CIRCUIT ANALYSIS II · Circuit Analysis II WRM MT11 9 If we want to calculate the average power...
Transcript of CIRCUIT ANALYSIS II · Circuit Analysis II WRM MT11 9 If we want to calculate the average power...
Will Moore
MT 11 & MT12
CIRCUIT ANALYSIS II (AC Circuits)
Syllabus Complex impedance, power factor, frequency response of AC networks
including Bode diagrams, second-order and resonant circuits, damping
and Q factors. Laplace transform methods for transient circuit analysis
with zero initial conditions. Impulse and step responses of second-order
networks and resonant circuits. Phasors, mutual inductance and ideal
transformers.
Learning Outcomes At the end of this course students should:
1. Appreciate the significance and utility of Kirchhoff’s laws.
2. Be familiar with current/voltage relationships for resistors, capacitors
and inductors.
3. Appreciate the significance of phasor methods in the analysis of AC
circuits.
4. Be familiar with use of phasors in node-voltage and loop analysis of
circuits.
5. Be familiar with the use of phasors in deriving Thévenin and Norton
equivalent circuits
6. Be familiar with power dissipation and energy storage in circuit
elements.
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7. Be familiar with methods of describing the frequency response of
AC circuits and in particular
8. Be familiar with the Argand diagram and Bode diagram methods
9. Be familiar with resonance phenomena in electrical circuits
10. Appreciate the significance of the Q factor and damping factor.
11. Appreciate the significance of the Q factor in terms of energy
storage and energy dissipation.
12. Appreciate the significance of magnetic coupling and mutual
inductance.
13. Appreciate the transformer as a means to transform voltage, current
and impedance.
14. Appreciate the importance of transient response of electrical
circuits.
15. Be familiar with first order systems
16. Be familiar with the use of Laplace transforms in the analysis of the
transient response of electrical networks.
17. Appreciate the similarity between the use of Laplace transform and
phasor techniques in circuit analysis.
Circuit Analysis II WRM MT11
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AC Circuits
1. Basic Ideas
Our development of the principles of circuit analysis in Circuit Analysis I
was in terms of DC circuits in which the currents and voltages were
constant and so did not vary with time. We will now extend this analysis
to consider time varying currents and voltages. In our initial discussions
we will limit ourselves to sinusoidal functions. We choose this special
case because, as you have now learnt in P1, it allows us to make use of
some very powerful and helpful mathematical techniques. It is a
common waveform in nature and it is easy to generate in the lab.
However as you have also learnt in P1, any waveform can be expressed
as a weighted superposition of sinusoids of different frequencies and
hence if we analyse a linear circuit for sinusoidal functions we can, by
appropriate superposition, handle any function of time.
Let's begin by considering a sinusoidal variation in voltage
tVv m ω= cos
4
in which ω is the angular frequency and is measured in radians/second.
Since the angle ωt must change by 2π radians in the course of one
period, T, it follows that
πω 2=T
However the time period f
T 1= where f is the frequency measured in
Hertz. Thus
fT
ππ
ω 22==
This is a simple and very important relationship. We naturally measure
frequency in Hz – the mains frequency in the UK is 50Hz – and it is easy
to measure the time period, ( )fT 1= from an oscilloscope screen.
However as we will soon see, it is mathematically more convenient to
work in terms of the angular frequency ω. Mistakes may be easily made
because in practice the word frequency is commonly used to refer to
both ω and f. It is important in calculations to make sure that if ω
appears, then the correct value for f = 50 Hz, say, is ω = 100π rads/sec.
A simple point to labour I admit, but if I had a pound for every time
someone forgets and substitutes ω = 50 . . . . . . . . !!
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In our example above, tVv m ω= cos , it was convenient that
0== tVv m at . In general this will not be the case and the waveform will
have an arbitrary relationship to the origin t = 0 or, equivalently the origin
may have been chosen arbitrarily and the voltage, say, may be written in
terms of a phase angle, φ, as
( )φω −= tVv m cos
Alternatively, in terms of a different phase angle, ψ, the same waveform
can be written
( )ψω += tVv m sin
where
φπψ −= 2
T
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The phase difference between two sinusoids is almost always measured
in angle rather than time and of course one cycle (i.e. one period)
corresponds to 2π or 360°. Thus we might say that the waveform above
is out of phase with the earlier sinusoid by φ. When 2π±=φ we say
that the two sinusoids are said to be in quadrature. When π=φ the
sinusoids are in opposite phase or in antiphase.
2. RMS Values
We refer to the maximum value of the sinusoid, Vm, as the “peak” value.
On the other hand, if we are looking at the waveform on an oscilloscope,
it is usually easier to measure the “peak-to-peak” value 2Vm, i.e. from the
bottom to the top. However, you will notice that most meters are
calibrated to measure the root-mean-square or rms value. This is found,
as the name suggests, for a particular function, f, by squaring the
function, averaging over a period and taking the (positive) square root of
the average. Thus the rms value of any function f(x), over the interval x
to x+X, where X denotes the period is
( )dyyfX
fXx
xrms
21∫+
=
For our sinusoidal function tVv m ω= cos
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The average of the square is given by
dttVT m
Tω∫
22
0cos1
where the time period ωπ=2T . At this point it's probably easiest to
change variables to tω=θ and to write ( )θ+=θ 2cos121cos2 . Thus the
mean square value becomes
( )2
2cos122
1 22
0
2mm VdV
=θθ+π ∫
π
The root mean square value, which is simply the positive square root of
this, may be written as
Vrms = Vm /√2 ≈ 0.7 Vm.
Since we nearly always use rms values in our AC analysis, we assume
rms quantities unless told otherwise so by convention we just call it V as
in:
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V = Vm /√2.
So, for example, when we say that the UK mains voltage is 230V what
we are really saying is that the rms value 230V. Its peak or maximum
value is actually 230√2 ≈ 325 V.
To see the real importance of the rms value let's calculate the power
dissipated in a resistor.
Here the current is given by Rvi =
tItRVi mm ω=ω= coscos
where RVI mm = and the power, vip= , is given by
tRVp m ω= 22
cos
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If we want to calculate the average power dissipated over a cycle we
must integrate from π=ω=ω=ω 20 Ttt to . If we again introduce tω=θ ,
the average power dissipated, P, is given by
θθπ
= ∫π
dRVP m 2
2
0
2
cos.21
( ) θθ+π
= ∫π
dRVP m 2cos1
21.
21 2
0
2
RVP m 22=
If we now introduce the rms value of the voltage 2mVV = then the
average power dissipated may be written as
RVP 2=
Indeed if the rms value of the current 2mII = is also introduced then
RIRVP 22 ==
which is exactly the same form of expression we derived for the DC
case.
Therefore if we use rms values we can use the same formula for the
average power dissipation irrespective of whether the signals are AC or
DC.
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Circuit Analysis II WRM MT11
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3. Circuit analysis with sinusoids
Let us begin by considering the following circuit and try to find an
expression for the current, i, after the switch is closed.
The Kirchhoff voltage law permits us to write
tVRidtdiL m ω=+ cos
This is a linear differential equation, which you know how to solve.
We begin by finding the complementary function, from the homogeneous
equation:
0=+RidtdiL
which yields the solution:
( )LRtAi −= exp
12
We now need to find the particular integral which, for the sinusoidal
"forcing function" tVm ωcos , will take the form tCtB ω+ω sincos . Thus
the full solution is given by
( ) ( ) tCtBLRtAti ω+ω+−= sincosexp
We see that the current consists of a "transient" term, ( )LRtA −exp ,
which eventually decays and becomes negligible in comparison with the
"steady state" response. The transient response arises because of the
sudden opening or closing of a switch but we will concentrate here on
the final sinusoidal steady state response. How long do we have to wait
for the steady state? If for example Ω=100R and mHL 25= then
13 sec104 −−×=LR and so after only 1ms ( ) 018.04expexp =−=− LRt
and so any measurements we are likely to make on this circuit will be
truly 'steady state' measurements. Thus our solution of interest reduces
to
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tCtBi ω+ω= sincos
In order to find B and C we need to substitute this expression back into
the governing differential equation to give
{ } { } tVtCtBRtBtCL m ω=ω+ω+ω−ωω cossincossincos
It is now a simple matter to compare coefficients of cosωt and sinωt to
obtain expressions for B and C which lead, after a little algebra, to
( ){ }tLtR
LRVi m ωω+ωω+
= sincos22
If we now introduce the inductive reactance ( )LXL ω= we can write this
equation as
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
ω+
+ω++
= tXR
XtXR
RXR
ViL
L
LL
m sincos222222
The expression in curly brackets is of the form
( )φ−ω=ωφ+ωφ ttt cossinsincoscos
and hence
i = VmR 2 +X L
2cos ωt −ϕ( )
where
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ϕ = tan−1X L
R"
#$
%
&'
Thus we see that the effect of the inductor has been to introduce a
phase lag φ between the current flowing in the circuit and the voltage
source. Similarly the ratio of the maximum voltage to the maximum
current is given by 22LXR + which since it is a combination of
resistance and reactance is given the new name of impedance.
It is apparent that we could solve all networks containing combinations
of resistors, inductors and capacitors in this way. We would end up with
a series of simultaneous equations to solve – just as we did when
analysing DC circuits – the problem is that they would be simultaneous
differential equations which, given the effort we went through to solve
one equation in the simple example above, would be very tedious and
therefore rather error-prone. Fortunately there is an easier way.
We are saved because the differential equations we have to solve are
linear and hence the principle of superposition applies. This tells us
that if a forcing function v1(t) produces current i1(t) and a forcing function
( )tv2 provides current ( )ti2 then ( ) ( )tvtv 21 + produces ( ) ( )titi 21 + . The
trick then is to choose a more general forcing function ( ) ( ) ( )tvtvtv 21 += in
which, say, ( )tv1 corresponds to tVm ωcos and which made the
differential equation easy to solve. We achieve this with complex
algebra.
You should know that
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exp j ωt =cosωt + j sinωt .
where j = √(-1), [electrical engineers like to use i for current]
so let’s solve the differential equation with the general forcing function
v t( )=Vm exp j ωt =Vm cosωt + j Vm sinωt
where
v1 t( )=Re v t( ){ }=Re Vm exp j ωt{ }=Vm cosωt .
The solution will be of the form
i (t ) =I exp jωt
where 𝐼 = 𝐼 exp −j∅ will, in general, be a complex number. Then in
order to find that part of the full solution corresponding to the real part of
the forcing function, tjVm ωexp we merely need to find the real part of
( )ti . Thus
i1 t( )=Re I exp jωt{ }=Re I exp j ωt −ϕ( ){ }= I cos ωt −ϕ( )
Let's illustrate this by returning to our previous example where we tried
to solve:
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tVRidtdiL m ω=+ cos
Now, instead, we solve the more general case:
tjVRidtdiL m ω=+ exp
and take the real part of the solution. As suggested above an
appropriate particular integral is tjIi ω= exp which leads to
tjVtjRItjILj m ω=ω+ωω expexpexp
The factor tj ωexp is common and hence
( ) mVILjR =ω+
in which LjR ω+ may be regarded as a complex impedance. The
complex current I is now given by
( )φ−
ω+=
ω+= j
LR
VLjR
VI mm exp22
with ( )RLω=φ −1tan and hence
( ) { }( )
( )φ−ωω+
=ω= tLR
VtjIti m cosexpRe22
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which, thankfully, is the same solution as before but arrived at with
considerably greater ease.
Let us be clear about the approach. We have
(i) introduced a complex forcing function tjVm ωexp knowing that in
reality the voltage source must be real i.e. { }tjVm ωexpRe .
(ii) We solved the equations working with complex voltages and
complex currents, tjV ωexp and tjI ωexp (or rather V and I since the
time dependence exp jωt cancelled out).
(iii) Since the actual voltage is given by { }tjVm ωexpRe the actual
current is given by { } { } ( )ψ+ω=ωψ=ω tItjjItjI cosexpexpReexpRe .
(iv) Since the differential of exp j𝜔𝑡 − ∅ = exp j𝜔𝑡 . exp −j∅ is simply
𝑗𝜔. exp j𝜔𝑡 . exp −j∅ and since we always take exp j𝜔𝑡 out as a
common factor, you may see now that our differential equations turn into
polynomial equations in jω (and you knew how to solve these at GCSE!)
This is a very powerful approach that will permit us to solve AC circuit
problems very easily.
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Example
Let's now do an example to show, formally, how we can solve AC
problems. Let's imagine we want to find the steady state current, i2,
flowing through the capacitor in the following example
The two KVL loop equations may be written
Ri1+Ldi1dt
+R i1− i 2( )=Em cos ωt +α( )
and
R i 2 − i1( )+ 1C i 2dt =0∫
Replacing ( )αω +cos tEm by ( ) tjEtjEm ωαω exp=+exp 1 where
αjEE m exp=1 and further introducing I1 and I2 via
tjIitjIi ωω exp=exp= 2211 and we obtain
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( )
01
2
12
121
=−⎟⎟⎠
⎞⎜⎜⎝
⎛
ω+
=−ω+
IRICj
R
EIRILjR
and, after a little algebra
NjMjE
CLj
CRLR
EI m
+
α=
⎥⎦
⎤⎢⎣
⎡ω
−ω++=
exp2
12
where CRLRM += and CLN ω−ω= 2 . We note that this may be
written, introducing MN=θtan
( )θ−α+
= jNM
EI m exp222
and hence the actual current ( )tjIi ω= expRe 22 may be written as
( ) ( )θ−α+ω+
= tNM
Eti m cos222
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4. Phasors
We have just introduced a very powerful method of circuit analysis. In
essence we have introduced the use of complex quantities to represent
sinusoidal functions of time. The complex number φjAexp (often written
φ∠A ) when used in this context to represent ( )φ+ωtAcos is called a
phasor. Since the phase angle φ must be measured relative to some
reference we may call the phasor 0∠A the reference phasor.
Since the phasor, φjAexp , is complex it may be represented in
Cartesian form jyx + just like any other complex quantity and thus
22 +=
sin=cos=
yxA
AyAx
φ
φ
Further since the phasor is a complex quantity it is very easy to display it
on an Argand diagram (also in this context called a phasor diagram).
Thus the phasor φjAexp is drawn as a line of length A at an angle φ to
the real axis.
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We emphasise that this is a graphical representation of an actual
sinusoid ( )φω +cos tA . The rules for addition, subtraction and
multiplication of phasors are identical to those for complex numbers.
Thus addition:
For multiplication it is easiest to multiply the magnitudes and add the
phases. Consider the effect of multiplying a phasor by j
( )2+exp=exp2exp=exp πφφπφ jAjAjjAj
which causes the phasor to be rotated by 90o.
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Similarly, dividing by j leads to
( ) ( )2expexp2expexp1π−φ=φπ−=φ jAjAjjA
j
i.e. a rotation of –90o.
We finally note that it is usual to use rms values for the magnitude of
phasors.
Circuit Analysis II WRM MT11
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5. Phasor relations in passive elements
Consider now a voltage tVm ωcos applied to a capacitor. As we have
indicated we elect to use the complex form tjVm ωexp and so omit the
"real part" as we calculate the current via
( ) tjVCjtjVdtd
Cdtdv
Ci mm ωωω exp=exp==
If we now drop the tj ωexp notation and write the voltage phasor Vm as
V and the current phasor as I we have
VCjI ω= or ICj
Vω1
=
In terms of a phasor diagram, taking the voltage V as the reference
from which we confirm two things we already knew
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(i) the ratio of the voltage to the current is Cω1
- the reactance.
(ii) the current leads the voltage by 90°. The pre-multiplying factor j
describes this.
For the inductance an analogous procedure leads to
ILjV ω=
Where the reactance is now jωL and, if we now take, say, the current as
the reference phasor we have
and here the current lags the voltage by 90°.
Circuit Analysis II WRM MT11
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[It is important to get these relationships the right way around and as a
check we may use the memory aid “CIVIL” – in a capacitor, the current
leads the voltage CIVIL and in an inductor, the current lags the voltage
CIVIL.]
Finally for a resistor we know that the current and voltage are in phase
and hence, in phasor terms
RIV =
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6. Phasors in circuit analysis
We are now in a position to summarise the method of analysis of AC
circuits.
(i) We include all reactances as imaginary quantities ( )LXjLj =ω for
an inductor and ( )cXjCj −=ω1 for a capacitor.
(ii) All voltages and currents are represented by phasors, which usually
have rms magnitude, and one is chosen as a reference with zero phase
angle.
(iii) All calculations are carried out in complex notation.
(iv) The magnitude and phase of, say, the current is obtained as
φjI exp . This can, if necessary, be converted back into a time varying
expression ( )φω +cos2 tI .
Circuit Analysis II WRM MT11
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Suppose we wish to find the current flowing through the inductor in the
circuit below
The reactances have been calculated and marked on the diagram. The
left hand voltage source has been chosen as reference and provides
10V rms. The right hand source produces 5V rms but at a phase angle
of 37° with respect to the 10V source. If we introduce phasor loop
currents I1 and I2 as shown then we may write KVL loop equations as
( )( ) ( ) 212
211
5103437exp5
10510
IjjIIjj
IIjIo −−−−=+=
−+=
where we have noted that 3+4=37sin5+37cos5=37exp5 jjj ooo . It
is routine to solve these simultaneous equations to give
( )5.12
71
jI −= and
5.128+5.6
=2j
I
and hence the current 21 III −= becomes
-
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oo jjI 8.86exp72.08.8672.05.1295.0
−=−∠==
Since rms values are involved, if we want to convert this into a function
of time we must multiply by 2 to obtain the peak value. Thus
( ) ( )otti 8.86cos02.1 −ω=
In our example we do not know the value of ω but it was accounted for in
the value of the reactances. Since everything is linear and the sources
are independent it would be a good exercise for you to check this result
by using the principle of superposition.
We have used mesh or loop analysis in our examples so far. It is, of
course equally appropriate to use node-voltage analysis if that looks like
an easier way to solve the problem.
As an example let's suppose we would like to find the voltage V in the
circuit below where the reactances have been calculated corresponding
to the frequency, ω, of the source
Circuit Analysis II WRM MT11
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It's probably as easy as anything to introduce two phasor node voltages
V1 and V. The two node voltage relationships may be written as
050
51010 111 =
−
−+
−+
−
jV
jVVV
and
01050
1050
51 =
+−
+−−
+−
jV
jV
jVV
We note that in writing these equations no thought was given to whether
currents flowing into or out of the nodes were being considered. As in
the DC case it is merely necessary to be consistent. It is now
straightforward to solve these two simultaneous equations to yield oV 6.71101 −∠= or, if the time domain result is required, remembering
that the voltage supply is 10V rms then ( ) ( )ottv 6.71cos201 −ω= .
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7. Combining impedances
As we have seen before the ratio of the voltage to the current phasors is
in general a complex quantity, Z, which generalises Ohms law, in terms
of phasors, to
IZV =
where Z in general takes the form
ee XjRZ +=
where the overall effect is equivalent to a resistance, Re, in series with a
reactance Xe. If Xe is positive the effective reactance is inductive
whereas negative values suggest that the effective reactance is
capacitative.
Consider the circuit below
Circuit Analysis II WRM MT11
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ICj
LjRV ⎟⎟⎠
⎞⎜⎜⎝
⎛
ω+ω+=
1
Thus the combined impedance Z is given by
XjRC
LjRZ +=⎟⎠
⎞⎜⎝
⎛ω
−ω+=1
This may be visualised on an Argand or phasor diagram
We note that the reactance may be positive or negative according to the
relative values of Lω and Cω1 . Indeed at a frequency LC=ω we see
that 0=X and that the impedance is purely resistive. We will return to
this point later.
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It is straightforward to show, and hopefully intuitive, that all the DC rules
for combining resistances in series and parallel carry over to
impedances. Thus if we have n elements in series, nZZZZ …321 ,,
Where
∑n
iineff ZZZZZZ
1321
=
=+++= …
and similarly for parallel elements
i
N
ieq ZZZZZ1
=+1
+1
+1
=1
1=321∑…
We note that the inverse of impedance, Z, is known as admittance, Y.
Thus, as in the DC case it is sometimes more convenient to write
∑n
iieq YY
1==
Circuit Analysis II WRM MT11
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and finally since Y is also a complex number it may be written
BjGY +=
where G is a conductance and B is known as the susceptance.
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Example
Find the equivalent impedance of the circuit below
Using the usual combination rules gives
( )
CjLjR
CjLjR
ZZZZ
Z
ωω
ωω
1++
1+
=+
=21
21
Which we can simplify to
RCjLCLjRZω+ω−
ω+= 21
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8. Operations on phasors
We have just introduced a method of analysing AC circuits in terms of
complex currents and voltages. This method inevitably involves the
manipulation of complex phasor quantities and so we list below the
results for manipulating these quantities which are, course, simply the
standard rules for complex numbers. Sometimes it is easier to use the
a+jb notation and sometimes the θ∠=θ rjr exp notation is easiest. We
summarise below the important relationships.
Addition and Subtraction
If
jdcIjbaI +=+= 21 and
then
( )dbjcaII ±+±=± 21
where the real and imaginary parts add/subtract
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Multiplication Here it is easiest by far to use the θ∠r rotation.
If 22211111 expexp θ=θ=θ∠= jrIjrrI and
then ( ) 2121212121 exp θ+θ∠=θ+θ= rrjrrII
when we see the amplitudes multiply and the arguments add
For division we have
( ) 212
121
2
1
22
11
2
1 expexpexp
θ−θ∠=θ−θ=θ
θ=
rrj
rr
jrjr
II
the amplitudes divide and the arguments subtract.
Circuit Analysis II WRM MT11
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Complex conjugates also appear.
If
θ∠=θ=+= rjrbjaI exp
then the complex conjugate I*, is given by
θ−∠=θ−=−= rjrbjaI exp*
from which we see
{} {}IjIIIII Im2;Re2 ** =−=+
where Re{ } denotes the real part the Im { } denotes the imaginary part.
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Rationalising
We are often confronted with expressions of the form
djcbja
++
And sometimes we wish to rationalise them. We do this in one of two
ways. The first is to multiply top and bottom by djc − . This gives
⎟⎟⎠
⎞⎜⎜⎝
⎛
+
−+⎟⎟⎠
⎞⎜⎜⎝
⎛
+
+=
−−
++
=++
2222 dcadbcj
dcbdca
djcdjc
djcbja
djcbja
Alternatively, we can write jba+ as 11 exp θjr with 221 += bar and
ab=tan 1θ . Similarly jdc+ may be written as 22 exp θjr and hence
( ) 212
121
2
1
22
11 expexpexp
θ−θ∠=θ−θ=θ
θ=
++
rrj
rr
jrjr
jdcjba
where the amplitudes divide and the arguments subtract.
There is no golden rule as to which approach to take – it is determined
by the problem at hand. However, if you do not have a pressing need to
rationalise the expression, we will see some quite good reasons why we
may often prefer to stick with the factorised form and not rationalise at
all.
Circuit Analysis II WRM MT11
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9. Phasor diagrams
Although direct calculation is easily carried out using phasors it is
sometimes useful to use a phasor (Argand) diagram to show the
relationships between say a voltage and current phasor graphically. In
this way it is easy to see their relative amplitudes and phases and hence
gain quick insight into how the circuit operates. As a simple example
consider the circuit below
Since the current, I, flows through both elements it is sensible to choose
this as the reference phasor. Having made this choice the voltage drop
across the resistor, IRVR = , whereas that across the capacitor,
ICjCjIVc ω−=ω= . The sum of these voltages must equal V. The
phasor diagram is easily drawn as
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from which it is clear that the voltage lags the current by an angle φ. The
angle φ may be obtained from the diagram as ( )CRωφ 1tan= 1 .
Let's consider another example
We could obtain the relationship between I and V by using the
equivalent impedance derived earlier. However we will use a phasor
diagram to show the various currents and voltages which appear across
the various components. Since the voltage, V, is the same across each
arm it is sensible to choose this as the reference phasor.
The relationships are
21112 IIIILjRIVVCjI +=ω+=ω= and
The two phasor diagrams are
Circuit Analysis II WRM MT11
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or, combining onto a single diagram
where φ denotes the phase angle between V and I. In the diagram
above I leads V by φ.
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10. Thévenin and Norton equivalent circuits
The Thévenin and Norton theorems apply equally well in the AC case.
Here we replace any arbitrarily complicated circuit containing resistors,
capacitors, inductors by a circuit whose behaviour, as far as the outside
world is concerned, is entirely equivalent.
The two choices are the Thévenin equivalent
and the Norton equivalent
The methods for determining V, I and Z are identical to those used in the
DC case. In general
(i) Calculate the open circuit voltage, Voc
(ii) Calculate the short circuit current, Isc
Circuit Analysis II WRM MT11
43
From which
sc
ocscoc I
VZIIVV ==,= and
We note, of course that in the absence of dependent sources it is often
easier to "set the sources to zero" and simply calculate the terminal
impedance Zab. We emphasise again that when a voltage source is "set
to zero" it is replaced by a short circuit whereas when a current source is
"set to zero", no current flows, and hence the source is replaced by an
open circuit.
Finally we note that the equivalent impedance Z is also frequently
referred to as either
(i) internal impedance
or
(ii) output impedance.
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Example
Find the Norton and Thévenin equivalents of
Noting that there are no dependent sources it is easiest to calculate Zab
directly with the voltage source replaced by a short circuit. This is easy
since the circuit then reduces to an inductor in parallel with a resistor.
Thus
8+16=40+2020.40
== jj
jZZ ab
We now need to calculate the open circuit voltage between a and b.
This is easy since the circuit is essentially a voltage divider
V
jjj
jj
Voc
4.533.22
4.53exp3.224.63exp5
10exp50
10exp54020
20
−∠=
−==
+=
Thus the Thévenin equivalent circuit takes the form
50
Circuit Analysis II WRM MT11
45
In order to find the Norton equivalent we need to find the current flowing
between the terminals a and b when they are shorted together. In this
case the circuit becomes
and
8025.180exp25.190exp4010exp50
4010exp50
−∠=−=== jjj
jjIsc
and hence the Norton equivalent becomes
46
We have elected to find the Norton equivalent directly. However it is
equally possible to transform between Thévenin and Norton equivalents
directly as we did in the DC case. It is left as an exercise to confirm that
Hence we could have worked out the Norton current source in our
example directly from the Thévenin equivalent as
8025.180exp25.16.26exp89.174.53exp36.22
81604.53exp36.22
−∠=−=−
=
+−
==
jjj
jj
ZVI oc
which, of course, is the value we previously calculated.