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33
Thévenin’s and
Norton’s theorems
At the end of this chapter you should be able to:
ž understand and use Thévenin’s theorem to analyse a.c. and
d.c. networks
ž understand and use Norton’s theorem to analyse a.c. and d.c.
networks
ž appreciate and use the equivalence of Thévenin and Norton
networks
33.1
33.2
Introduction
Thévenin’s theorem
Many of the networks analysed in Chapters 30, 31 and 32 using
Kirchhoff’s laws, mesh-current and nodal analysis and the superposition
theorem can be analysed more quickly and easily by using Thévenin’s
or Norton’s theorems. Each of these theorems involves replacing what
may be a complicated network of sources and linear impedances with a
simple equivalent circuit. A set procedure may be followed when using
each theorem, the procedures themselves requiring a knowledge of basic
circuit theory. (It may be worth checking some general d.c. circuit theory
in Section 13.4. page 174, before proceeding)
Thévenin’s theorem states:
‘The current which flows in any branch of a network is the same as that
which would flow in the branch if it were connected across a source of
electrical energy, the e.m.f. of which is equal to the potential difference
which would appear across the branch if it were open-circuited, and the
internal impedance of which is equal to the impedance which appears
across the open-circuited branch terminals when all sources are replaced
by their internal impedances.’
The theorem applies to any linear active network (‘linear’ meaning that the
measured values of circuit components are independent of the direction
and magnitude of the current flowing in them, and ‘active’ meaning that
it contains a source, or sources, of e m f.)
The above statement of Thévenin’s theorem simply means that
a complicated network with output terminals AB, as shown in
Figure 33.1(a), can be replaced by a single voltage source E in series
with an impedance z, as shown in Figure 33.1(b). E is the open-circuit
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voltage measured at terminals AB and z is the equivalent impedance of
the network at the terminals AB when all internal sources of e.m.f. are
made zero. The polarity of voltage E is chosen so that the current flowing
through an impedance connected between A and B will have the same
direction as would result if the impedance had been connected between A
and B of the original network. Figure 33.1(b) is known as the Thévenin
equivalent circuit, and was initially introduced in Section 13.4, page 174
for d.c. networks.
The following four-step procedure can be adopted when determining,
by means of Thévenin’s theorem, the current flowing in a branch
containing impedance ZL of an active network:
remove the impedance ZL from that branch;
determine the open-circuit voltage E across the break;
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 shortcircuit), and then determine the internal impedance, z, ‘looking in’
at the break;
(iv) determine the current from the Thévenin equivalent circuit shown
in Figure 33.2, i.e.
(i)
(ii)
(iii)
current iL =
Figure 33.1 The Thévenin
equivalent circuit
E
.
ZL Y z
A simple d.c. network (Figure 33.3) serves to demonstrate how the above
procedure is applied to determine the current flowing in the 5 resistance by using Thévenin’s theorem. This is the same network as used in
Chapter 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)
(ii)
the 5 resistor is removed, as shown in Figure 33.4(a).
The open-circuit voltage E across the break is now required. The
network of Figure 33.4(a) is redrawn for convenience as shown in
Figure 33.4(b), where current,
I1 D
Figure 33.2
E1 E2
83
5
D
D
r1 C r2
1C2
3
or
1
2
A
3
Hence the open-circuit voltage E is given by
E D E1 I1 r1 i.e., E D 8 1 23 1
D 6 13 V
(Alternatively, E D E2 I1 r2 D 3 C 1 23 2
D 6 13 V.
(iii)
Figure 33.3
Removing each 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
2
3
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Figure 33.4
(iv)
The Thévenin equivalent circuit is shown in Figure 33.6, where
current iL is given by
iL D
6 31
E
D 1.1177
D
ZL C z
5 C 23
D 1.12 A, correct to two decimal places
Figure 33.5
To determine the currents flowing in the other two branches of
the circuit of Figure 33.3, basic circuit theory is used. Thus, from
Figure 33.7, voltage V D 1.1177
5
D 5.5885 V.
Then V D E1 IA r1 , i.e., 5.5885 D 8 IA 1
, from which
current IA D 8 5.5885 D 2.41 A.
Similarly, V D E2 IB r2 , i.e., 5.5885 D 3 IB 2
, from which
current IB D
Figure 33.6
3 5.5885
D −1.29 A
2
(i.e., flowing in the direction opposite to that shown in Figure 33.7).
The Thévenin theorem procedure used above may be applied to a.c. as
well as d.c. networks, as shown below.
An a.c. network is shown in Figure 33.8 where it is required to find the
current flowing in the 6 C j8
impedance by using Thévenin’s theorem.
Using the above procedure
Figure 33.7
(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. The
network is redrawn for convenience as shown in Figure 33.9(b),
where current
I1 D
Figure 33.8
5 C j0
C 2 C j4
7 C j4
D
3 C j4
C 2 j5
5 j
D 1.5816 41.05° A
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Figure 33.9
Hence open-circuit voltage across AB,
E D E1 I1 3 C j4
, i.e.,
E D 5 C j0
1.5816 41.05° 56 53.13° Figure 33.10
from which E D 9.5676 54.73° V
(iii)
From Figure 33.10, the impedance z ‘looking in’ at terminals AB
is given by
zD
3 C j4
2 j5
3 C j4
C 2 j5
D 5.2816 3.76° (iv)
Figure 33.11
or
5.270 j0.346
The Thévenin equivalent circuit is shown in Figure 33.11, from
which current
iL D
E
9.6576 54.73°
D
ZL C z
6 C j8
C 5.270 j0.346
Thus, current in 6 C j8
impedance,
iL D
Figure 33.12
9.6576 54.73°
D 0.71 6 −88.91° A
13.6236 34.18°
The network of Figure 33.8 is analysed using Kirchhoff’s laws in
problem 3, page 539, and by the superposition theorem in problem 4,
page 568. The above analysis using Thévenin’s theorem is seen to be
much quicker.
Problem 1. For the circuit shown in Figure 33.12, use Thévenin’s
theorem to determine (a) the current flowing in the capacitor, and
(b) the p.d. across the 150 k resistor.
(a)
Figure 33.13
(i) Initially the 150 j120
k impedance is removed from the
circuit as shown in Figure 33.13.
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Note that, to find the current in the capacitor, only the capacitor
need have been initially removed from the circuit. However,
removing each of the components from the branch through
which the current is required will often result in a simpler
solution.
(ii) From Figure 33.13,
current I1 D
2006 0°
D 8 mA
5000 C 20000
The open-circuit e m f. E is equal to the p.d. across the 20 k
resistor, i.e.
E D 8 ð 103 20 ð 103 D 160 V.
Figure 33.14
(iii) Removing the 2006 0° V source gives the network shown in
Figure 33.14.
The impedance, z, ‘looking in’ at the open-circuited terminals
is given by
z D
5 ð 20
k D 4 kZ
5 C 20
(iv) The Thévenin equivalent circuit is shown in Figure 33.15,
where current iL is given by
iL D
Figure 33.15
E
160
D
ZL C z
150 j120
ð 103 C 4 ð 103
D
160
195.23 ð 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 iL R D 0.82 ð 103 150 ð 103 D 123 V
Figure 33.16
Figure 33.17
Problem 2. Determine, for the network shown in Figure 33.16,
the value of current I. Each of the voltage sources has a frequency
of 2 kHz.
(i)
The impedance through which current I is flowing is initially
removed from the network, as shown in Figure 33.17.
(ii)
From Figure 33.17,
current, I1 D
20 10
D2A
2C3
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Hence the open circuit e.m.f. E D 20 I1 2
D 20 22
D 16 V.
(Alternatively, E D 10 C I1 3
D 10 C 2
3
D 16 V.)
(iii)
When the sources of e m f. are removed from the circuit, the
impedance, z, ‘looking in’ at the break is given by
zD
(iv)
Figure 33.18
2ð3
D 1.2 Z
2C3
The Thévenin equivalent circuit is shown in Figure 33.18, where
inductive reactance,
XL D 2fL D 22000
235 ð 106 D 2.95 Hence current
ID
16
16
D
1.2 C 1.5 C j2.95
4.06 47.53°
D 4.06 −47.53° A
or
.2.70 − j 2.95/ A
Problem 3. Use Thévenin’s theorem to determine the power dissipated in the 48 resistor of the network shown in Figure 33.19.
Figure 33.19
The power dissipated by a current I flowing through a resistor R is given
by I2 R, hence initially the current flowing in the 48 resistor is required.
(i)
(ii)
The 48 C j144
impedance is initially removed from the network
as shown in Figure 33.20.
From Figure 33.20,
current, i D
Figure 33.20
506 0°
D 0.16 53.13° A
300 j400
Hence the open-circuit voltage
E D i300
D 0.16 53.13° 300
D 306 53.13° V
(iii)
When the 506 0° V source shown in Figure 33.20 is removed, the
impedance, z, is given by
zD
j400
300
4006 90° 300
D
300 j400
5006 53.13°
D 2406 36.87° (iv)
.192 − j 144/Z
The Thévenin equivalent circuit is shown in Figure 33.21 connected
to the 48 C j144
load.
Current I D
Figure 33.21
or
306 53.13°
306 53.13°
D
192 j144
C 48 C j144
2406 0°
D 0.1256 53.13° A
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Hence the power dissipated in the 48 Z resistor
D I2 R D 0.125
2 48
D 0.75 W
Problem 4. For the network shown in Figure 33.22, use
Thévenin’s theorem to determine the current flowing in the 80 resistor.
Figure 33.22
One method of analysing a multi-branch network as shown in
Figure 33.22 is to use Thévenin’s theorem on one part of the network
at a time. For example, the part of the circuit to the left of AA may be
reduced to a Thévenin equivalent circuit.
From Figure 33.23,
E1 D
and z1 D
20
100 D 80 V, by voltage division
20 C 5
20 ð 5
D4
20 C 5
Thus the network of Figure 33.22 reduces to that of Figure 33.24. The
part of the network shown in Figure 33.24 to the left of BB may be
reduced to a Thévenin equivalent circuit, where
E2 D
Figure 33.23
50
80
D 40 V
50 C 46 C 4
Figure 33.24
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and current I6 D
6 C j8
0.062 C j0.964
3 C j4 C 6 C j8
D 0.041 C j0.643A
(v) If Figure 32.22 is superimposed on Figure 32.20, the resultant
currents 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 .0.573 Y j 0.618/A or 0.8436 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 .0.560 Y j 1.327/A or 1.4406 67.12° A
Resultant current flowing through the 6 C j8 impedance is
given 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 0.7096 88.92° 106 53.13° D 7.096 35.79° V
i.e., the magnitude of the voltage across the 6 C j8 impedance
is 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 the
phase angle between E2 and (I3 C I4 ), i.e.,
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and z2 D
50 ð 50
D 25 50 C 50
Thus the 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 be
reduced to a Thévenin equivalent circuit, where
E3 D
Figure 33.25
and z3 D
60
40
D 24 V
60 C 25 C 15
60
40
D 24 60 C 40
Thus the original network reduces to that of Figure 33.26, from which
the current in the 80 Z resistor is given by
ID
24
80 C 16 C 24
D 0.20 A
Problem 5. Determine the Thévenin equivalent circuit with
respect to terminals AB of the circuit shown in Figure 33.27.
Hence determine (a) the magnitude of the current flowing in a
3.75 C j11
impedance connected across terminals AB, and
(b) the magnitude of the p.d. across the 3.75 C j11
impedance.
Figure 33.26
Current I1 shown in Figure 33.27 is given by
I1 D
246 0°
246 0°
D 6 6 0° A
D
4 C j3 j3
46 0°
The Thévenin equivalent voltage, i.e., the open-circuit voltage across
terminals AB, is given by
Figure 33.27
E D I1 4 C j3
D 66 0° 56 36.87° D 306 36.87° V
When the 246 0° V source is removed, the impedance z ‘looking in’ at
AB is given by
z D
4 C j3
j3
9 j12
D
D .2.25 − j 3.0/Z
4 C j3 j3
4
Thus the Thévenin equivalent circuit is as shown in Figure 33.28.
(a)
When a 3.75 C j11
impedance is connected across terminals AB,
the current I flowing in the impedance is given by
ID
306 36.87°
306 36.87°
D
3.75 C j11
C 2.25 j3.0
106 53.13°
D 36 16.26° A
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Figure 33.28
Figure 33.29
Hence the current flowing in the .3.75 Y j 11/Z impedance
is 3 A.
P.d. across the 3.75 C j11
impedance is given by
(b)
V D 36 16.26° 3.75 C j11
D 36 16.26° 11.626 71.18° D 34.866 54.92° V
Hence the magnitude of the p.d. across the impedance is 34.9 V.
Problem 6. Use Thévenin’s theorem to determine the current
flowing in the capacitor of the network shown in Figure 33.29.
Figure 33.30
(i)
The capacitor is removed from branch AB, as shown in
Figure 33.30.
(ii)
The open-circuit voltage, E, shown in Figure 33.30, is given by
I2 5
. I2 may be determined by current division if I1 is known.
(Alternatively, E may be determined by the method used in
problem 4.)
Current I1 D V/Z, where Z is the total circuit impedance and
V D 16.556 22.62° V.
Impedance, Z D 4 C
j2
8 C j6
12 C j16
D4C
j2 C 8 C j6
8 C j8
D 4.5966 22.38° Hence I1 D
16.556 22.62°
D 3.606 45° A
4.5966 22.38°
and
I2 D
j2
26 90° 3.606 45° I1 D
j2 C 3 C j6 C 5
11.3146 45°
D 0.6366 0° A
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(An alternative method of finding I2 is to use Kirchhoff’s laws or
mesh-current or nodal analysis on Figure 33.30.)
E D I2 5
D 0.6366 0° 5
D 3.186 0° V
Hence
(iii)
If the 16.556 22.62° V source is removed from Figure 33.30, the
impedance, z, ‘looking in’ at AB is given by
zD
z D 3.6546 22.61° i.e.
(iv)
5[4 ð j2
/4 C j2
C 3 C j6
]
53.8 C j7.6
D
5 C [4 ð j2
/4 C j2
C 3 C j6]
8.8 C j7.6
or
.3.373 Y j 1.405/Z
The Thévenin equivalent circuit is shown in Figure 33.31, where
the current flowing in the capacitor, I, is given by
ID
3.186 0°
3.186 0°
D
3.373 C j1.405
j8
7.4086 62.91°
D 0.436 62.91° A in the direction from A to B
Figure 33.31
Problem 7. For the network shown in Figure 33.32, derive the
Thévenin equivalent circuit with respect to terminals PQ, and
hence determine the power dissipated by a 2 resistor connected
across PQ.
Figure 33.32
Current I1 shown in Figure 33.32 is given by
I1 D
Figure 33.33
106 0°
D 1.0546 18.43° A
5 C 4 C j3
Hence the voltage drop across the 5 resistor is given by VX D I1 5
D
5.276 18.43° V, and is in the direction shown in Figure 33.32, i.e., the
direction opposite to that in which I1 is flowing.
The open-circuit voltage E across PQ is the phasor sum of V1 , Vx and
V2 , as shown in Figure 33.33.
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Thus
E D 106 0° 56 45° 5.276 18.43°
D 1.465 j1.869
V
or
2.375 6 −51.91° V
The impedance, z, ‘looking in’ at terminals PQ with the voltage sources
removed is given by
z D8C
Figure 33.34
54 C j3
D 8 C 2.6356 18.44° D .10.50 Y j 0.833/Z
5 C 4 C j3
The Thévenin equivalent 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
ID
2.3756 51.91°
D 0.18966 55.72° A
10.50 C j0.833
C 2
The power P dissipated in the 2 resistor is given by
P D I2 R D 0.1896
2 2
D 0.0719 W 72 mW, correct to two
significant figures.
Problem 8. For the a.c. bridge network shown in Figure 33.35,
determine the current flowing in the capacitor, and its direction,
by using Thévenin’s theorem. Assume the 306 0° V source to have
negligible internal impedance.
Figure 33.35
(i)
(ii)
The j25 capacitor is initially removed from the network, as
shown in Figure 33.36.
P.d. between A and C,
VAC D
Z1
Z 1 C Z4
VD
15
306 0° 15 C 5 C j5
D 21.836 14.04° V
P.d. between B and C,
Figure 33.36
VBC D
Z2
Z 2 C Z3
VD
40
306 0° 40 C 20 C j20
D 18.976 18.43° V
Assuming that point A is at a higher potential than point B, then
the p.d. between A and B is
21.836 14.04° 18.976 18.43°
D 3.181 C j0.701
V 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. Between C and A is a volt
drop of 21.836 14.04° V. Hence the voltage at point A is
306 0° 21.836 14.04° D 10.29 6 30.98° V
Between points 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° D
13.426 26.55° V.
Since the magnitude of the voltage at B is higher than at A, current
must flow in the direction B to A.
(iii)
Replacing the 306 0° V source with a short-circuit (i.e., zero internal
impedance) gives the network shown in Figure 33.37(a). The
network is shown redrawn in Figure 33.37(b) and simplified in
Figure 33.37(c). Hence the impedance, z, ‘looking in’ at terminals
AB is given by
zD
15
5 C j5
40
20 C j20
C
15 C 5 C j5
40 C 20 C j20
D 5.1456 30.96° C 17.8896 26.57°
i.e.,
z D 20.41 C j10.65
Figure 33.37
(iv)
The Thévenin equivalent circuit is shown in Figure 33.38, where
current I is given by
ID
3.2576 12.43°
3.2576 12.43°
D
20.41 C j10.65
j25
24.956 35.11°
D 0.1316 47.54° A
Figure 33.38
Thus a current of 131 mA flows in the capacitor in a direction
from B to A.
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Further problems on Thévenin’s theorem may be found in Section 33.5,
problems 1 to 10, page 598.
33.3 Norton’s theorem
A source of electrical energy can be represented by a source of e.m.f. in
series with an impedance. In Section 33.2, the Thévenin constant-voltage
source 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 form
of representation. A source of electrical energy can also be represented by
a constant-current source, which may be alternating or direct, in parallel
with an impedance. It is shown in Section 33.4 that the two forms are in
fact equivalent.
Norton’s theorem states:
Figure 33.39 The Norton
equivalent circuit
Figure 33.40
‘The current that flows in any branch of a network is the same as that which
would flow in the branch if it were connected across a source of electrical
energy, the short-circuit current of which is equal to the current that would
flow in a short-circuit across the branch, and the internal impedance of
which is equal to the impedance which appears across the open-circuited
branch terminals.’
The above statement simply means that any linear active network
with output terminals AB, as shown in Figure 33.39(a), can be replaced
by a current source in parallel with an impedance z as shown in
Figure 33.39(b). The equivalent current source ISC (note the symbol in
Figure 33.39(b) as per BS 3939:1985) is the current through a short-circuit
applied to the terminals of the network. The impedance z is the equivalent
impedance of the network at the terminals AB when all internal sources of
e.m.f. are made zero. Figure 33.39(b) is known as the Norton equivalent
circuit, and was initially introduced in Section 13.7, page 181 for d.c.
networks.
The following four-step procedure may be adopted when determining
the current flowing in an impedance ZL of a branch AB of an active
network, using Norton’s theorem:
(i)
(ii)
(iii)
(iv)
short-circuit branch AB;
determine the short-circuit current ISC ;
remove each source of e m f. and replace it by its internal impedance
(or, if a current source exists, replace with an open circuit), then
determine the impedance, z, ‘looking in’ at a break made between
A and B;
determine the value of the current iL flowing in impedance ZL from
the Norton equivalent network shown in Figure 33.40, i.e.,
iL D
Figure 33.41
z
ISC
ZL C z
A simple d.c. network (Figure 33.41) serves to demonstrate how the above
procedure is applied to determine the current flowing in the 5 resistance
by using Norton’s theorem:
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(i)
The 5 branch is short-circuited, as shown in Figure 33.42.
(ii)
From Figure 33.42, ISC D I1 C I2 D
(iii)
If each source of e m f. is removed the impedance ‘looking in’ at
a break made between A and B is given by z D 1 ð 2
/1 C 2
D
2
.
3
8
1
C
3
2
D 9.5 A
(iv)
Figure 33.42
From the Norton equivalent network shown in Figure
33.43,
the
2
2
5 C 3 9.5 D
current in the 5 resistance is given by IL D 3
1.12 A, as obtained previously using Kirchhoff’s laws, the superposition theorem and by Thévenin’s theorem.
As with Thévenin’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 the
current flowing in the 6 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 in
Figure 33.45.
(ii)
From Figure 33.45,
ISC D I1 C I2 D
5 C j0
2 C j4
C
3 C j4
2 j5
4.4726 63.43°
5.3856 68.20°
D 1.152 j1.421
A or 1.8296 −50.97° A
D 16 53.13° Figure 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
zD
3 C j4
2 j5
3 C j4
C 2 j5
D 5.286 −3.76° Z
(iv)
Figure 33.46
or
.5.269 − j 0.346/Z
From the Norton equivalent network shown in Figure 33.46, the
current is given by
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Figure 33.47
Figure 33.48
iL D
D
Figure 33.49
z
ISC
ZL C z
5.286 3.76°
1.8296 50.97°
6 C j8
C 5.269 j0.346
i.e., current in .6 Y j 8/Z impedance, iL = 0.71 6 −88.91° A
Figure 33.50
Problem 9. Use Norton’s theorem to determine the value of
current I in the circuit shown in Figure 33.47.
(i)
(ii)
Figure 33.51
(iii)
(iv)
The branch containing the 2.8 resistor is short-circuited, as shown
in Figure 33.48.
The 3 resistor in parallel with a short-circuit is the same
as 3 in parallel with 0 giving an equivalent impedance of
3 ð 0
/3 C 0
D 0. Hence the network reduces to that shown in
Figure 33.49, where ISC D 5/2 D 2.5 A.
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 gives
z D 2 ð 3
/2 C 3
D 1.2 Z (see Figure 33.50).
The Norton equivalent network is shown in Figure 33.51, where
current I is given by
ID
Figure 33.52
1.2
3
2.5
D
D 0.606 36.87° A
1.2 C 2.8 j3
4 j3
Problem 10. For the circuit shown in Figure 33.52 determine the
current flowing in the inductive branch by using Norton’s theorem.
(i)
(ii)
Figure 33.53
The inductive branch is initially short-circuited, as shown in
Figure 33.53.
From Figure 33.53,
ISC D I1 C I2 D
20 10
C
D 13.3̇ A
2
3
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(iii)
(iv)
If the voltage sources are removed, the impedance, z, ‘looking in’
at a break made in AB is given by z D 2 ð 3
/2 C 3
D 1.2 Z.
The Norton equivalent network is shown in Figure 33.54, where
current I is given by
ID
Figure 33.54
1.2
16
P D
13.3
1.2 C 1.5 C j2.95
2.7 C j2.95
D 4.06 −47.53° A
or
.2.7 − j 2.95/A
Problem 11. Use Norton’s theorem to determine the magnitude
of the p.d. across the 1 resistance of the network shown in
Figure 33.55.
(i)
Figure 33.55
(ii)
(iii)
The branch containing the 1 resistance is initially short-circuited,
as shown in Figure 33.56.
4 in parallel with j2 in parallel with 0 (i.e., the
short-circuit) is equivalent to 0, giving the equivalent circuit of
Figure 33.57. Hence ISC D 10/4 D 2.5 A.
The 10 V source is removed from the network of Figure 33.55, as
shown in Figure 33.58, and the impedance z, ‘looking in’ at a break
made in AB is given by
1
1 1
1
j j C 2
2 j2
D C C
D
D
z
4 4 j2
j4
j4
Figure 33.56
from which
zD
(iv)
j4
8 j8
j42 C j2
D
D
D .1 − j 1/Z
2 j2
22 C 22
8
The Norton equivalent network is shown in Figure 33.59, from
which current I is given by
ID
Figure 33.57
1 j1
2.5
D 1.586 18.43° A
1 j1
C 1
Hence the magnitude of the p.d. across the 1 Z resistor is
given by
IR D 1.58
1
D 1.58 V.
Figure 33.58
Figure 33.59
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Figure 33.60
Figure 33.61
Problem 12. For the network shown in Figure 33.60, obtain the
Norton equivalent network at terminals AB. Hence determine the
power dissipated in a 5 resistor connected between A and B.
(i)
(ii)
Terminals AB are initially short-circuited, as shown in Figure 33.61.
The circuit impedance Z presented to the 206 0° V source is
given by
ZD2C
9 j12
4 C j3
j3
D2C
4 C j3
C j3
4
D 4.25 j3
or
5.2026 35.22° Thus current I in Figure 33.61 is given by
ID
206 0°
D 3.8456 35.22° A
5.2026 35.22°
Hence
ISC D
4 C j3
3.8456 35.22° 4 C j3
j3
D 4.8066 72.09° A
(iii)
Figure 33.62
Removing the 206 0° V source of Figure 33.60 gives the network
of Figure 33.62.
Impedance, z, ‘looking in’ at terminals AB is given by
z D j3 C
24 C j3
D j3 C 1.4916 10.3°
2 C 4 C j3
D .1.467 − j 2.733/Z
(iv)
or
3.1026 −61.77° Z
The Norton equivalent network is shown in Figure 33.63.
Current IL D
3.1026 61.77°
4.8066 72.09° 1.467 j2.733 C 5
D 2.1236 33.23° A
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Figure 33.63
Hence the power dissipated in the 5 Z resistor is
I2L R D 2.123
2 5
D 22.5 W
Problem 13. Derive the Norton equivalent network with respect
to terminals PQ for the network shown in Figure 33.64 and hence
determine the magnitude of the current flowing in a 2 resistor
connected across PQ.
Figure 33.64
This is the same problem as problem 7 on page 584 which was solved
by Thévenin’s theorem.
A comparison of methods may thus be made.
(i)
Terminals PQ are initially short-circuited, as shown in Figure 33.65.
(ii)
Currents I1 and I2 are shown labelled. Kirchhoff’s laws are used.
For loop ABCD, and moving anticlockwise,
106 0° D 5I1 C 4 C j3
I1 C I2 ,
i.e.,
Figure 33.65
9 C j3
I1 C 4 C j3
I2 10 D 0
1
For 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
4 C j3
8
I1
D 9 C j3
5
10
6
°
5 45 10
I2
10
6
°
5 45 10
I
9
C
j3
4 C j3
5
8 D 19
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from which
9 C j3
I2 D
D
10
[9 C j3
56 45° 10
C 50]
5
56 45° 10
D
9 C j3
4 C j3
[72 j24 20 j15]
5
8 [22.526 146.50° ]
D 0.2256 303.53° or 0.2256 56.47°
[99.9256 157.03° ]
Hence the short-circuit current ISC D 0.2256 56.47° A flowing
from P to Q.
(iii)
The impedance, z, ‘looking in’ at a break made between P and Q
is given by
z D 10.50 C j0.833
(see problem 7, page 584).
(iv)
The Norton equivalent circuit is shown in Figure 33.66, where
current I is given by
ID
10.50 C j0.833
0.2256 56.47° 10.50 C j0.833 C 2
D 0.196 55.74° A
Figure 33.66
Hence the magnitude of the current flowing in the 2 Z resistor
is 0.19 A.
Further problems on Norton’s theorem may be found in Section 33.5,
problems 11 to 15, page 600
33.4 Thévenin and
Norton equivalent
networks
It is seen in Sections 33.2 and 33.3 that when Thévenin’s and Norton’s
theorems are applied to the same circuit, identical results are obtained.
Thus the Thévenin and Norton networks shown in Figure 33.67 are equivalent to each other. The impedance ‘looking in’ at terminals AB is the
same in each of the networks; i.e., z.
If terminals AB in Figure 33.67(a) are short-circuited, the short-circuit
current is given by E/z.
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Figure 33.67
Equivalent Thévenin and Norton circuits
If terminals AB in Figure 33.67(b) are short-circuited, the short-circuit
current is ISC .
Thus
Figure 33.68
ISC D E =z .
Figure 33.68 shows a source of e.m.f. E in series with an impedance z
feeding a load impedance ZL . From Figure 33.68,
E
E/z
D
IL D
D
z C ZL
z C ZL /z
i.e.,
Figure 33.69
IL D
z
z Y ZL
z
z C ZL
E
z
ISC , from above.
From Figure 33.69 it can be seen that, when viewed from the load, the
source appears as a source of current ISC which is divided between z and
ZL connected in parallel.
Thus it is shown that the two representations shown in Figure 33.67 are
equivalent.
Problem 14. (a) Convert the circuit shown in Figure 33.70(a) to
an equivalent Norton network. (b) Convert the network shown in
Figure 33.70(b) to an equivalent Thévenin circuit.
Figure 33.70
(a)
If the terminals AB of Figure 33.70(a) are short circuited, the shortcircuit current, ISC D 20/4 D 5 A. The impedance ‘looking in’ at
terminals AB is 4 . Hence the equivalent Norton network is as
shown in Figure 33.71(a).
(b)
The open-circuit voltage E across terminals AB in Figure 33.70(b)
is given by E D ISC z
D 3
2
D 6 V. The impedance ‘looking
in’ at terminals AB is 2 .
Hence the equivalent
Figure 33.71(b).
Thévenin
circuit
is
as
shown
in
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Figure 33.71
Figure 33.72
Problem 15. (a) Convert the circuit to the left of terminals AB
in Figure 33.72 to an equivalent Thévenin circuit by initially
converting to a Norton equivalent circuit. (b) Determine the
magnitude of the current flowing in the 1.8 C j4
impedance
connected between terminals A and B of Figure 33.72.
(a)
For the branch containing the 12 V source, conversion to a Norton
equivalent network gives ISC1 D 12/3 D 4 A and z1 D 3 . For
the branch containing the 24 V source, conversion to a Norton
equivalent circuit gives ISC2 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 4 C 12 D 16 A,
and the total impedance is given by 3 ð 2
/3 C 2
D 1.2 . Thus
Figure 33.73 simplifies to Figure 33.74.
Figure 33.73
Figure 33.74
The open-circuit voltage across AB of Figure 33.74, E D
16
1.2
D 19.2 V, and the impedance ‘looking in’ at AB, z D
1.2 . Hence the Thévenin equivalent circuit is as shown in
Figure 33.75.
(b)
When the 1.8 C j4
impedance is connected to terminals AB of
Figure 33.75, the current I flowing is given by
ID
Figure 33.75
19.2
D 3.846 53.13° A
1.2 C 1.8 C j4
Hence the current flowing in the .1.8Yj 4/Z impedance is 3.84 A.
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Problem 16. Determine, by successive conversions between
Thévenin’s and Norton’s equivalent networks, a Thévenin
equivalent circuit for terminals AB of Figure 33.76. Hence
determine the magnitude of the current flowing in the capacitive
branch connected to terminals AB.
Figure 33.76
For the branch containing the 5 V source, converting to a Norton
equivalent network gives ISC D 5/1000 D 5 mA and z D 1 k. For the
branch containing the 10 V source, converting to a Norton equivalent
network gives ISC D 10/4000 D 2.5 mA and z D 4 k. Thus the circuit
of 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 may
be combined, since the total short-circuit current is 5 C 2.5
D 7.5 mA
and the total impedance z is given by 1 ð 4
/1 C 4
D 0.8 k. This
results in the network of Figure 33.78.
Both of the Norton equivalent networks shown in Figure 33.78 may
be converted to Thévenin equivalent circuits. Open-circuit voltage across
CD is
7.5 ð 103 0.8 ð 103 D 6 V
Figure 33.79
and the impedance ‘looking in’ at CD is 0.8 k. Open-circuit voltage
across EF is 1 ð 103 2 ð 102 D 2 V and the impedance ‘looking in’
at EF is 2 k. Thus Figure 33.78 converts to Figure 33.79.
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Combining the two Thévenin circuits gives e.m.f. E D 6 2 D 4 V,
and impedance z D 0.8 C 2
D 2.8 kZ. Thus the Thévenin equivalent
circuit for terminals AB of Figure 33.76 is as shown in Figure 33.80.
If an impedance 200 j4000
is connected across terminals AB,
then the current I flowing is given by
ID
Figure 33.80
4
4
D 0.806 53.13° mA
D
2800 C 200 j4000
50006 53.13°
i.e., the current in the capacitive branch is 0.80 mA.
Problem 17. (a) Determine an equivalent Thévenin circuit for
terminals AB of the network shown in Figure 33.81. (b) Calculate
the power dissipated in a 600 j800
impedance connected
between A and B of Figure 33.81.
(a)
Figure 33.81
Converting the Thévenin circuit to a Norton network gives
ISC D
5
D j5 mA or
j1000
56 90° mA and z D j1 k
Thus Figure 33.81 converts to that shown in Figure 33.82. The two
Norton equivalent networks may be combined, giving
ISC D 4 C 56 90° D 4 j5
mA or 6.4036 51.34° mA
and z D
Figure 33.82
2
j1
D 0.4 C j0.8
k or
2 C j1
0.8946 63.43° k
Figure 33.83
This results in the equivalent network shown in Figure 33.83.
Converting to an equivalent Thévenin circuit gives open circuit
e.m.f. across AB,
E D 6.403 ð 103 6 51.34° 0.894 ð 103 6 63.43° D 5.724 6 12.09° V
and
impedance z D 0.8946 63.43° k
Figure 33.84
or
.400 Y j 800/Z
Thus the Thévenin equivalent circuit is as shown in Figure 33.84.
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(b)
When a 600 j800
impedance is connected across AB, the
current I flowing is given by
ID
5.7246 12.09°
D 5.7246 12.09° mA
400 C j800
C 600 j800
Hence the power P dissipated in the 600 j800
impedance is
given by
P D I2 R D 5.724 ð 103 2 600
D 19.7 mW
Further problems on Thévenin and Norton equivalent networks may be
found in Section 33.5 following, problems 16 to 21, page 600
33.5
Further problems
on Thévenin’s and
Norton’s theorem
Thévenin’s theorem
1
Use Thévenin’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 Thévenin’s theorem, the values of currents I1 , I2
and I3 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 Thévenin equivalent circuit with respect to terminals
AB of the network shown in Figure 33.87. Hence determine
the magnitude of the current flowing in a 4 j7
impedance
connected across terminals AB and the power delivered to this
impedance.
[E D 15.376 38.66° ,
z D 3.20 C j4.00
; 1.97 A; 15.5 W]
4
For the network shown in Figure 33.88 use Thévenin’s theorem to
determine the current flowing in the 3 resistance.
[1.17 A]
5
Derive for the network shown in Figure 33.89 the Thévenin
equivalent circuit at terminals AB, and hence determine the current
flowing 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 Thévenin
equivalent circuit with respect to terminals AB, and hence determine
the current flowing in the 5 C j6
impedance connected between
A and B.
[E D 14.326 6.38° , z D 3.99 C j0.55
; 1.29 A]
7
For the network shown in Figure 33.91, derive the Thévenin
equivalent circuit with respect to terminals AB, and hence determine
the magnitude of the current flowing in a 2 C j13
impedance
connected between A and B.
[1.157 A]
8
Use Thévenin’s theorem to determine the power dissipated in the
4 resistance of the network shown in Figure 33.92.
[0.24 W]
Figure 33.85
Figure 33.86
Figure 33.87
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Figure 33.88
Figure 33.91
Figure 33.89
Figure 33.90
Figure 33.92
9
For the bridge network shown in Figure 33.93 use Thévenin’s
theorem to determine the current flowing in the 4 C j3
impedance
and its direction. Assume that the 206 0° V source has negligible
internal 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 11
to 15, page 559 of Chapter 31 and 2 to 8, page 573 of Chapter 32,
using Thévenin’s theorem and compare the method of solution with
that used for Kirchhoff’s laws, mesh-current and nodal analysis and
the 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 theorem
instead of Thévenin’s theorem.
12
Determine the current flowing in the 10 resistance of the network
shown 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 to
determine the current flowing in the 10 resistance.
[1.08 A]
14
Determine for the network shown in Figure 33.96 the Norton equivalent network at terminals AB. Hence determine the current flowing
in a 2 C j4
impedance connected between A and B.
[ISC D 2.1856 43.96° A, z D 2.40 C j1.47
; 0.88 A]
15
Repeat problems 1 to 10, page 542 of Chapter 30 and 2 and 3 and
12 to 15, page 559 of Chapter 31, using Norton’s theorem.
Figure 33.96
Figure 33.97
Thévenin and Norton equivalent networks
16
Convert the circuits shown in Figure 33.97 to Norton equivalent
networks.
[(a) ISC D 2.5 A, z D 2 (b) ISC D 26 30° , z D 5 ]
17
Convert the networks shown in Figure 33.98 to Thévenin equivalent
circuits.
[(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 33.99
to an equivalent Thévenin circuit by initially converting to a
Norton equivalent network.
(b) Determine the current flowing in the 2.8 j3
impedance
connected between A and B in Figure 33.99.
[(a) E D 18 V, z D 1.2 (b) 3.6 A]
19
Figure 33.98
Determine, by successive conversions between Thévenin and Norton
equivalent networks, a Thévenin equivalent circuit for terminals
AB of Figure 33.100. Hence determine the current flowing in a
2 C j4
impedance connected
between A and B.
E D 9 13 V, z D 1 I 1.876 53.13° A
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Figure 33.99
Figure 33.101
Figure 33.100
20
Derive an equivalent Thévenin circuit for terminals AB of the
network shown in Figure 33.101. Hence determine the p.d. across
AB when a 3 C j4
k impedance is connected between these
terminals.
[E D 4.826 41.63° V, z D 0.8 C j0.4
k; 4.15 V]
21
For the network shown in Figure 33.102, derive (a) the Thévenin
equivalent circuit, and (b) the Norton equivalent network. (c) A 6 resistance is connected between A and B. Determine the current
flowing in the 6 resistance by using both the Thévenin and Norton
equivalent circuits.
[(a) E D 6.716 26.57° V, z D 4.50 C j3.75
(b) ISC D 1.156 66.38° , z D 4.50 C j3.75
(c) 0.60 A
Figure 33.102
28