ttv 
A.2000cos
2000cos5020001010
2000sin50
1010)(
6
6
t
t
dt
t)d(
dt
dv
Cti






11. 11
Example
Determine the voltage across a 2-F capacitor if the
current through it is
Assume that the initial capacitor voltage is zero.
Solution
mA.6)( 3000t
eti 

V.)1(
)1(
0)3000(102
106
106
102
11
3000
30003000
6
3
0
30003
6
0
t
tt
t
t
t
e
e
t
e
dteidt
C
v













 

12. 12
Example
Obtain the energy in each capacitor under dc
conditions

13. 13
Solution
• Under dc conditions, replaced each capacitor with
an open circuit.
• Current through 2-k and 4k resistor is
mA
mAi
2
)6(
423
3




14. 14
Solution
• Voltage v1 across capacitor 2 mF is
• Voltage v2 across capacitor 4 mF is
• Energy stored in capacitor 2 mF is
• Energy stored in capacitor 4 mF is
mJvCw 104102
2
1
2
1 232
111  
mJvCw 1288104
2
1
2
1 232
222  
V.420001  iv
V.840002  iv

15. 15
Series and parallel capacitors
• A equivalent capacitance of N parallel-connected
capacitors is the sum of the individual capacitors.
Ceq = C1 + C2 + C3 + … + Cn.
• The equivalent capacitance of series-connected
capacitors is the reciprocal of the sum of the
reciprocal of the individual capacitors.
• For n = 2,
21
21
21
111
CC
CC
C
CCC
eq
eq



16. 16
Example
• Find the equivalent capacitance seen
between terminal a and b of the circuit.
Ceq
a
b
20F
5F
6F20F
60F

17. 17
Solution
• 20F and 5F are in series, their equivalent
capacitance is
• This 4F is parallel with the 6F and 20F
capacitors, their combined capacitance is
4+6+20 = 30F.
• This 30 F is in series with 60 F capacitor. Hence
the equivalent capacitance is
.4
520
520
F


.20
3060
3060
FCeq 




18. 18
Inductors
• An inductor is a passive element designed to store
energy in its magnetic field.
• An inductor consists of a coil of conducting wire.
• If current is allowed to pass through an inductor, it
is found that the voltage across the inductor is
directly proportional to the time rate of change of
the current.
where L is the
inductance. dt
di
Lv 

19. 19
Inductance
• The unit of inductance is the Henry (H).
• Inductance is the property whereby an inductor
exhibits opposition to the change of current
flowing through it measured in Henry (H).
• The inductance of an inductor depends on its
physical dimension and construction.
• For solenoid where N is the
number of turns, l is the length, A is the cross-
sectional area, and  is the permeability of the
core.
l
AN
L
2


20. 20
Circuit symbol for inductors
(a) air-core (b) Iron-core (c) Variable

21. 21
Current-voltage relationship for
inductor
• The current-voltage relationship is obtained from
where i(to) is the current at time to.






t
t
o
t
o
tidttv
L
i
dttv
L
i
vdt
L
di
)()(
1
)(
1
1

22. 22
Power delivered to the inductor
• Power delivered is
• Energy stored is
i
dt
di
Lvip 






2
2
1
LiidiL
idt
dt
di
Lpdtw
i
tt













23. 23
Properties of inductor
1. Voltage across an inductor is zero when the
current is constant.
An inductor acts like a short circuit to dc.
2. Its opposition to the change in current flowing
through it.
The current through an inductor cannot change
instantaneously.
3. Ideal inductor does not dissipate energy.
4. Real inductor has winding resistance and
capacitance.

24. 24
Example
The current through a 0.1-H inductor is
Find the voltage across the inductor and the
energy stored in it.
Solution
Voltage
Energy stored
A.10)( 5t
teti 

 
V.)51(
)5(
10
1.0
5
55
5
t
tt
t
et
ete
dt
ted
v





J.5t)10)(1.0(
2
1 10225 tt
etew 


25. 25
Example
Consider the circuit as shown, under dc
condition, find (a) I, vc and iL,, (b) the energy
stored in the capacitor and inductor.

26. 26
Solution
(a) Under dc condition, replace the capacitor with an
open circuit and inductor with a short circuit.
i = iL = 12/(1+5) = 2 A.
vC = 5 i = 5x2 = 10V
(a) Energy in capacitor is
Energy in inductor is
J.50)10)(1(
2
1 2
Cw
J.4)2)(2(
2
1 2
Lw

27. 27
Series Inductors
• Consider a series of N inductors, the equivalent
inductor
• The equivalent inductance of series-connected
inductors is the sum of the individual inductors.
Neq LLLLL  ...321

28. 28
Parallel Inductors
• Consider a parallel–connection of N inductors, the
equivalent inductor
• The equivalent inductance of parallel inductors is
the reciprocal of the sum of the reciprocal of the
individual inductors.
Neq LLLLL
1
...
1111
321


29. 29
Two inductors in parallel
• For two inductors in parallel combination,
Or
• The combination is the same way as resistors in
parallel.
21
111
LLLeq

.
21
21
LL
LL
Leq



30. 30
Example
Find the equivalent inductance of the circuit.

31. 31
Solution
• The 10-H, 12-H and 20-H inductors are in series,
combining them gives 42-H inductance.
• This 42-H inductor is in parallel to 7-H inductor
so that they are combined, to give
• This 6-H inductor is in series with 4-H and 8-H
inductors. Hence Leq = 4+6+8 = 18H.
.6
427
427
H



32. 32
Example
For the circiut shown,
If i2(0) = -1 mA,
Find: (a) i1(o), (b) v(t), v1(t) and
v2(t); (c) i1(t) and i2(t).
.)2(4)( 10
mAeti t


33. 33
Solution
(a)
(b) Equivalent inductance is
Leq = 2 + 4║12 = 2+3 = 5H
.5)1(4)0()0()0(
.4)12(4)0(
21
2121
mAiii
iiiiii
mAmAi



mV.200)10)(1(45
))2(4(
5)(
1010
10
tt
t
eq
ee
dt
ed
dt
di
Ltv






34. 34
Solution
Since v = v1 +v2.
v2.= v - v1 = 120e-10t mV.
(c) Current i1 is obtained as
similarly
mV.80)10)(1)(4(22)( 1010
1
tt
ee
dt
di
tv 

.385335
0
-3e
mA5
4
120
)0(
4
1
)(
101010t-
0 0
10
121
mAeemA
t
dteidtvti
tt
t t
t



  
.111
0
-e
mA1
12
120
)0(
12
1
)(
101010t-
0 0
10
222
mAeemA
t
dteidtvti
tt
t t
t



  

35. 35
***AC current

36. 36
)2cos(
2
1
)cos(
2
1
)( ivmmivmm tIVIVtp  
• Instantaneous power changes with time and is therefore difficult to
measure. The average power is more convenient to measure.
• Wattmeter is measuring average power.
Average power (watts) – average power of the instantaneous power over the period.
The average power is given as:

T
avg dttp
T
P
0
)(
1
Average power

37. 37
Average Power
)cos(
2
1
ivmmavg IVP  
)2cos(
2
1
)cos(
2
1
)( ivmmivmm tIVIVtp  

T
avg dttp
T
P
0
)(
1
From the,
It can be derived that the average power:
(constant) (sinusoid)
Since, the average of constant is constant and the average of sinusoid = 0.
So,
dttIV
T
dtIV
T
P ivmm
T
ivmm
T
avg )2cos(
2
11
)cos(
2
11
00
  

38. 38
Average power of purely resistive load, R
090cos
2
1
 o
mmavg IVP
For purely resistive load, v = i, so that:
Average power of reactive circuit: C, L
# The resistive load absorb power all the time.
For purely reactive elements, v - i =  90o, so that:
RRIIVP mmmavg
22
|I|
2
1
2
1
2
1

# The purely reactive load (L or C) absorbs zero power.

39. 39
• Effective value determine the effectiveness of voltage or current source in delivering
power to the load.
• For any periodic function x(t) in general, the rms value is given by
• For a sinusoid i(t) = Im cos t Similarly for v(t) = Vm cos t

T
rms dtx
T
X
0
21
2
cos
1
0
22 m
T
mrms
I
dttI
T
I   
2
cos
1
0
22 m
T
mrms
V
dttV
T
V   
)cos()cos(
2
1
ivrmsrmsivmm IVIVP  
R
V
RIP rms
rms
2
2

The average power can be written in terms of rms values.
The average power absorbed by a resistor, R is,
Effective or RMS Power

40. 40
The voltage V produced between the
terminals of an ac generator fluctuates
sinusoidally in time. Why?

41. 41
What’s the average voltage in an AC circuit?
Average current?

42. 42
Does zero average current and voltage imply
zero average power?
# NO, it depends on your load is R, C or L.

43. 43
What power is dissipated in a resistor R
connected across an AC voltage source?
V = V 0 sin 2p f t I = (V0/R) sin 2p f t = I 0 sin 2p f t
P = I0*V 0 sin2 2p f t Pave = I0*V0/2

44. 44
***In this three-phase circuit there is 266 V rms between line
1 and ground. What is the rms voltage between line 2 and 3?
VVVV
tV
ttVVV
VVV
rms
rms
460
3
sin2
2
1
)(
)2(cos)(sin2
)]3/2sin()3/4[sin(
2662/
023
2
1
3
2
2
1
0
023
0




p

pp
p

45. 45
AC vs DC: The battle of the
currents

46. 46
Example
A stereo receiver applies a peak ac
voltage of 34 V to a speaker. The
speaker behaves approximately as if it
has a resistance of 8.0 , as shown.
Determine (a) the rms voltage, (b) the
rms current, and (c) the average power
for this circuit.

47. 47
Example

48. 48
Example

49. 49
In a resistive load, the current and voltage are in
phase
RC and RL circuit

50. 50
In a purely capacitive load, the current leads the
voltage by 90º.

51. 51
In a purely capacitive load, the current leads the
voltage by 90º.
)(
1
)/1(
2
cossin
cos
cos
0
max
max
maxmax
max
max
reactancecapacitive
C
X
C
I
tCtC
dt
dQ
I
t
C
Q
tV
V
C
C
C



p
















CCC XIV 

52. 52
What is the average power dissipated
in the capacitor?
On the average, the power is zero and a capacitor uses no energy in an ac
circuit.
V=V0 sin (2pft)
I=I0 sin (2pft + p/2) = I0 cos (2pft)
Pave = (I·V)ave = 0
Average power, P = 0
But, Reactive power, Q ≠ 0
c
c
X
V
XIQ
2
2


53. 53
In a purely inductive circuit, the current lags the
voltage by 90º
)(
2
cos
2
cos
)0(sincos
cos
cos
0
max
max
max
max
maxmax
max
max
reactanceinductiveLX
L
I
tIt
L
I
CCt
L
dtt
L
I
t
dt
dI
L
tV
V
L
L
L



p

p
































54. 54
In a purely inductive circuit, the current lags the
voltage by 90º
c
c
X
V
XIQ
2
2

For purely inductive load, again
Average power, P = 0
But, Reactive power, Q ≠ 0

55. 55
RC circuit diagram

56. 56
RL circuit diagram

57. 57
RLC circuit diagram

58. 58
Phasors for a series RLC circuit
Max or rms: V0
2 = VR
2 + (VL - VC)2
Vrms = IrmsZ
VR = IrmsR, VC = IrmsXC, and VL = IrmsXL

59. 59
Power factor for a series RLC circuit

60. 60
Summary of the AC phasor relations

61. 61
Summary of impedance relations
= 1/ωC
= ωL