1
Topic 2a :AC Circuits Analysis
• Capacitors network
• Inductors network
• AC analysis – average power, rms power
• Frequency response of RC and RL circuits.
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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.
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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
AC
4
Symbols for capacitors
Variable capacitor
Fixed capacitor
Fixed capacitor
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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
dqi .
dt
dvCi
)(1
or 1
o
t
t
t
tvidtC
vidtC
vo
6
Power delivered to capacitor
• The instantaneous power delivered to a capacitor
is
• Energy stored in the capacitor is
• v(-) = 0 then
.dt
dvCvvip
t
tCvvdvCdt
dt
dvvCpdtw
tt2
2
1
.2
or 2
1 22
C
qwCvw
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Important properties of capacitor1. 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.
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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
.600201032
1
2
1 2122 pJCvw
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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
6000cos10105)(
6
6
t
t
dt
t)d(
dt
dvCti
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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
2000sin501010)(
6
6
t
t
dt
t)d(
dt
dvCti
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ExampleDetermine the voltage across a 2-F capacitor if the
current through it is
Assume that the initial capacitor voltage is zero.
Solution
mA. 6)( 3000teti
V. )1(
)1(0)3000(102
106
106102
11
3000
300030006
3
0
300036
0
t
tt
tt
t
e
et
e
dteidtC
v
12
Example
Obtain the energy in each capacitor under dc conditions
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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
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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 1041022
1
2
1 232111
mJvCw 12881042
1
2
1 232222
V.420001 iv
V.840002 iv
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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
CCC
CCC
eq
eq
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Example• Find the equivalent capacitance seen
between terminal a and b of the circuit.
Ceq
a
b
20F
5F
6F20F
60F
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Solution• 20F and 5F are in series, their equivalent
capacitance is
• This 4F is parallel with the 6F and 20F capacitors, their combined capacitance is
4+6+20 = 30F.• This 30 F is in series with 60 F capacitor. Hence
the equivalent capacitance is
.4520
520F
.203060
3060FCeq
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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
diLv
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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
ANL
2
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Circuit symbol for inductors
(a) air-core (b) Iron-core (c) Variable
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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
tidttvL
i
dttvL
i
vdtL
di
)()(1
)(1
1
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Power delivered to the inductor
• Power delivered is
• Energy stored is
idt
diLvip
2
2
1LiidiL
idtdt
diLpdtw
i
tt
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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.
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ExampleThe current through a 0.1-H inductor is
Find the voltage across the inductor and the energy stored in it.
SolutionVoltage
Energy stored
A. 10)( 5tteti
V. )51(
)5(10
1.0
5
555
t
ttt
et
etedt
tedv
J. 5t) 10)(1.0(2
1 10225 tt etew
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Example Consider the circuit as shown, under dc
condition, find (a) I, vc and iL,, (b) the energy stored in the capacitor and inductor.
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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
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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
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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
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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
LLLeq
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Example
Find the equivalent inductance of the circuit.
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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.
.6427
427H
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ExampleFor 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
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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
diLtv
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Solution
Since v = v1 +v2.
v2.= v - v1 = 120e-10t mV.
(c) Current i1 is obtained as
similarly
mV. 80)10)(1)(4(22)( 10101
tt eedt
ditv
.3853350
-3e
mA 54
120)0(
4
1)(
101010t-
0 0
10121
mAeemAt
dteidtvti
tt
t tt
.1110
-e
mA 112
120)0(
12
1)(
101010t-
0 0
10222
mAeemAt
dteidtvti
tt
t tt
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***AC current
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)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 dttpT
P0
)(1
Average power
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Average Power
)cos(2
1ivmmavg IVP
)2cos(2
1)cos(
2
1)( ivmmivmm tIVIVtp
T
avg dttpT
P0
)(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,
dttIVT
dtIVT
P ivmm
T
ivmm
T
avg )2cos(2
11)cos(
2
1100
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Average power of purely resistive load, R
090cos2
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 mmmavg22 |I|
2
1
2
1
2
1
# The purely reactive load (L or C) absorbs zero power.
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• 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 dtxT
X0
21
2cos
10
22 mT
mrms
IdttI
TI
2cos
10
22 mT
mrms
VdttV
TV
)cos()cos(2
1ivrmsrmsivmm IVIVP
R
VRIP rms
rms
22
The average power can be written in terms of rms values.
The average power absorbed by a resistor, R is,
Effective or RMS Power
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The voltage V produced between the terminals of an ac generator fluctuates
sinusoidally in time. Why?
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What’s the average voltage in an AC circuit? Average current?
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Does zero average current and voltage imply zero average power?
# NO, it depends on your load is R, C or L.
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What power is dissipated in a resistor R connected across an AC voltage source?
V = V 0 sin 2 f t I = (V 0/R) sin 2 f t = I 0 sin 2 f t
P = I0*V 0 sin2 2 f t Pave = I0*V 0/2
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***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
4603
sin22
1)(
)2(cos)(sin2
)]3/2sin()3/4[sin(
2662/
023
2
1
3
2
2
10
023
0
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AC vs DC: The battle of the currents
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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.
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Example
48
Example
49
In a resistive load, the current and voltage are in phase
RC and RL circuit
50
In a purely capacitive load, the current leads the voltage by 90º.
51
In a purely capacitive load, the current leads the voltage by 90º.
)(1
)/1(
2cossin
cos
cos
0
maxmax
maxmax
max
max
reactancecapacitiveC
X
CI
tCtCdt
dQI
tC
Q
tV
V
C
C
C
CCC XIV
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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 (2ft)
I=I0 sin (2ft + /2) = I0 cos (2ft)
Pave = (I·V)ave = 0
Average power, P = 0 But, Reactive power, Q ≠ 0
cc X
VXIQ
22
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In a purely inductive circuit, the current lags the voltage by 90º
)(
2cos
2cos
)0(sincos
cos
cos
0
maxmax
maxmax
maxmax
max
max
reactanceinductiveLXL
I
tItL
I
CCtL
dttL
I
tdt
dIL
tV
V
L
L
L
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In a purely inductive circuit, the current lags the voltage by 90º
cc X
VXIQ
22
For purely inductive load, again
Average power, P = 0 But, Reactive power, Q ≠ 0
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RC circuit diagram
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RL circuit diagram
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RLC circuit diagram
58
Phasors for a series RLC circuit
Max or rms: V02 = VR
2 + (VL - VC)2
Vrms = IrmsZ
VR = IrmsR, VC = IrmsXC, and VL = IrmsXL
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Power factor for a series RLC circuit
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Summary of the AC phasor relations
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Summary of impedance relations
= 1/ωC
= ωL