CIRCUIT ANALYSIS II · Circuit Analysis II WRM MT11 9 If we want to calculate the average power dissipated over a cycle we must integrate from ωt =0 to ωt = ωT =2π. If we again

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.

2

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

3

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 . . . . . . . . !!


5

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

6

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


7

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:

8

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


9

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.

10


11

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


13

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

14

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


15

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:

16

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


17

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.

18

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


19

( )

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

20

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.


21

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.

22

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.


23

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

24

(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°.


25

[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 =

26

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 .


27

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

-

28

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


29

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= .

30

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


31

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.

32

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==


33

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.

34

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


35

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

36

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.


37

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.

38

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.


39

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

40

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


41

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.

42

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


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.

44

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


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.

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