33 Thevenin’s andNorton’s theorems

At the end of this chapter you should be able to:

ž understand and use Thevenin’s theorem to analyse a.c. andd.c. networks

ž understand and use Norton’s theorem to analyse a.c. and d.c.networks

ž appreciate and use the equivalence of Thevenin and Nortonnetworks

33.1 Intr oduction Many of the networks analysed in Chapters 30, 31 and 32 usingKirchhoff’s laws, mesh-current and nodal analysis and the superpositiontheorem can be analysed more quickly and easily by using Thevenin’sor Norton’s theorems. Each of these theorems involves replacing whatmay be a complicated network of sources and linear impedances with asimple equivalent circuit. A set procedure may be followed when usingeach theorem, the procedures themselves requiring a knowledge of basiccircuit theory. (It may be worth checking some general d.c. circuit theoryin Section 13.4. page 174, before proceeding)

33.2 Thevenin’s theorem Thevenin’stheoremstates:

‘The current which flows in any branch of a network is the same as thatwhich would flow in the branch if it were connected across a source ofelectrical energy, the e.m.f. of which is equal to the potential differencewhich would appear across the branch if it were open-circuited, and theinternal impedance of which is equal to the impedance which appearsacross the open-circuited branch terminals when all sources are replacedby their internal impedances.’

The theorem applies to any linear active network (‘linear’ meaning that themeasured values of circuit components are independent of the directionand magnitude of the current flowing in them, and ‘active’ meaning thatit contains a source, or sources, of e m f.)

The above statement of Thevenin’s theorem simply means thata complicated network with output terminals AB, as shown inFigure 33.1(a), can be replaced by a single voltage sourceE in serieswith an impedancez, as shown in Figure 33.1(b).E is the open-circuit

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Figure 33.1 The Theveninequivalent circuit

voltage measured at terminals AB andz is the equivalent impedance ofthe network at the terminals AB when all internal sources of e.m.f. aremade zero. The polarity of voltageE is chosen so that the current flowingthrough an impedance connected between A and B will have the samedirection as would result if the impedance had been connected between Aand B of the original network. Figure 33.1(b) is known as theTheveninequivalent circuit, and was initially introduced in Section 13.4, page 174for d.c. networks.

The following four-stepprocedure can be adopted when determining,by means of Thevenin’s theorem, the current flowing in a branchcontaining impedanceZL of an active network:

(i) remove the impedanceZL from that branch;

(ii) determine the open-circuit voltageE across the break;

(iii) remove each source of e m f. and replace it by its internal impedance(if it has zero internal impedance then replace it by a short-circuit), and then determine the internal impedance,z, ‘looking in’at the break;

(iv) determine the current from the Thevenin equivalent circuit shownin Figure 33.2, i.e.

current iL =E

ZL Y z.

A simpled.c. network (Figure 33.3) serves to demonstrate how the aboveprocedure is applied to determine the current flowing in the 5� resis-tance by using Thevenin’s theorem. This is the same network as used inChapter 30 when it was solved using Kirchhoff’s laws (see page 535),and by means of the superposition theorem in Chapter 32 (see page 562).

A comparison of methods may be made.

Using the above procedure:

(i) the 5 � resistor is removed, as shown in Figure 33.4(a).

(ii) The open-circuit voltageE across the break is now required. Thenetwork of Figure 33.4(a) is redrawn for convenience as shown inFigure 33.4(b), where current,

I1 D E1 � E2

r1 C r2D 8 � 3

1 C 2D 5

3or 1

2

3A

Figure 33.2Hence the open-circuit voltageE is given by

E D E1 � I1r1 i.e., E D 8 �(12

3

)1 D 61

3 V

(Alternatively,E D E2 � �I1r2 D 3 C(12

3

)2 D 61

3 V.

(iii) Removingeach source of e m f. gives the network of Figure 33.5.The impedance,z, ‘looking in’ at the break AB is given by

z D 1 ð 2/1 C 2 D 23 �Figure 33.3

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Figure 33.4

(iv) The Thevenin equivalent circuit is shown in Figure 33.6, wherecurrentiL is given by

iL D E

ZL C zD 61

3

5 C 23

D 1.1177

D 1.12 A, correct to two decimal placesFigure 33.5

To determine the currents flowing in the other two branches ofthe circuit of Figure 33.3, basic circuit theory is used. Thus, fromFigure 33.7, voltageV D 1.11775 D 5.5885 V.

ThenV D E1 � IAr1, i.e., 5.5885D 8 � IA1, from which

currentIA D 8 � 5.5885D 2.41 A.

Similarly, V D E2 � IBr2, i.e., 5.5885D 3 � IB2, from which

currentIB D 3 � 5.5885

2D −1.29 A

Figure 33.6(i.e., flowing in the direction opposite to that shown in Figure 33.7).

The Thevenin theorem procedure used above may be applied to a.c. aswell as d.c. networks, as shown below.

An a.c. network is shown in Figure 33.8 where it is required to find thecurrent flowing in the6 C j8� impedance by using Thevenin’s theorem.

Figure 33.7

Using the above procedure

(i) The 6 C j8� impedance is removed, as shown in Figure 33.9(a).

(ii) The open-circuit voltage across the break is now required. Thenetwork is redrawn for convenience as shown in Figure 33.9(b),where current

I1 D 5 C j0 C 2 C j4

3 C j4 C 2 � j5D 7 C j4

5 � j

D 1.5816 41.05° AFigure 33.8

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Figure 33.9

Hence open-circuit voltage across AB,

E D E1 � I13 C j4, i.e.,

E D 5 C j0 � 1.5816 41.05°56 53.13°

from which E D 9.5676 �54.73° VFigure 33.10

(iii) From Figure 33.10, the impedancez ‘looking in’ at terminals ABis given by

z D 3 C j42 � j5

3 C j4 C 2 � j5

D 5.2816 �3.76° � or 5.270� j0.346�

(iv) The Thevenin equivalent circuit is shown in Figure 33.11, fromwhich current

iL D E

ZL C zD 9.6576 �54.73°

6 C j8 C 5.270� j0.346

Figure 33.11

Thus, current in6 C j8� impedance,

iL D 9.6576 �54.73°

13.6236 34.18°D 0.716 −88.91° A

The network of Figure 33.8 is analysed using Kirchhoff’s laws inproblem 3, page 539, and by the superposition theorem in problem 4,page 568. The above analysis using Thevenin’s theorem is seen to bemuch quicker.

Figure 33.12

Problem1. For the circuit shown in Figure 33.12, use Thevenin’stheorem to determine (a) the current flowing in the capacitor, and(b) the p.d. across the 150 k� resistor.

(a) (i) Initially the 150� j120k� impedance is removed from thecircuit as shown in Figure 33.13.Figure 33.13

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Notethat,to find the current in the capacitor, only the capacitorneed have been initially removed from the circuit. However,removing each of the components from the branch throughwhich the current is required will often result in a simplersolution.

(ii) From Figure 33.13,

currentI1 D 2006 0°

5000C 20000D 8 mA

Theopen-circuit e m f.E is equal to the p.d. across the 20 k�resistor, i.e.

E D 8 ð 10�320ð 103 D 160 V.

(iii) Removing the 2006 0° V sourcegives the network shown inFigure 33.14.

Figure 33.14

The impedance,z, ‘looking in’ at the open-circuited terminalsis given by

z D 5 ð 20

5 C 20k� D 4 kZ

Figure 33.15

(iv) The Thevenin equivalent circuit is shown in Figure 33.15,where currentiL is given by

iL D E

ZL C zD 160

150� j120 ð 103 C 4 ð 103

D 160

195.23ð 103 6 �37.93°

D 0.826 37.93° mA

Thus the current flowing in the capacitor is 0.82 mA.

(b) P.d. across the 150 k� resistor,

V0 D iLR D 0.82ð 10�3150ð 103 D 123 V

Figure 33.16 Problem2. Determine, for the network shown in Figure 33.16,the value of currentI. Each of the voltage sources has a frequencyof 2 kHz.

(i) The impedance through which currentI is flowing is initiallyremoved from the network, as shown in Figure 33.17.

(ii) From Figure 33.17,

current,I1 D 20� 10

2 C 3D 2 A

Figure 33.17

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Figure 33.18

Hence the open circuit e.m.f.E D 20� I12 D 20� 22 D 16 V.

(Alternatively,E D 10C I13 D 10C 23 D 16 V.)

(iii) When the sources of e m f. are removed from the circuit, theimpedance,z, ‘looking in’ at the break is given by

z D 2 ð 3

2 C 3D 1.2 Z

(iv) The Thevenin equivalent circuit is shown in Figure 33.18, whereinductive reactance,

XL D 2�fL D 2�2000235ð 10�6 D 2.95 �Hencecurrent

I D 16

1.2 C 1.5 C j2.95D 16

4.06 47.53°

D 4.06 6 −47.53° A or .2.70− j 2.95/ A

Problem 3. Use Thevenin’s theorem to determine the power dissi-pated in the 48� resistor of the network shown in Figure 33.19.

Figure 33.19

The power dissipated by a currentI flowing through a resistorR is givenby I2R, henceinitially the current flowing in the 48� resistor is required.

(i) The 48C j144� impedance is initially removed from the networkas shown in Figure 33.20.

(ii) From Figure 33.20,

current,i D 506 0°

300� j400D 0.16 53.13° A

Figure 33.20Hence the open-circuit voltage

E D i300 D 0.16 53.13°300 D 306 6 53.13° V

(iii) When the 506 0° V sourceshown in Figure 33.20 is removed, theimpedance,z, is given by

z D �j400300

300� j400D 4006 �90°300

5006 �53.13°

D 2406 �36.87° � or .192− j 144/Z

(iv) TheThevenin equivalent circuit is shown in Figure 33.21 connectedto the48C j144� load.

Current I D 306 53.13°

192� j144 C 48C j144D 306 53.13°

2406 0°

D 0.1256 53.13° AFigure 33.21

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Hencethe power dissipated in the 48Z resistor

D I2R D 0.125248 D 0.75 W

Problem4. For the network shown in Figure 33.22, useThevenin’s theorem to determine the current flowing in the 80�resistor.

Figure 33.22

One method of analysing a multi-branch network as shown inFigure 33.22 is to use Thevenin’s theorem on one part of the networkat a time. For example, the part of the circuit to the left of AA may bereduced to a Thevenin equivalent circuit.From Figure 33.23,

E1 D(

20

20C 5

)100D 80 V, by voltage division

and z1 D 20ð 5

20C 5D 4 �

Thus the network of Figure 33.22 reduces to that of Figure 33.24. Thepart of the network shown in Figure 33.24 to the left of BB may bereduced to a Thevenin equivalent circuit, where

E2 D(

50

50C 46C 4

)80 D 40 V

Figure 33.23 Figure 33.24

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and current I6 D[

6 C j8

�3 C j4� C �6 C j8�

]��0.062 C j0.964�

D ��0.041 C j0.643�A

(v) If Figure 32.22 is superimposed on Figure 32.20, the resultantcurrents are as shown in Figure 32.24.

Figure 32.24

(vi) Resultant current flowing from (5 C j0)V source is given by

I1 C I6 D �0.614 � j0.025� C ��0.041 C j0.643�

D .0.573Y j 0.618/A or 0.843 6 6 47.16° A

Resultant current flowing from (2 C j4)V source is given by

I3 C I4 D �0.622 C j0.363� C ��0.062 C j0.964�

D .0.560Y j 1.327/A or 1.440 6 6 67.12° A

Resultant current flowing through the �6 C j8� impedance isgiven by

I2 � I5 D ��0.00731 � j0.388� � ��0.0207 C j0.321�

D .0.0134 − j 0.709/A or 0.709 6 −88.92° A

(b) Voltage across �6 C j8� impedance is given by

�I2 � I5��6 C j8� D �0.709 6 �88.92°��10 6 53.13°�

D 7.09 6 �35.79° V

i.e., the magnitude of the voltage across the �6 C j8� impedanceis 7.09 V

(c) Total active power P delivered to the network is given by

P D E1�I1 C I6� cos �1 C E2�I3 C I4� cos �2

where �1 is the phase angle between E1 and (I1 C I6) and �2 is thephase angle between E2 and (I3 C I4), i.e.,

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Figure 33.25

and z2 D 50ð 50

50C 50D 25 �

Thusthe original network reduces to that shown in Figure 33.25.The part of the network shown in Figure 33.25 to the left of CC may bereduced to a Thevenin equivalent circuit, where

E3 D(

60

60C 25C 15

)40 D 24 V

and z3 D 6040

60C 40D 24 �

Thus the original network reduces to that of Figure 33.26, from whichthe current in the 80 Z resistor is given by

I D(

24

80C 16C 24

)D 0.20 A

Figure 33.26

Problem5. Determine the Thevenin equivalent circuit withrespect to terminals AB of the circuit shown in Figure 33.27.Hence determine (a) the magnitude of the current flowing in a3.75C j11� impedance connected across terminals AB, and(b) the magnitude of the p.d. across the3.75C j11� impedance.

Figure 33.27

CurrentI1 shownin Figure 33.27 is given by

I1 D 246 0°

4 C j3 � j3D 246 0°

46 0°D 66 0° A

The Thevenin equivalent voltage, i.e., the open-circuit voltage acrossterminals AB, is given by

E D I14 C j3 D 66 0°56 36.87° D 306 6 36.87° V

When the 246 0° V source is removed, the impedancez ‘looking in’ atAB is given by

z D 4 C j3�j3

4 C j3 � j3D 9 � j12

4D .2.25− j 3.0/Z

Thusthe Thevenin equivalent circuit is as shown in Figure 33.28.

(a) When a3.75C j11� impedance is connected across terminals AB,the currentI flowing in the impedance is given by

I D 306 36.87°

3.75C j11 C 2.25� j3.0D 306 36.87°

106 53.13°

D 36 �16.26° A

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Figure 33.28 Figure 33.29

Hence the current flowing in the .3.75Y j 11/Z impedanceis 3 A.

(b) P.d. across the3.75C j11� impedance is given by

V D 36 �16.26°3.75C j11 D 36 �16.26°11.626 71.18°

D 34.866 54.92° V

Hencethe magnitude of the p.d. across the impedance is 34.9 V.

Problem6. Use Thevenin’s theorem to determine the currentflowing in the capacitor of the network shown in Figure 33.29.

(i) The capacitor is removed from branch AB, as shown inFigure 33.30.

(ii) The open-circuit voltage,E, shown in Figure 33.30, is given byI25. I2 may be determined by current division ifI1 is known.(Alternatively, E may be determined by the method used inproblem 4.)Figure 33.30Current I1 D V/Z, where Z is the total circuit impedance andV D 16.556 �22.62° V.

Impedance,Z D 4 C j28 C j6

j2 C 8 C j6D 4 C �12C j16

8 C j8

D 4.5966 22.38° �

Hence I1 D 16.556 �22.62°

4.5966 22.38°D 3.606 �45° A

and I2 D(

j2

j2 C 3 C j6 C 5

)I1 D 26 90°3.606 �45°

11.3146 45°

D 0.6366 0° A

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(An alternative method of findingI2 is to use Kirchhoff’s laws ormesh-current or nodal analysis on Figure 33.30.)

Hence E D I25 D 0.6366 0°5 D 3.186 6 0° V

(iii) If the 16.556 �22.62° V source is removed from Figure 33.30, theimpedance,z, ‘looking in’ at AB is given by

z D 5[4 ð j2/4 C j2 C 3 C j6]

5 C [4 ð j2/4 C j2 C 3 C j6]D 53.8 C j7.6

8.8 C j7.6

i.e. z D 3.6546 22.61° � or .3.373Y j 1.405/Z

(iv) The Thevenin equivalent circuit is shown in Figure 33.31, wherethe current flowing in the capacitor,I, is given by

I D 3.186 0°

3.373C j1.405 � j8D 3.186 0°

7.4086 �62.91°

D 0.436 6 62.91° A in the direction from A to BFigure 33.31

Problem7. For the network shown in Figure 33.32, derive theThevenin equivalent circuit with respect to terminals PQ, andhence determine the power dissipated by a 2� resistor connectedacross PQ.

Figure 33.32

CurrentI1 shownin Figure 33.32 is given by

I1 D 106 0°

5 C 4 C j3D 1.0546 �18.43° A

Hencethe voltage drop across the 5� resistor is given byVX D I15 D5.276 �18.43° V, and is in the direction shown in Figure 33.32, i.e., thedirection opposite to that in whichI1 is flowing.

The open-circuit voltageE across PQ is the phasor sum ofV1, Vx andV2, as shown in Figure 33.33.Figure 33.33

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Figure 33.34

Thus E D 106 0° � 56 45° � 5.276 �18.43°

D 1.465� j1.869V or 2.3756 −51.91° V

The impedance,z, ‘looking in’ at terminals PQ with the voltage sourcesremoved is given by

z D 8 C 54 C j3

5 C 4 C j3D 8 C 2.6356 18.44° D .10.50Y j 0.833/Z

The Theveninequivalent circuit is shown in Figure 33.34 with the 2�resistance connected across terminals PQ.The current flowing in the 2� resistance is given by

I D 2.3756 �51.91°

10.50C j0.833 C 2D 0.18966 �55.72° A

The powerP dissipatedin the 2� resistor is given by

P D I2R D 0.189622 D 0.0719W � 72 mW, correct to twosignificant figures.

Problem8. For the a.c. bridge network shown in Figure 33.35,determine the current flowing in the capacitor, and its direction,by using Thevenin’s theorem. Assume the 306 0° V sourceto havenegligible internal impedance.

Figure 33.35(i) The �j25 � capacitor is initially removed from the network, as

shown in Figure 33.36.(ii) P.d. between A and C,

VAC D(

Z1

Z1 C Z4

)V D

(15

15C 5 C j5

)306 0°

D 21.836 �14.04° V

Figure 33.36

P.d. between B and C,

VBC D(

Z2

Z2 C Z3

)V D

(40

40C 20C j20

)306 0°

D 18.976 �18.43° V

Assumingthat point A is at a higher potential than point B, thenthe p.d. between A and B is

21.836 � 14.04° � 18.976 �18.43°

D 3.181C j0.701V or 3.2576 12.43° V,

i.e., the open-circuit voltage across AB is given by

E D 3.2576 12.43° V.

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Point C is at a potential of 306 0° V. BetweenC and A is a voltdrop of 21.836 �14.04° V. Hencethe voltage at point A is

306 0° � 21.836 �14.04° D 10.296 30.98° V

Betweenpoints C and B is a voltage drop of 18.976 �18.43° V.Hence the voltage at point B is 306 0° � 18.976 �18.43° D13.426 6 26.55° V.

Sincethemagnitude of the voltage at B is higher than at A, currentmust flow in the direction B to A.

(iii) Replacing the 306 0° V sourcewith a short-circuit (i.e., zero internalimpedance) gives the network shown in Figure 33.37(a). Thenetwork is shown redrawn in Figure 33.37(b) and simplified inFigure 33.37(c). Hence the impedance,z, ‘looking in’ at terminalsAB is given by

z D 155 C j5

15C 5 C j5C 4020C j20

40C 20C j20

D 5.1456 30.96° C 17.8896 26.57°

i.e., z D 20.41C j10.65�

Figure 33.37

(iv) The Thevenin equivalent circuit is shown in Figure 33.38, wherecurrentI is given by

I D 3.2576 12.43°

20.41C j10.65 � j25D 3.2576 12.43°

24.956 �35.11°

D 0.1316 47.54° A

Thus a current of 131 mA flows in the capacitor in a directionfrom B to A.Figure 33.38

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Further problems on Thevenin’s theorem may be found in Section 33.5,problems 1 to 10, page 598.

33.3 Norton’s theorem A sourceof electrical energy can be represented by a source of e.m.f. inseries with an impedance. In Section 33.2, the Thevenin constant-voltagesource consisted of a constant e.m.f.E, which may be alternating or direct,in series with an internal impedance,z. However, this is not the only formof representation. A source of electrical energy can also be represented bya constant-current source, which may be alternating or direct, in parallelwith an impedance. It is shown in Section 33.4 that the two forms are infact equivalent.

Norton’s theorem states:

‘The current that flows in any branch of a network is the same as that whichwould flow in the branch if it were connected across a source of electricalenergy, the short-circuit current of which is equal to the current that wouldflow in a short-circuit across the branch, and the internal impedance ofwhich is equal to the impedance which appears across the open-circuitedbranch terminals.’

The above statement simply means that any linear active networkwith output terminals AB, as shown in Figure 33.39(a), can be replacedby a current source in parallel with an impedancez as shown inFigure 33.39(b). The equivalent current sourceISC (note the symbol inFigure 33.39(b) as per BS 3939:1985) is the current through a short-circuitapplied to the terminals of the network. The impedancez is the equivalentimpedance of the network at the terminals AB when all internal sources ofe.m.f. are made zero. Figure 33.39(b) is known as theNorton equivalentcircuit , and was initially introduced in Section 13.7, page 181 for d.c.networks.

Figure 33.39 The Nortonequivalent circuit

The following four-step procedure may be adopted when determiningthe current flowing in an impedanceZL of a branch AB of an activenetwork, using Norton’s theorem:

(i) short-circuit branch AB;(ii) determine the short-circuit currentISC;

(iii) removeeach source of e m f. and replace it by its internal impedance(or, if a current source exists, replace with an open circuit), thendetermine the impedance,z, ‘looking in’ at a break made betweenA and B;

(iv) determine the value of the currentiL flowing in impedanceZL fromthe Norton equivalent network shown in Figure 33.40, i.e.,

iL D(

z

ZL C z

)ISC

Figure 33.40

A simple d.c. network (Figure 33.41) serves to demonstrate how the aboveprocedure is applied to determine the current flowing in the 5� resistanceby using Norton’s theorem:Figure 33.41

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Figure 33.42

(i) The 5 � branch is short-circuited, as shown in Figure 33.42.

(ii) From Figure 33.42,ISC D I1 C I2 D 81 C 3

2 D 9.5 A

(iii) If each source of e m f. is removed the impedance ‘looking in’ ata break made between A and B is given byz D 1 ð 2/1 C 2 D23 �.

(iv) From the Norton equivalent network shown in Figure 33.43, the

current in the 5� resistance is given byIL D(

23

/ (5 C 2

3

))9.5 D

1.12 A, as obtained previously using Kirchhoff’s laws, the super-position theorem and by Thevenin’s theorem.

As with Thevenin’s theorem, Norton’s theorem may be used with a.c.as well as d.c. networks, as shown below.

An a.c. network is shown in Figure 33.44 where it is required to find thecurrent flowing in the6 C j8� impedance by using Norton’s theorem.Using the above procedure:

Figure 33.43 Figure 33.44

(i) Initially the 6 C j8� impedance is short-circuited, as shown inFigure 33.45.

(ii) From Figure 33.45,

I SC D I1 C I2 D 5 C j0

3 C j4C �2 C j4

2 � j5

D 16 �53.13° � 4.4726 63.43°

5.3856 �68.20°

D 1.152� j1.421A or 1.8296 6 −50.97° AFigure 33.45

(iii) If each source of e.m.f. is removed, the impedance,z, ‘looking in’at a break made between A and B is given by

z D 3 C j42 � j5

3 C j4 C 2 � j5

D 5.286 6 −3.76° Z or .5.269− j 0.346/Z

(iv) From the Norton equivalent network shown in Figure 33.46, thecurrent is given byFigure 33.46

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Figure 33.47 Figure 33.48 Figure 33.49

iL D(

z

ZL C z

)ISC

D(

5.286 �3.76°

6 C j8 C 5.269� j0.346

)1.8296 �50.97°

i.e., curr ent in .6Y j 8/Z impedance,iL = 0.716 −88.91° A

Problem9. Use Norton’s theorem to determine the value ofcurrentI in the circuit shown in Figure 33.47.

(i) Thebranch containing the 2.8� resistor is short-circuited, as shownin Figure 33.48.

(ii) The 3 � resistor in parallel with a short-circuit is the sameas 3� in parallel with 0 giving an equivalent impedance of3 ð 0/3 C 0 D 0. Hence the network reduces to that shown inFigure 33.49, whereI SC D 5/2 D 2.5 A.

(iii) If the 5 V source is removed from the network the input impedance,z, ‘looking-in’ at a break made in AB of Figure 33.48 givesz D 2 ð 3/2 C 3 D 1.2 Z (see Figure 33.50).

Figure 33.50

Figure 33.51

(iv) The Norton equivalent network is shown in Figure 33.51, wherecurrentI is given by

I D(

1.2

1.2 C 2.8 � j3

)2.5 D 3

4 � j3D 0.606 6 36.87° A

Problem10. For the circuit shown in Figure 33.52 determine thecurrent flowing in the inductive branch by using Norton’s theorem.

Figure 33.52

(i) The inductive branch is initially short-circuited, as shown inFigure 33.53.

(ii) From Figure 33.53,

ISC D I1 C I2 D 20

2C 10

3D 13.3 A

Figure 33.53

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Figure 33.54

(iii) If the voltage sources are removed, the impedance,z, ‘looking in’at a break made in AB is given byz D 2 ð 3/2 C 3 D 1.2 Z.

(iv) The Norton equivalent network is shown in Figure 33.54, wherecurrentI is given by

I D(

1.2

1.2 C 1.5 C j2.95

)13.P3 D 16

2.7 C j2.95

D 4.06 6 −47.53° A or .2.7 − j 2.95/A

Problem11. Use Norton’s theorem to determine the magnitudeof the p.d. across the 1� resistance of the network shown inFigure 33.55.

Figure 33.55

(i) The branch containing the 1� resistance is initially short-circuited,as shown in Figure 33.56.

Figure 33.56

(ii) 4 � in parallel with �j2 � in parallel with 0� (i.e., theshort-circuit) is equivalent to 0, giving the equivalent circuit ofFigure 33.57. HenceISC D 10/4 D 2.5 A.

Figure 33.57

(iii) The 10 V source is removed from the network of Figure 33.55, asshown in Figure 33.58, and the impedancez, ‘looking in’ at a breakmade in AB is given by

1

zD 1

4C 1

4C 1

�j2D �j � j C 2

�j4D 2 � j2

�j4

from which

z D �j4

2 � j2D �j42 C j2

22 C 22D 8 � j8

8D .1 − j 1/Z

(iv) The Norton equivalent network is shown in Figure 33.59, fromwhich currentI is given by

I D(

1 � j1

1 � j1 C 1

)2.5 D 1.586 �18.43° A

Hence the magnitude of the p.d. across the 1Z resistor isgiven by

IR D 1.581 D 1.58 V.

Figure 33.58 Figure 33.59

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Figure 33.60 Figure 33.61

Problem12. For the network shown in Figure 33.60, obtain theNorton equivalent network at terminals AB. Hence determine thepower dissipated in a 5� resistor connected between A and B.

(i) Terminals AB are initially short-circuited, as shown in Figure 33.61.(ii) The circuit impedanceZ presented to the 206 0° V source is

given by

Z D 2 C 4 C j3�j3

4 C j3 C �j3D 2 C 9 � j12

4

D 4.25� j3� or 5.2026 �35.22° �

ThuscurrentI in Figure 33.61 is given by

I D 206 0°

5.2026 �35.22°D 3.8456 35.22° A

Hence

ISC D(

4 C j3

4 C j3 � j3

)3.8456 35.22°

D 4.8066 6 72.09° A

(iii) Removing the 206 0° V source of Figure 33.60 gives the networkof Figure 33.62.Figure 33.62Impedance,z, ‘looking in’ at terminals AB is given by

z D �j3 C 24 C j3

2 C 4 C j3D �j3 C 1.4916 10.3°

D .1.467− j 2.733/Z or 3.1026 6 −61.77°Z

(iv) The Norton equivalent network is shown in Figure 33.63.

CurrentIL D(

3.1026 �61.77°

1.467� j2.733C 5

)4.8066 72.09°

D 2.1236 33.23° A

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Figure 33.63

Hence the power dissipated in the 5Z resistor is

I2LR D 2.12325 D 22.5 W

Problem13. Derive the Norton equivalent network with respectto terminals PQ for the network shown in Figure 33.64 and hencedetermine the magnitude of the current flowing in a 2� resistorconnected across PQ.

Figure 33.64This is the same problem as problem 7 on page 584 which was solvedby Thevenin’s theorem.

A comparison of methods may thus be made.

(i) Terminals PQ are initially short-circuited, as shown in Figure 33.65.

(ii) CurrentsI1 andI2 areshown labelled. Kirchhoff’s laws are used.

For loop ABCD, and moving anticlockwise,

106 0° D 5I1 C 4 C j3I1 C I2,

i.e., 9 C j3I1 C 4 C j3I2 � 10 D 0 1

Figure 33.65For loop DPQC, and moving clockwise,

106 0° � 56 45° D 5I1 � 8I2,

i.e., 5I1 � 8I2 C 56 45° � 10 D 0 2

Solving Equations (1) and (2) by using determinants gives

I1∣∣∣∣ 4 C j3 �10�8 56 45° � 10

∣∣∣∣D �I2∣∣∣∣ 9 C j3 �10

5 56 45° � 10

∣∣∣∣D I∣∣∣∣ 9 C j3 4 C j3

5 �8

∣∣∣∣

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from which

I2 D�

∣∣∣∣ 9 C j3 �105 56 45°�10

∣∣∣∣∣∣∣∣ 9 C j3 4 C j35 �8

∣∣∣∣D �[9 C j356 45°�10 C 50]

[�72� j24� 20� j15]

D �[22.526 146.50°]

[99.9256 �157.03°]D �0.2256 303.53° or � 0.2256 �56.47°

Hence the short-circuit currentISC D 0.2256 �56.47° A flowingfrom P to Q.

(iii) The impedance,z, ‘looking in’ at a break made between P and Qis given by

z D 10.50C j0.833� (see problem 7, page 584).

(iv) The Norton equivalent circuit is shown in Figure 33.66, wherecurrentI is given by

I D(

10.50C j0.833

10.50C j0.833C 2

)0.2256 �56.47°

D 0.196 �55.74° A

Figure 33.66

Hence the magnitude of the current flowing in the 2Z resistoris 0.19 A.

Further problems on Norton’s theorem may be found in Section 33.5,problems 11 to 15, page 600

33.4 Thevenin andNorton equivalent

networks

It is seen in Sections 33.2 and 33.3 that when Thevenin’s and Norton’stheorems are applied to the same circuit, identical results are obtained.Thus the Thevenin and Norton networks shown in Figure 33.67 are equiv-alent to each other. The impedance ‘looking in’ at terminals AB is thesame in each of the networks; i.e.,z.

If terminals AB in Figure 33.67(a) are short-circuited, the short-circuitcurrent is given byE/z.

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Figure 33.67 Equivalent Thevenin and Norton circuits

If terminals AB in Figure 33.67(b) are short-circuited, the short-circuitcurrent isISC.

Thus I SC D E=z.

Figure 33.68Figure 33.68 shows a source of e.m.f.E in series with an impedancez

feeding a load impedanceZL. From Figure 33.68,

IL D E

z C ZLD E/z

z C ZL/zD

(z

z C ZL

)E

z

i.e., I L D(

zz Y ZL

)I SC, from above.

From Figure 33.69 it can be seen that, when viewed from the load, thesource appears as a source of currentISC which is divided betweenz andZL connectedin parallel.

Figure 33.69

Thus it is shown that the two representations shown in Figure 33.67 areequivalent.

Problem14. (a) Convert the circuit shown in Figure 33.70(a) toan equivalent Norton network. (b) Convert the network shown inFigure 33.70(b) to an equivalent Thevenin circuit.

(a) If the terminals AB of Figure 33.70(a) are short circuited, the short-circuit current,ISC D 20/4 D 5 A. The impedance ‘looking in’ atterminals AB is 4�. Hence the equivalent Norton network is asshown in Figure 33.71(a).

(b) The open-circuit voltageE across terminals AB in Figure 33.70(b)is given byE D ISCz D 32 D 6 V. The impedance ‘lookingin’ at terminals AB is 2�.

Hence the equivalent Thevenin circuit is as shown inFigure 33.71(b).Figure 33.70

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Figure 33.71 Figure 33.72

Problem15. (a) Convert the circuit to the left of terminals ABin Figure 33.72 to an equivalent Thevenin circuit by initiallyconverting to a Norton equivalent circuit. (b) Determine themagnitude of the current flowing in the1.8 C j4� impedanceconnected between terminals A and B of Figure 33.72.

(a) For the branch containing the 12 V source, conversion to a Nortonequivalent network givesISC1 D 12/3 D 4 A and z1 D 3 �. Forthe branch containing the 24 V source, conversion to a Nortonequivalent circuit givesISC2 D 24/2 D 12 A and z2 D 2 �.

Thus Figure 33.73 shows a network equivalent to Figure 33.72.From Figure 33.73, the total short-circuit current is 4C 12 D 16 A,and the total impedance is given by3 ð 2/3 C 2 D 1.2 �. ThusFigure 33.73 simplifies to Figure 33.74.

Figure 33.73 Figure 33.74

The open-circuit voltage across AB of Figure 33.74,E D161.2 D 19.2 V, and the impedance ‘looking in’ at AB,z D1.2 �. Hence the Thevenin equivalent circuit is as shown inFigure 33.75.

(b) When the1.8 C j4� impedance is connected to terminals AB ofFigure 33.75, the currentI flowing is given by

I D 19.2

1.2 C 1.8 C j4D 3.846 �53.13° A

Hencethe current flowing in the .1.8Yj 4/Z impedance is 3.84 A.Figure 33.75

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Problem16. Determine, by successive conversions betweenThevenin’s and Norton’s equivalent networks, a Theveninequivalent circuit for terminals AB of Figure 33.76. Hencedetermine the magnitude of the current flowing in the capacitivebranch connected to terminals AB.

Figure 33.76

For the branch containing the 5 V source, converting to a Nortonequivalent network givesISC D 5/1000D 5 mA and z D 1 k�. For thebranch containing the 10 V source, converting to a Norton equivalentnetwork givesISC D 10/4000D 2.5 mA and z D 4 k�. Thus the circuitof Figure 33.76 converts to that of Figure 33.77.

Figure 33.77 Figure 33.78

The above two Norton equivalent networks shown in Figure 33.77 maybe combined, since the total short-circuit current is5 C 2.5 D 7.5 mAand the total impedancez is given by 1 ð 4/1 C 4 D 0.8 k�. Thisresults in the network of Figure 33.78.

Both of the Norton equivalent networks shown in Figure 33.78 maybe converted to Thevenin equivalent circuits. Open-circuit voltage acrossCD is

7.5 ð 10�30.8 ð 103 D 6 V

and the impedance ‘looking in’ at CD is 0.8 k�. Open-circuit voltageacross EF is1 ð 10�32 ð 102 D 2 V and the impedance ‘looking in’at EF is 2 k�. Thus Figure 33.78 converts to Figure 33.79.Figure 33.79

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Figure 33.80

Combining the two Thevenin circuits gives e.m.f.E D 6 � 2 D 4 V,and impedancez D 0.8 C 2 D 2.8 kZ. Thus the Thevenin equivalentcircuit for terminals AB of Figure 33.76 is as shown in Figure 33.80.

If an impedance200� j4000� is connected across terminals AB,then the currentI flowing is given by

I D 4

2800C 200� j4000D 4

50006 �53.13°D 0.806 53.13° mA

i.e., the curr ent in the capacitive branch is 0.80 mA.

Problem17. (a) Determine an equivalent Thevenin circuit forterminals AB of the network shown in Figure 33.81. (b) Calculatethe power dissipated in a600� j800� impedance connectedbetween A and B of Figure 33.81.

Figure 33.81

(a) Converting the Thevenin circuit to a Norton network gives

ISC D 5

j1000D �j5 mA or 56 �90° mA andz D j1 k�

ThusFigure 33.81 converts to that shown in Figure 33.82. The twoNorton equivalent networks may be combined, giving

ISC D 4 C 56 �90° D 4 � j5mA or 6.4036 �51.34° mA

and z D 2j1

2 C j1D 0.4 C j0.8k� or 0.8946 63.43° k�

Figure 33.82 Figure 33.83

This results in the equivalent network shown in Figure 33.83.Converting to an equivalent Thevenin circuit gives open circuite.m.f. across AB,

E D 6.403ð 10�3 6 �51.34°0.894ð 103 6 63.43°

D 5.7246 12.09° V

and

impedancez D 0.8946 63.43° k� or .400Y j 800/Z

Thusthe Thevenin equivalent circuit is as shown in Figure 33.84.Figure 33.84

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(b) When a 600� j800� impedance is connected across AB, thecurrentI flowing is given by

I D 5.7246 12.09°

400C j800 C 600� j800D 5.7246 12.09° mA

Hencethe power P dissipated in the600� j800� impedance isgiven by

P D I2 R D 5.724ð 10�32600 D 19.7 mW

Further problems on Thevenin and Norton equivalent networks may befound in Section 33.5 following, problems 16 to 21, page 600

33.5 Further problemson Thevenin’s andNorton’s theorem

Thevenin’s theorem

1 Use Thevenin’s theorem to determine the current flowing in the 10�resistor of the d.c. network shown in Figure 33.85. [0.85 A]

2 Determine, using Thevenin’s theorem, the values of currentsI1, I2

andI3 of the network shown in Figure 33.86.[I1 D 2.8 A, I2 D 4.8 A, I3 D 7.6 A]

3 Determine the Thevenin equivalent circuit with respect to terminalsAB of the network shown in Figure 33.87. Hence determinethe magnitude of the current flowing in a4 � j7� impedanceconnected across terminals AB and the power delivered to thisimpedance. [E D 15.376 �38.66°,

z D 3.20C j4.00�; 1.97 A; 15.5 W]

4 For the network shown in Figure 33.88 use Thevenin’s theorem todetermine the current flowing in the 3� resistance. [1.17 A]

5 Derive for the network shown in Figure 33.89 the Theveninequivalent circuit at terminals AB, and hence determine the currentflowing in a 20� resistance connected between A and B.

[E D 2.5 V, r D 5 �; 0.10 A]

6 Determine for the network shown in Figure 33.90 the Theveninequivalent circuit with respect to terminals AB, and hence determinethe current flowing in the5 C j6� impedance connected betweenA and B. [E D 14.326 6.38°, z D 3.99C j0.55�; 1.29 A]

Figure 33.85

Figure 33.86

7 For the network shown in Figure 33.91, derive the Theveninequivalent circuit with respect to terminals AB, and hence determinethe magnitude of the current flowing in a2 C j13� impedanceconnected between A and B. [1.157 A]

8 Use Thevenin’s theorem to determine the power dissipated in the4 � resistance of the network shown in Figure 33.92. [0.24 W]

Figure 33.87

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Figure 33.88 Figure 33.89 Figure 33.90

Figure 33.91 Figure 33.92

9 For the bridge network shown in Figure 33.93 use Thevenin’stheorem to determine the current flowing in the4 C j3� impedanceand its direction. Assume that the 206 0° V source has negligibleinternal impedance. [0.12 A from Q to P]

Figure 33.93

10 Repeat problems 1 to 10, page 542 of Chapter 30, 2 and 3 and 11to 15, page 559 of Chapter 31 and 2 to 8, page 573 of Chapter 32,using Thevenin’s theorem and compare the method of solution withthat used for Kirchhoff’s laws, mesh-current and nodal analysis andthe superposition theorem.

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Figure 33.94 Figure 33.95

Norton’s theorem

11 Repeat problems 1 to 4 and 6 to 8 above using Norton’s theoreminstead of Thevenin’s theorem.

12 Determine the current flowing in the 10� resistance of the networkshown in Figure 33.94 by using Norton’s theorem. [3.13 A]

13 For the network shown in Figure 33.95, use Norton’s theorem todetermine the current flowing in the 10� resistance. [1.08 A]

14 Determine for the network shown in Figure 33.96 the Norton equiv-alent network at terminals AB. Hence determine the current flowingin a 2 C j4� impedance connected between A and B.

[ISC D 2.1856 �43.96° A, z D 2.40C j1.47�; 0.88 A]

Figure 33.96

15 Repeat problems 1 to 10, page 542 of Chapter 30 and 2 and 3 and12 to 15, page 559 of Chapter 31, using Norton’s theorem.

Thevenin and Norton equivalent networks

16 Convert the circuits shown in Figure 33.97 to Norton equivalentnetworks. [(a)ISC D 2.5 A, z D 2 � (b) ISC D 26 30°, z D 5 �]

Figure 33.97

17 Convert the networks shown in Figure 33.98 to Thevenin equivalentcircuits. [(a)E D 20 V, z D 4 �; (b) E D 126 50° V, z D 3 �]

18 (a) Convert the network to the left of terminals AB in Figure 33.99to an equivalent Thevenin circuit by initially converting to aNorton equivalent network.

(b) Determine the current flowing in the2.8 � j3� impedanceconnected between A and B in Figure 33.99.

[(a) E D 18 V, z D 1.2 � (b) 3.6 A]

19 Determine, by successive conversions between Thevenin and Nortonequivalent networks, a Thevenin equivalent circuit for terminalsAB of Figure 33.100. Hence determine the current flowing in a2 C j4� impedance connected between A and B.[

E D 913 V, z D 1 �I 1.876 �53.13° A

]Figure 33.98

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Figure 33.99 Figure 33.100

20 Derive an equivalent Thevenin circuit for terminals AB of thenetwork shown in Figure 33.101. Hence determine the p.d. acrossAB when a 3 C j4k� impedance is connected between theseterminals.

[E D 4.826 �41.63° V, z D 0.8 C j0.4 k�; 4.15 V]

Figure 33.101

21 For the network shown in Figure 33.102, derive (a) the Theveninequivalent circuit, and (b) the Norton equivalent network. (c) A 6�resistance is connected between A and B. Determine the currentflowing in the 6� resistance by using both the Thevenin and Nortonequivalent circuits.

[(a) E D 6.716 �26.57° V, z D 4.50C j3.75�(b) ISC D 1.156 �66.38°, z D 4.50C j3.75�(c) 0.60 A

Figure 33.102

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