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CIRCUIT ANALYSIS
CHAPTER 4
CIRCUIT ANALYSIS 4.0
INTRODUCTION
The process of circuit analysis consists of determining the currents in, and potential differences across, all branches of a network or circuit. The data normally comprise all source voltages and/or currents and all branch resistances/impedances. O h m ' s and Kirchhoff's Laws form the basis of all methods of analysis, either directly or in modified form, according to circumstances. Thus, in circuits containing only voltage sources, mesh or loop analysis is generally to be preferred to Kirchhoff's Laws themselves since fewer simultaneous equations result and, when only current sources are present, nodal analysis is generally best. In addition, many network theorems exist which, although not strictly falling within the above definition of circuit analysis, are nevertheless useful aids in the process. A few of these will be dealt with in this chapter. 4.1 N E T W O R K
TERMINOLOGY
First a few definitions. (a) A branch of a circuit is that part of the circuit containing a single active or passive element. (b) A node is the point of interconnection of two or more branches. The junction of three or more branches is referred to as a principal node; a secondary node is the junction of two branches. 200
CIRCUIT ANALYSIS
201
(c) A planar network is one whose schematic may be drawn on a plane surface without branch crossings. (d) A non-planar network is one whose schematic may not be so drawn. (e) A loop is any closed path in a network which does not pass through any branch or node more than once. (f) A mesh is a loop which contains no other loops. It is defined only for planar networks.
4.2 KIRCHHOFF'S L A W S
The reader is referred to section 1.3 for a statement of the laws, which are valid for both planar and non-planar networks. A n unknown current is assigned to every branch and the first law applied at as many nodes as will provide independent equations. The second law is then applied to as many independent loops as are necessary to form, in total, the same n u m b e r of simultaneous equations as there are unknowns. They are solved for the branch currents by substitution, determinants or matrix methods. All branch voltages are then easily found by applying O h m ' s Law. With simple circuits it is relatively easy to choose nodes and loops which yield independent equations. A rule of t h u m b , satisfactory in simple cases, is to choose each new node or loop so as to involve at least one branch current not already involved, either explicitly or implicitly. For more complicated circuits some care is needed and for guaranteed independence of the equations, the methods of section 4.5 must be used. The method of application of the laws is most readily understood by dealing initially with steady (direct current) conditions. P R O B L E M 4.1
Calculate, using Kirchhoff's Laws, all the branch currents and voltages in the network represented by Fig. 4 . 1 . Solution The circuit has six branches and four nodes and we begin by assigning six unknown currents, one to each branch (Fig. 4.1(a)).
202
250
ELECTRIC CIRCUIT THEORY
a
250 &
FIG. 4.1. (a) Diagram of the network for Problem 4.1.
250
a
FIG. 4.1. (b) The network for Fig. 4.1(a) with the number of unknowns reduced to three.
95 a
FIG. 4.1. (c) Analysing the circuit of Fig. 4.1(a) using the loop-current method.
Applying the first Law at node A :
/i = / + 2
h-
(4.1)
A t node B : A t node D :
h - U = /6.
(4.2)
h + h = /5-
(4.3)
CIRCUIT ANALYSIS
203
This is the limit of i n d e p e n d e n t equations obtainable from the first Law since, at n o d e C, the equation would be: h + h = /,
(4.4)
which is already obtainable from eqns. (4.1), (4.2) and (4.3). Applying the second Law to loop (or mesh) A B D , taking the clockwise direction as positive: 2 5 0 / 2 + 1200/4 - 2OOO/3 = 0.
(4.5)
h and U flow clockwise round the loop and / 3, anticlockwise (and hence is negative in eqn. (4.5)) whilst the total e.m.f. in the loop is zero. Similarly, in loop (or mesh) B C D : 9 5 / 6 - IOOO/5 - 1200/4 = - 0 . 1 .
(4.6)
In the loop containing A D C and the 6-V source: 2OOO/3 + 1000/ 5 = 6.
(4.7)
Solving eqns. (4.1), (4.2), (4.3), (4.5), (4.6) and (4.7) gives: A = 19.11mA,
/ 4= - 0 . 1 4 m A ,
h = 17.06 m A ,
/ 5= 1.91mA,
/ 3 = 2.05 m A ,
/ 6= 17.20 m A .
O h m ' s Law is used to calculate the potential differences for all branches. Thus: VAB = 2 5 0 / 2
= 4.265 V ,
VAD = 2OOO/3
= 4.100 V,
K BD = 1200/ 4
= - 0 . 1 6 8 V,
VDC = 1000/ 5
= 1.91 V,
VBC = 9 5 / 6+ 0.1 = 1.734 V.
204
ELECTRIC CIRCUIT THEORY
4.2.1 Reducing the number of unknowns The computational work may be reduced considerably, for this problem, by a more skilful application of the two Laws at the outset. T h u s , referring to Fig. 4.1(b), if at node A , /i is m a d e to divide into i2 and (i\ - / 2) , three branches have been covered with the introduction of only two u n k n o w n currents. Similarly at node B , if / 3 flows along B D , then (i2 - / 3) flows along B C , and (i\ - i2 + / 3) flows along D C . N o d e C adds no further information (as before) since (i2 - / 3) and (i\ - i2 + 13) combine to give i\, A total of only three unknown currents has been introduced for the whole circuit and only three independent equations are required for their solution. Using the same loops as before: 250/ + I2OO/3 - 2000(/! - i ) = 0,
(4.8)
95(i2 - i 3) - 1000(/i - i2 + i 3) - 1200/3 = - 0 . 1
(4.9)
2000(/j - / 2) + 1000(/i - i2 + / 3) = 6
(4.10)
2
2
and
which may be rearranged as follows: 2000/j - 2250/ 2 - I2OO/3 = 0,
(4.8a)
1000/, - 1095/2 + 2295/3 = 0 . 1 ,
(4.9a)
3000ii ~ 3000/2 + IOOO/3 = 6
(4.10a)
and subsequently solved for i u i2 and / 3. T h e current in any branch may be found by combining, as appropriate, the values for i u i2 and / 3. 4.3 T H E L O O P - A N D M E S H - C U R R E N T M E T H O D S
Either of these m e t h o d s replaces both of Kirchhoff s Laws by one rule, and is therefore generally simpler to apply in practice. T h e two methods differ only in the choice of closed path in a network round which the total voltage d r o p is equated to the e.m.f. In the mesh-current m e t h o d t h e paths are t h e obvious simple meshes and t h e method is therefore restricted to planar networks. T h e loop-current m e t h o d is
CIRCUIT ANALYSIS
205
m o r e general in that it applies to both planar and non-planar networks, but, as with the loops chosen for Kirchhoff's second Law, care must be taken to ensure a sufficient n u m b e r of independent equations in order to solve for all the u n k n o w n currents. In either m e t h o d a circulating current is associated with each selected closed p a t h . O n e particular direction is arbitrarily chosen as positive and the total voltage d r o p round each path is calculated by taking the algebraic sum of the products of (a) the path circulating current and the total path resistance, and (b) such other circulating currents (from adjoining paths) as impinge upon the chosen path and the resistances through which they flow. T h e result is equal to the total e.m.f. acting r o u n d the p a t h . Referring to Fig. 4.1(c), three possible loops (which in this case are meshes) are chosen, as shown, and the currents / A, h and IC associated with t h e m . R o u n d loop A : / A( 2 5 0 + 1200 4- 2000) - 1 2 0 0 / B- 2 0 0 0 / c = 0 or 3450/A - 1200/B - 2 0 0 0 / c = 0.
(4.11)
Comparing Figs. 4.1 (b) and (c), clearly for the m e t h o d to be valid, IA = h, h = ( ' 2 ~ *3) and IC = / i - Substituting these into eqn. (4.11) gives: 3450/2 ~ 1200/ 2 + I2OO/3 - 2000/, = 0 or 2000/! - 2250/ 2 - I2OO/3 = 0
(4.12)
which is seen to be identical with eqn. (4.8a), and the m e t h o d is therefore justified by Kirchhoff's Laws. R o u n d loop B : 2295/B - 1 2 0 0 / A- 1000/C = - 0 . 1 .
(4.13)
3000/C - 2 0 0 0 / A- 1000/B - 6.
(4.14)
R o u n d loop C:
ELECTRIC CIRCUIT THEORY
206
The three loop equations (4.11), (4.13) and (4.14) are solved to give: 7 I
A = 17.06 m A ,
B = 17.20 m A ,
/ c = 19.11 m A and the branch currents calculated as appropriate. Thus: IAB
= 17.06 m A ,
^BD
= (IA - h)
IAD
= (/c - IA) = 2.05 m A ,
IDC
= (/c -
ICA
= 19.11 m A ,
/B)
=
=
7
BC = 17.20 m A ,
-0.14 mA,
1.91 m A ,
i.e. the same result as for Kirchhoff s Laws. O t h e r loops might have been chosen—for the network of Fig. 4.1 there are seven such possible. They cannot all be independent as there are only three u n k n o w n currents, but in a circuit as simple as that of Fig. 4.1 the most obvious choices are the three mesh currents. 4.4 T H E N O D E - V O L T A G E M E T H O D
O n e principal node in the given network is chosen as a reference node and an unknown p . d . (with respect to the reference node) is assigned to every other principal n o d e . Kirchhoff s first Law is applied to each node in turn, each branch current meeting at a node being written as the product of the branch voltage and the branch conductance. If a current source feeds the node it must, of course, be included in the Kirchhoff Law summation. The method is the exact dual of the loop-current method and is valid for both planar and non-planar networks. P R O B L E M 4.2
Calculate, by the node-voltage m e t h o d , all branch currents and node potential differences for the circuit represented by Fig. 4 . 1 .
CIRCUIT ANALYSIS
207
Solution The circuit diagram is redrawn in Fig. 4.2, and it will be seen that node C has been chosen as the reference n o d e . Since VA (with respect to the reference) is 6 V , as determined by the voltage source, this leaves only two node voltages to be determined. B
FIG. 4.2. Analysing the circuit of Fig.4.1(a) using the node-voltage method (see Problem 4.2).
Designating these VB and VD , we apply Kirchhoff's first Law to n o d e D: Vn-6\
. /VD Vn\ 2000 / \ 1200 /
Vn-0 1000
(4.15)
Each term in t h e equation is a current flowing away from the n o d e —calculated by multiplying a branch conductance by the potential difference between the ends of the branch. Since there is no current source connected to the n o d e , their sum is equal to zero. Similarly, at n o d e B : V B- 6 \ . /VB-VD \ . 250 / V 1200 ) \
/Vu-0.V 95
(4.16)
There is no need to include the secondary node E in the analysis si
ETC - O
ELECTRIC CIRCUIT THEORY 208 _ (VB VE) is uniquely determined by the 0.1 V source, and, as soon as VB is found from (4.15) and (4.16), VE follows. Rearranging (4.15) and (4.16) gives:
(4.15a)
and (4.16a) an arrangement which suggests an alternative, more easily r e m e m b e r e d form of the node-voltage rule. A t each n o d e , calculate the product of the voltage (to the reference node) and the sum of all conductances meeting at the n o d e . For every other node connected to the chosen node through one conductance, calculate the product of its voltage and the conductance and subtract from the first product. T h e result is equal to the total current flowing into the n o d e . Equations (4.15a) and (4.16a) reduce to: 2 3 . 3 3 3 V D- 8.333KB = 30
(4.15b)
- 8 . 3 3 3 V D+ 153.597KB = 250.526
(4.16b)
and
which may be solved to give: VD = 1.905 volts
and
VB = 1.734 volts.
T h e other potential differences follow: yAB = 6 - 1.734 = 4.266 V;
VAD = 6 - 1.905 = 4.095 V,
VBD = 1.734 - 1.905 = - 0 . 1 7 1 V;
VEC = 1.734 - 0.1 - 1.634 V.
A n d the current in each branch calculated by O h m ' s Law:
/AD
CIRCUIT ANALYSIS
^BC ^DC
1.734 - 0.1 95
209
17.20 m A ,
1.905 m A , -0.171 1200
0.143 m A .
T h e network of Fig. 4.2 has four nodes, o n e being chosen as t h e reference. In general, therefore, three node voltages would be u n k n o w n and three equations would be required for their determination. This would have been the case if the 6-V source had been a current source. However, in cases where a voltage source is connected between two principal nodes, one of these nodes is chosen as the reference and this immediately determines the voltage of the other, thus reducing the number of unknown voltages by o n e . If more than o n e pair of principal nodes are connected to voltage sources, a solution by the node-voltage method, directly, is not possible. Source transformation or mesh (or loop) analysis must be used. 4.5 N E T W O R K
TOPOLOGY
A method of determining the correct n u m b e r of independent loops in a network consists of first drawing the network graph. This is a line diagram showing all principal nodes and branches of the original network but omitting the character of each branch. A s an example, the diagram of Fig. 4.3(a) shows the graph of Fig. 4.2. Branches are then removed from the graph so that all nodes remain interconnected but n o loops remain. T h e remaining diagram is called a TREE. For the graph of Fig. 4.3(a) there are several possible trees, three of which are shown by the full lines in 4.3 ( b ) , (c) and (d). T h e full lines are referred to as TREE BRANCHES and the dotted lines (which convert the tree to the original graph) as LINKS.
Clearly, no loop current can flow in a tree, but for each link added (in the process of reforming the graph from the tree) an extra loop current can flow. T h u s , in the tree of Fig. 4.3(b) the reinsertion of link B C would allow the loop A B C D to be formed; link B D would allow the loop A B D and link A C allows the loop A D C F . Each of these loops
ELECTRIC CIRCUIT THEORY
210
B
(d)
(c)
(b)
FIG. 4.3. (b-d) Three possible trees of the graph of Fig. 4.3(a).
is the path of an independent loop current. Obviously, as soon as each link current is known, all branch currents can be calculated from various linear combinations of the link currents. If a network has TV principal nodes, then the tree must have (TV - 1) branches, since the first branch connects two nodes and each subsequent branch adds one n o d e . If there are B branches altogether then L , the number of links, must be: L = B-(N-\)
= B-N+\
(4.17)
and this is the minimum n u m b e r of independent loop equations for solution of the original network. For the network represented by Fig. 4.3(a) B = 6 and /V = 4. H e n c e : L = 6 - 4 + l = 3 so that three equations are required. However, the trees of Fig. 4.3 (b), (c) and (d) show that at least three different sets of three equations are possible. In (b), the three loops containing the link currents are A B D A
CIRCUIT ANALYSIS
211
(which includes the link B D ) , A B C D A (link BC) and A D C F A (link A C ) . In (c) the three loops are A B D A , B C D B and A B D C F A ; and in (d) the three loops correspond to the three meshes chosen for Fig. 4.1(c). Several others are possible. T h e choice of suitable loops can also depend on the type and n u m b e r of energy sources in the network. Thus, for a network containing current sources the tree should be chosen so that, as far as possible, the current sources appear as links, so reducing the n u m b e r of unknown currents to less than that given by eqn. (4.17). P R O B L E M 4.3
Calculate, by both the node-voltage and the loop-current methods, all branch currents and node potential differences for the non-planar network represented in Fig. 4.4(a). Solution (a) Node-voltage method. For the given network N = 4, hence a sufficient n u m b e r of equations to solve for the unknown voltages is three. Choosing node C as the reference, the three equations are set up as follows: For node A : (4.18) ( 1 + i + i) VA~ V -hV =5
B
D
B
5A F i g . 4.4. (a) Diagram of the network for Problem 4.3.
F i g . 4.4. (b) Tree for the network of Fig. 4.4(a).
ELECTRIC CIRCUIT THEORY
212
At node B : ( l + i + i ) V B- V A- ( i + J ) V D= 0.
(4.19)
(i + i + i + i) vD - ivA - (j + i ) vB = o.
(4.20)
At node D :
These m a y b e rearranged as follows: IVA - 4VB - 2 K D= 20,
(4.18a)
- 2 0 V A + 3 9 K B- 9V D= 0,
(4.19a)
-30VA - 27VB + 11VD= 0
(4.20a)
and solved by substitution, determinants o r matrix methods t o give V
A
V and
V
B
= 6 . 3 7 4 1 volts,
(4.21)
= 4 . 1 8 0 1 volts,
(4.22)
D=
(4.23)
3 . 9 4 9 2 volts.
Thus, t h e voltages between nodes a r e : V and
AB =
V
A
-
V
A
-
V
VAD = V
BD — VB —
V
B
= 2 . 1 9 4 0 volts,
(4.24)
2 . 4 2 4 9 volts,
D=
Vo = 0 . 2 3 0 9 volts,
enabling t h e currents to be calculated, by O h m ' s Law, as follows: 2.194 1
he
=^4^
2
2.4249 »AD
2.194 A ,
0.2309 ^BD — '
4
6.3741
= 2.0901 A ,
/ AC=
= 1.2125 A ,
/ BD(through 5 Q )
= 0.0577 A , = 1.5935 A , 0.2309 0.0462 A ,
3 9492 /DC = ^ ^ = 1 . 3 1 6 4 A .
(4.25)
CIRCUIT ANALYSIS
(b) Loop-current N = 4,
213
method. T h e n u m b e r of branches, 5 , is 8 , and since L =
8
- 4 + 1
= 5. H e n c e five loop currents are n e e d e d , requiring five equations. W e begin by drawing the line graph of the network, and proceed by reducing to a tree as shown in Fig. 4.4(b). The links are shown dotted, as before, and, by making the current source one of the links, we can reduce the u n k n o w n currents to four. The loop currents are chosen, as shown, and the equations are set up as follows: R o u n d loop h 8/i + 2 / 2 + 3 / 3 + 3 / 4 - 2 x 5 = 0, ^ R o u n d loop I2
211 + 9 / 2 + 5 / 3 + 2 / 4 - 5 x 5 = 0,
R o u n d loop h
3/i + 5 / 2 + 8 / 3 4- 3 / 4 - 5 x 5 = 0,
(4.26)
R o u n d loop / 4 3/i + 2 / 2 + 3 / 3 + 7 / 4 - 2 x 5 = 0. The unknowns are best found by matrix m e t h o d s , and the result is: /, - 0.04619 A , 12 = 1.5935 A , 13 = 2.0901 A , / 4 = 0.05774 A . By combining these currents in the manners indicated in Fig. 4.4(b), the current in any branch may be calculated. For example, 7 AB = h + h + h = 0.04619 + 2.0901 + 0.05774 = 2.194 A which agrees with / AB in eqns. (4.25). T h e other currents may be found similarly. T h e n o d e potential differences can be found using O h m ' s Law. For example, = 4 x 1.5935 = 6.374 volts.
214
ELECTRIC CIRCUIT THEORY
In general, the m e t h o d chosen to solve a particular network should be that which gives the smaller n u m b e r of unknowns, as calculated from:
and
L = B - N + 1 N - 1
for the loop-current m e t h o d , for the node-voltage m e t h o d .
However, as Problem 4.3 shows, the n u m b e r of unknowns may be reduced by a judicious choice of links (or reference n o d e ) . 4.6 S T E A D Y - S T A T E S I N U S O I D A L
CONDITIONS
T h e application of source voltages and/or currents which are sinusoidal functions of time to a network composed of linear elements will always result in sinusoidal steady-state responses at every point in the network. Hence the m e t h o d s of analysis discussed in the present chapter are applicable to such networks providing complex phasors are substituted for the constant values hitherto used. PROBLEM
4.4
For the circuit of Fig. 4.5 determine the power delivered by the source. 25 a
FIG.
4.5.
Solution T h e r e are two principal nodes and three branches, hence B = 3, N 2. F o r the loop-current m e t h o d , L = 3 — 2 + 1 = 2 (unknowns). F o r the node-voltage m e t h o d , unknowns = N - 1 = 1. H e n c e , choose the latter. All data are frequency-domain quantities, hence the frequency is not required (it is implicit in the reactances). T h e reference n o d e is indicated in the diagram. Let V be the voltage at the other n o d e .
CIRCUIT ANALYSIS
215
Then
v 25
j20
10-J30/
- — •240 Z_0° = 0 25
or V(0.04 - jO.05 + 0.01 + J0.03) - 9.6 zL0° or V
9.6 Z_0°
9.6 Z.0°
- = 178.273 ^ 2 1 ° 4 8 ' 0.05385 ZL-21°48' —
0.05 - jO.02
= (165.522 + J66.209) volts. The current in the 25-Q branch is: 240 - 165.522 - J66.209 25
-(2.979 -j2.648)A.
So that the power delivered by the source, which is: P = 9fce[VP] - 240 x 2.979 = 714.989 W. P R O B L E M 4.5
For the circuit of Fig. 4.6, determine the current in each branch and the power delivered to the circuit by each source.
FIG.
4.6.
Solution In this circuit, B = 5 and N = 3, hence L =5 - 3 + 1 = 3. However, if the three loops (meshes in this case) are chosen as shown, the current h is already known—thus reducing the unknowns to 2. The node-voltage m e t h o d would also produce two unknowns (and two equations) but since we require the branch currents we choose the loop-current m e t h o d .
216
ELECTRIC CIRCUIT THEORY
Let Ii, I 2 and I 3 be the three loop currents, as shown in Fig. 4.6. Then, I 2(23 + j l 5 - j l 2 ) - 30 ^ 2 0 ° - I 3(8 + j l 5 ) = 0
(4.27)
-I 2(8 + J15) + I 3( 8 + j5) = - 6 0 Z45°
(4.28)
and
since
Ii = 3 ^ 2 0 ° . Solving for I 2 by C r a m e r ' s Rule:
I2 =
30^20° - 6 0 zL45°
- ( 8 + J15) (8 + j5)
23 + J3 - ( 8 + J15)
- ( 8 + J15) (8 + J5)
0
30Z.20 - 6 0 Z45°
- 1 7 ^61°56' 9.434 Z.32°
23.195 /J°26' -17
-17^61°56' 9.434 ^ 3 2 °
283.019 /L52° - 1020 ^ 1 0 6 ° 5 6 ' 218.819 ^ 3 9 ° 2 6 ' - 289 ^ 1 2 3 ° 5 2 ' 2.573 Z . - 4 0 ° 5 6 ' A = (1.944 - jl.686) A . Likewise for h (23 + j3) -(8 + J15) (23 + j3) -(8 + J 1 5 )
30 Z.20° -60-/45° I - ( 8 + J15) (8 + j5)
23.195 z!7°26' - 1 7 Z61°56' 23.195 Z.7°26' -17zL61°56'
30 ^ 2 0 ° - 6 0 ^45° -17Z61°56' 9.434 Z32°
CIRCUIT ANALYSIS
217
= 2.841 Z . - 1 2 5 ° 2 3 ' A = (-1.645 - j 2 . 3 1 6 ) A . Current in the 10-Q resistor = Ii — I 2 = 3 Z 2 0 ° - 1.944 + j l . 6 8 6 A = 2.819 + j l . 0 2 6 - 1.944 - j l . 6 8 6 A = (0.875 + J2.712) A . C u r r e n t in the (8 + j l 5 ) - Q impedance = h -
h
= 1 . 9 4 4 - j l . 6 8 6 + 1.645 + J2.316A = 3.589 + J 0 . 6 3 0 A .
0
C u r r e n t in t h e J20-Q impedance = 3 Z.20 A . C u r r e n t in t h e (5 - j 12)-Q impedance = I 2= 2.573 Z - 4 ( ) ° 5 6 ' A . C u r r e n t in the - j l O - Q impedance = I 3= 2.841 Z.—125°23' A.
Power from each source (a) Current source. We first calculate the voltage across the source. Voltage d r o p across the 10-Q resistor = 10(0.875 + J2.712) volts = 8.75 + j27.12 volts. Voltage d r o p across the J20-Q reactor = 3z.20°.20z.90°volts = 60 z. 110° volts = - 2 0 . 5 2 1 + J56.382 volts.
218
ELECTRIC CIRCUIT THEORY
Voltage across the source = (8.75 - 20.521) + j(56.382 + 27.12) volts = - 1 1 . 7 7 1 + j83.502 volts = 84.328 Z.90°l' volts. A n d power delivered
o = 9fte[VI*] = Sfte[84.328 Z98° 1'.3 z l - 2 0 ] W = 2fte[252.984 z.78° 1'] = 52.529 W. (b) Voltage source. T h e current through the source in the same sense as its e.m.f. is ( — 1 3 ) . H e n c e , power = & e [ V r ] = 0 t e [ V ( - I 3) * ] = 2fte[60 zl45° ( - 2 . 8 4 1 ) Z.125°23'] W = 2fte[-170.460 Z.170°23'] - 168.065 W. 4.7
MUTUAL
INDUCTANCE
Mutual inductance is said to exist between two circuits when a changing current in one induces, by electromagnetic induction, an e.m.f. in the
L1
FIG.
L
z
4.7.
other. Since this implies mutual magnetic flux, which is a fraction of the self-flux set up by each circuit, each circuit must also possess selfinductance. The idealised equivalent circuit of a mutual inductor is therefore as shown in Fig. 4.7. L\ and L 2 are the self-inductances of the two circuits and M the mutual inductance between them. The mutual inductance M is defined by the relationships: e2(t)
=
A/12
d/i(0 dt
(4.29)
CIRCUIT ANALYSIS
or
di2(t) *i(0 •M2[ dt
219 (4.30)
where e2 is the e.m.f. in circuit 2 due to i{ in circuit 1, and e{ that in circuit 1 due to i2 in circuit 2. A s in eqn. (1.11), when the units of / and v are amperes and volts, respectively, the units of MX2 and M2{ are henrys. It may be shown* that Ml2 = M2\. 4.7A The dot notation The polarity of the induced e.m.f. due to mutual inductance is identified by placing a dot, on the diagram, adjacent to that end of each equivalent winding which bears the same relationship to the magnetic flux. T h u s , if a current in one winding changes so as to induce an e.m.f. positive at the dot end of that winding, an e.m.f. will be induced in the other winding which is also positive at its dot end. Using this notation, the mesh- or loop-current m e t h o d may be applied to circuits having separate parts coupled by mutual inductance. For steady-state sinusoidal conditions eqns. (4.29) and (4.30) may be transformed, in the m a n n e r used to derive eqn. (3.11), into and
PROBLEM
E 2 - jcvMh
(4.29a)
Ei - ]a)Ml2.
(4.30a)
4.6
A mutual inductor is used to couple a 10-Q resistive load to a 100-V generator, as shown in Fig. 4.8. T h e generator has an internal resistance
FIG.
L,
L
R,
R
4.8.
2 2
* See, for example, E. G. Cullwick, The Fundamentals of Electromagnetism,
C.U.P.
ELECTRIC CIRCUIT THEORY
220
of 5 Q and the parameters of the mutual inductor are: Lx = 0.1 H , # ! = 1 0 Q , L 2= 0 . 2 H , R2=15Q and M = 0 . 1 H . Calculate the generator and load currents for (a) the two windings having the same sense (as shown by the position of the dots in Fig. 4.8) and (b) the windings having the opposite sense. Given co = 100 rad/sec. Solution (a) Referring to the diagram there are two clear mesh currents, as shown. T h e n , for mesh 1: + ](oL ) - ](oM\ = 100 ZL0°
x
2
or 1.(15 + jlO) - j l 0 I 2 = 100 Z0°
(4.31a)
and for mesh 2: -]wMh or
+ h(Ri + 10 + }toL2) = 0
- j l O l ! + I 2(25 + J20) = 0.
(4.32a)
In e q n . (4.31a) the term ]coMl2 is negative in mesh 1, since the flux produced by I 2induces an e.m.f. in L2 away from the d o t , which means that it induces an e.m.f. in Lx also away from the dot—i.e. in the same direction as Ii. This is in the same sense as the 100-V source and is therefore positive on the R . H . S . of eqn. (4.31a) or negative on the L . H . S . Similarly, in e q n . (4.32a) the ]ioM\\ term can be thought of as being the sole e.m.f. in mesh 2 and hence responsible for the current. It is therefore positive on the R . H . S . of e q n . (4.32a) or negative on the L.H.S. Solving eqns. (4.31a) and (4.32a) by Cramer's Rule:
= 1.626 z^26°34' A .
CIRCUIT ANALYSIS
221
And
Ii =
100 0
-J10 (25 + J20)
2500 + J2000
275 + j550
275 + j550
3201.562 Z.38°40' 614.919 Z.63°26' = 5.206 / . - 2 4 ° 4 6 ' A . (b) If the secondary winding (that feeding mesh 2) is reversed, so that the dots a p p e a r at opposite ends of the windings, the two mesh currents can still be applied clockwise. T h e n , eqns. (4.31a) and (4.32a) b e c o m e : 1,(15 + jlO) + j l 0 I 2 = 100
(4.31b)
+ j l 0 I i + I 2(25 + j20) = 0,
(4.32b)
and
giving the solutions: I, = 5.206 L-24°46'
A
and I 2 = 1.626 Z . - 1 5 3 ° 2 6 ' A . I 2 is 180° out of phase with the solution for part (a) since the winding has been reversed. However, I, is unchanged, since I 2 always flows in that direction which opposes the flux set up by I f. H e n c e , the winding direction is immaterial. However, if a mutual inductor forms one of several c o m m o n impedances between two meshes of a circuit, reversal of either winding will affect both primary and secondary currents. PROBLEM
4.7
T h e mutual inductor of Problem 4.6 is connected in the circuit of Fig. 4.9. Calculate the source and load currents for (a) the windings as shown (dots adjacent), (b) one winding reversed (dots at opposite ends). a) = 100 rad/sec.
ELECTRIC CIRCUIT THEORY
222
5 ii
.
M .
10 ii
FIG.
4.9.
Solution (a) Using the two mesh currents, as shown, the mesh equations are: Mesh 1 Ii(20 + J15) - j l O I 2 - I 2(5 + j5) = 100 Z.0°
(4.33a)
1,(20 + j l 5 ) - I 2(5 + j l 5 ) = 100.
(4.33b)
- J 1 0 I , - 1,(5 + j5) + I 2(30 + J25) = 0
(4.34a)
- 1 , ( 5 + j l 5 ) + I2(30 + J25) = 0.
(4.34b)
or
Mesh 2
or
Solving eqns. (4.33b) and (4.34b) by Cramer's Rule:
I, =
100 0
- ( 5 + jl5) (30 + j25)
(20 + J15) -(5 + J15)
- ( 5 + J15) (30 + J25)
= 4.311 Z . - 2 2 ° 1 3 ' A
3000 + J2500 425 + j800
CIRCUIT ANALYSIS
223
and
:
(20 + J15) -(5 + J15)
100 0
500 + J1500 425 + J800
425 + J800 = 1.745
ZL9°33'A.
(b) W h e n one of the windings of the mutual inductor is reversed, the ]coMl term in eqns. (4.33a) and (4.34a) changes sign but not those due to the voltage drop in the (5 + j5)-Q impedance. H e n c e , the two equations become:
o
Ii(20 + j l 5 ) - I 2(5 - j5) - 100 Z 0
(4.35)
- I i ( 5 - j5) + I 2(30 + j25) - 0,
(4.36)
and
whence
o l! = 3 . 8 1 0 z i - 3 7 3 r A
and
o I 2 = 0.690 z i - 1 2 2 1 9 ' A .
4.8 C I R C U I T T H E O R E M S
4.8.1 The superposition theorem T h e t h e o r e m is the most general expression of the linearity of a circuit or network. It states that the current in or voltage across any branch of a linear network containing m o r e than one source is the algebraic sum of the separate currents or voltages (as the case may be) due to each source acting alone, all other sources being temporarily set to zero. The t h e o r e m is useful in comparatively simple circuits, but containing several sources, and in which one branch is of particular interest. It enables the contribution of each source to the current or voltage in the branch to be assessed readily.
ETC - P
224 PROBLEM
ELECTRIC CIRCUIT THEORY 4.8
For the circuit of Fig. 4.10 d e t e r m i n e , using the superposition t h e o r e m , the current I and the component of I due to each source.
FIG. 4 . 1 0 .
Solution Consider the circuit with Ei acting alone, source E 2 being replaced by a short circuit. T h e n Ei acts upon the impedance of branch 1 in series with a parallel combination of branches 2 and 3. T h u s , the current through Z 2 due to Ei is:
Now consider E 2 acting alone with Ei replaced by a short circuit. I acts upon Z 2 in series with a parallel combination of Zi and Z 3. Henc( the current through Z? due to E? is:
CIRCUIT ANALYSIS
225
O r the current through Z 2 due to Ei is m o r e than 7 times greater than that due to E 2. I = (Ii + I 2) = 0.0862 - J4.628 - 0.181 + jO.629 A = 4.000 Z . - 9 1 ° 2 2 ' A . 4.8.2 Thevenin's and Norton's theorems W h e n interest is focused on one particular branch of a circuit or network containing active and passive elements, the theorems of Thevenin and N o r t o n can often provide a much simpler solution than either loop-current or node-voltage analysis. The theorems achieve this by replacing the whole of the rest of the network—viewed from the branch—by a voltage source in series with an impedance (Thevenin's t h e o r e m ) or by a current source in parallel with an impedance (Norton's t h e o r e m ) . T h e e.m.f. of the voltage source is equal to the open circuit p.d. at the terminals of the network to which the branch was connected, and the series impedance is equal to the impedance looking into the opencircuited terminals with all voltage sources in the network replaced by short circuits and all current sources replaced by open circuits. The result is called the Thevenin Equivalent G e n e r a t o r . For N o r t o n ' s t h e o r e m , the current of the current source is equal to the short-circuit current which would flow from the network at the terminals to which the branch was connected and the shunting impedance is defined in exactly the same way as for Thevenin's t h e o r e m . T h e result is called the N o r t o n Equivalent G e n e r a t o r . T h e theorems are therefore duals of each other and either could be deduced from the other by the source transformation process. PROBLEM
4.9
A 600/150-V volt-ratio box having a total resistance of 30 k Q is used in the m e a s u r e m e n t of the voltage of a d.c. source. T h e whole box is connected across the source and a voltmeter is applied to the 150-V tapping. Using Thevenin's t h e o r e m , calculate the lowest resistance which the voltmeter may have so that the m e a s u r e m e n t error shall not exceed 0.1%.
ELECTRIC CIRCUIT THEORY
226
Solution T h e volt-ratio box is a 30,000-Q resistor with a tapping at: 30,000 x ^ = 7500 Q 600
(see Fig. 4.11(a)).
The Thevenin equivalent of the circuit seen from the voltmeter terminals is as shown in Fig. 4.11(b), where R' is given by: R' =
7500 x 22,500 30,000
5625 Q.
R'=5625 G A
Voltmeter Voltmeter
(b)
(a) FIG. 4 . 1 1 .
If R were infinite, n o error would result, but for finite R a current is drawn through R' and the voltage at terminals A B falls. H e n c e , for 0 . 1 % error: R
(VIA)
R + 5625
- (1/1000)
-(VIA)
(VIA)
5625
1
R
999
1
999
1000
1000
giving R = 5,619,375 Q.
P R O B L E M 4.10
A simple d.c. slide-wire potentiometer is used to measure the voltage of a source, the circuit arrangement being as shown in Fig. 4.12(a). T h e slide wire current is 50 m A and its 100-cm length, A B C , has a volt-drop across it of 1.5 V. Balance (indicated by zero galvanometer current) is
227
CIRCUIT ANALYSIS
(a)
(b) FIG. 4 . 1 2 .
achieved when the slider at B is 20 cm from C. Using Thevenin's t h e o r e m calculate the galvanometer current when the slider is moved 1 m m , towards C. Solution A m o v e m e n t of the slider by 1 m m , to B ' , represents an out-of-balance voltage of 1.5 m V (since 100 cm represents 1.5 V ) . H e n c e , from the viewpoint of the galvanometer circuit, the e.m.f. of the Thevenin equivalent generator is 1.5 m V and its internal resistance is the resistance of B ' C shunted by 50 Q and A B ' in series. T h e resistance of B ' C is: 19.9
1.5
100 0.05 and that of A B ' is:
_ _ 5.97 Q
L5 0.05
- 5.97 = 24.03 Q
so that R' (see Fig. 4.12(b)) is given by:
Hence
228
ELECTRIC CIRCUIT THEORY
In fact there would have been negligible error involved had it been assumed that the internal resistance of the Thevenin equivalent generator does not change for this small movement of the slider. PROBLEM
4.11
Calculate, using (a) Thevenin's t h e o r e m , (b) Norton's t h e o r e m , the current in the branch A B of the circuit represented in Fig. 4.13(a). 3ii
j 6 a
A
-j 20
a
-o F i g . 4.13. (a) Diagram of the circuit for Problem 4.11.
j 20
a
(240 + j 120) V
Fig. 4.13. (b) The branch AB removed and the current source replaced by its equivalent voltage source.
6 . 6 6 7 - j 13.333 A
j 12 A
F i g . 4.13. (c) The branch AB removed and the voltage source replaced by its equivalent current source.
CIRCUIT ANALYSIS
229
Solution R e m o v e the branch A B , short-circuit the voltage source and opencircuit the current source. T h e internal impedance of the circuit viewed from A B is that of (3 + j6) Q and (10 - j20) Q in parallel, or Z
= (5.342 + J5.753) Q = 7.851 ^ 4 7 ° 7 ' Q. (a) By Thevenin's
theorem
First change the current source shunted by (10 - j20) Q into a voltage source of e.m.f.: E - j l 2 ( 1 0 - j20) = (240 + j l 2 0 ) V in series with (10 - j20) Q. T h e circuit is now as shown in Fig. 4.13(b). The circulating current is: I.- ^ 5
- (7.507
+, 1 7 . 3 . 5 ) A .
A n d the open-circuit voltage at A B is: 100 Z0° - (7.507 + J17.315) (3 + j6) V = (181.370 - J96.986) V. H e n c e , the current through the branch A B , on replacing it, is:
(b) By Norton's
theorem
First change the voltage source in series with (3 + j6) Q to a current source of generated current I Nshunted by (3 + j6) Q , as shown in Fig. 4.13(c), In
100 3 + J6
(6.667 - j ! 3 . 3 3 3 ) A.
230
ELECTRIC CIRCUIT THEORY
If terminals A B are short-circuited, t h e c u r r e n t which flows, which is also t h e g e n e r a t e d c u r r e n t of t h e N o r t o n equivalent s o u r c e , is: (6.667 - j'13.333) - j l 2 = (6.667 - J25.333) A and this is s h u n t e d by t h e i m p e d a n c e looking into A B — w h i c h is: (5.342 + J5.753) Q. H e n c e , t h e c u r r e n t t h r o u g h the b r a n c h A B , w h e n r e p l a c e d , is: ;6.667 - J25.333) A - 15.112 Z - 8 0 ° 2 0 ' A
the s a m e result as was o b t a i n e d using T h e v e n i n ' s t h e o r e m .
PROBLEMS IV
1. A Wheatstone-bridge circuit whose four arms have resistances of 3, 4, 5 and 6 Q (when taken in cyclic order) is fed by a 5-A current source connected between opposite corners. Calculate (a) the voltage across the other opposite corners, (b) the current taken by a 10-Q resistor connected to these corners. 2. For the circuit represented by Fig. P.IV.l, calculate the current through and the power absorbed by the 10-Q resistor. 3. Calculate the voltage appearing between the terminals A and B in the circuit represented by Fig. P.IV.2. Use the node-voltage rule. 4. For the circuit of Fig. P. IV.3, calculate the current in the 6-Q resistor by (a) Kirchhoff s Laws, (b) the superposition theorem. 5. Using Thevenin's theorem, calculate the current through the 10-Q resistor in the circuit of Fig. P.IV.4. 6. Deduce the conditions under which maximum power is delivered to a load of resistance R by a generator of e.m.f. E and internal resistance r. LFor the circuit of Fig. P.IV.5, calculate the resistance R for which maximum power is absorbed and the magnitude of this power. 7. For the circuit of Fig. P.IV.6, containing current-controlled voltage sources, calculate the mesh currents i i and / . u2 3 8. Use the node-voltage rule to obtain the currents in all the branches of the circuit of Fig. P.IV.7. 9. In the circuit represented by Fig. P.IV.8, the switch S is arranged to short-circuit half the 12-Q resistor. When S is closed the current through its contacts is 2 A. Calculate the current through the 12-Q resistor when S is open. 10. Define the terms graph, tree, link-branch in network topology. For the circuit of Fig. P.IV.9, draw the graph and choose a tree which allows all branch currents to be evaluated with just one equation. Calculate the branch currents and the power delivered by each source.
CIRCUIT ANALYSIS
231 3A
2a
FIG. P.IV.l (Problem IV.2). 3A
O A
5ft FIG. P.IV.2 (Problem IV.3). 4ft
10ft
8ft
FIG. P.IV.3 (Problem IV.4). 4ft
6ft
—I
I
f
FIG. P.IV.4 (Problem IV.5).
24 ft
FIG. P.IV.5 (Problem IV.6).
1
232
CIRCUIT ANALYSIS 3&
233 i,
FIG. P.IV.10 (Problem IV.11).
A
^
j 20
I
a io
30 a
is
aI
J
FIG. P.IV.ll (Problem IV.12).
FIG. P.IV.12 (Problem IV.13).
FIG. P.IV.13 (Problem IV.14).
a
234
ELECTRIC CIRCUIT THEORY
11. For the circuit of Fig. P.IV. 10, draw the graph and choose a suitable tree for evaluating all branch currents. Calculate each branch current and hence determine the power delivered by the 6-A source. 12. For the frequency-domain circuit of Fig. P.IV. 11, determine the voltage V (magnitude and phase) such that no current flows through the 18-Q resistor. 13. Determine the voltage—in polar form—across the source and the power delivered by the source in the circuit of Fig. P.IV. 12. Determine the Thevenin equivalent generator at the terminals AB and hence the current in the 10-Q resistor. 14. Determine the Thevenin equivalent generator for the network represented by Fig. P.IV. 13. Then apply a source-transformation to obtain the Norton equivalent generator. 15. In the bridge circuit represented by Fig. P.IV. 14, a voltage source is connected to D and B. Derive expressions for the values of R and L, in terms of r, C, C R and co, which u x cause the points A and C to be at the same potential (i.e. to balance the bridge).
C
A
B
D
C
FIG. P.IV.14 (Problem IV. 15). 16. For the bridged-T network of Fig. P.IV.15, derive the conditions for which the voltage across the terminals A and B is zero.
R
-i—y
FIG. P.IV.15 (Problem IV. 16).
CIRCUIT ANALYSIS
235
17. The circuit represented by Fig. P.IV. 16 is supplied by the voltage source for which V = 100 Z.0 at co = 1000 rad/sec. Calculate the current Ii, in polar form. M=O.OIH
FIG. P.IV.16 (Problem IV.17).
CIRCUIT ANALYSIS 4.0
INTRODUCTION
The process of circuit analysis consists of determining the currents in, and potential differences across, all branches of a network or circuit. The data normally comprise all source voltages and/or currents and all branch resistances/impedances. O h m ' s and Kirchhoff's Laws form the basis of all methods of analysis, either directly or in modified form, according to circumstances. Thus, in circuits containing only voltage sources, mesh or loop analysis is generally to be preferred to Kirchhoff's Laws themselves since fewer simultaneous equations result and, when only current sources are present, nodal analysis is generally best. In addition, many network theorems exist which, although not strictly falling within the above definition of circuit analysis, are nevertheless useful aids in the process. A few of these will be dealt with in this chapter. 4.1 N E T W O R K
TERMINOLOGY
First a few definitions. (a) A branch of a circuit is that part of the circuit containing a single active or passive element. (b) A node is the point of interconnection of two or more branches. The junction of three or more branches is referred to as a principal node; a secondary node is the junction of two branches. 200
CIRCUIT ANALYSIS
201
(c) A planar network is one whose schematic may be drawn on a plane surface without branch crossings. (d) A non-planar network is one whose schematic may not be so drawn. (e) A loop is any closed path in a network which does not pass through any branch or node more than once. (f) A mesh is a loop which contains no other loops. It is defined only for planar networks.
4.2 KIRCHHOFF'S L A W S
The reader is referred to section 1.3 for a statement of the laws, which are valid for both planar and non-planar networks. A n unknown current is assigned to every branch and the first law applied at as many nodes as will provide independent equations. The second law is then applied to as many independent loops as are necessary to form, in total, the same n u m b e r of simultaneous equations as there are unknowns. They are solved for the branch currents by substitution, determinants or matrix methods. All branch voltages are then easily found by applying O h m ' s Law. With simple circuits it is relatively easy to choose nodes and loops which yield independent equations. A rule of t h u m b , satisfactory in simple cases, is to choose each new node or loop so as to involve at least one branch current not already involved, either explicitly or implicitly. For more complicated circuits some care is needed and for guaranteed independence of the equations, the methods of section 4.5 must be used. The method of application of the laws is most readily understood by dealing initially with steady (direct current) conditions. P R O B L E M 4.1
Calculate, using Kirchhoff's Laws, all the branch currents and voltages in the network represented by Fig. 4 . 1 . Solution The circuit has six branches and four nodes and we begin by assigning six unknown currents, one to each branch (Fig. 4.1(a)).
202
250
ELECTRIC CIRCUIT THEORY
a
250 &
FIG. 4.1. (a) Diagram of the network for Problem 4.1.
250
a
FIG. 4.1. (b) The network for Fig. 4.1(a) with the number of unknowns reduced to three.
95 a
FIG. 4.1. (c) Analysing the circuit of Fig. 4.1(a) using the loop-current method.
Applying the first Law at node A :
/i = / + 2
h-
(4.1)
A t node B : A t node D :
h - U = /6.
(4.2)
h + h = /5-
(4.3)
CIRCUIT ANALYSIS
203
This is the limit of i n d e p e n d e n t equations obtainable from the first Law since, at n o d e C, the equation would be: h + h = /,
(4.4)
which is already obtainable from eqns. (4.1), (4.2) and (4.3). Applying the second Law to loop (or mesh) A B D , taking the clockwise direction as positive: 2 5 0 / 2 + 1200/4 - 2OOO/3 = 0.
(4.5)
h and U flow clockwise round the loop and / 3, anticlockwise (and hence is negative in eqn. (4.5)) whilst the total e.m.f. in the loop is zero. Similarly, in loop (or mesh) B C D : 9 5 / 6 - IOOO/5 - 1200/4 = - 0 . 1 .
(4.6)
In the loop containing A D C and the 6-V source: 2OOO/3 + 1000/ 5 = 6.
(4.7)
Solving eqns. (4.1), (4.2), (4.3), (4.5), (4.6) and (4.7) gives: A = 19.11mA,
/ 4= - 0 . 1 4 m A ,
h = 17.06 m A ,
/ 5= 1.91mA,
/ 3 = 2.05 m A ,
/ 6= 17.20 m A .
O h m ' s Law is used to calculate the potential differences for all branches. Thus: VAB = 2 5 0 / 2
= 4.265 V ,
VAD = 2OOO/3
= 4.100 V,
K BD = 1200/ 4
= - 0 . 1 6 8 V,
VDC = 1000/ 5
= 1.91 V,
VBC = 9 5 / 6+ 0.1 = 1.734 V.
204
ELECTRIC CIRCUIT THEORY
4.2.1 Reducing the number of unknowns The computational work may be reduced considerably, for this problem, by a more skilful application of the two Laws at the outset. T h u s , referring to Fig. 4.1(b), if at node A , /i is m a d e to divide into i2 and (i\ - / 2) , three branches have been covered with the introduction of only two u n k n o w n currents. Similarly at node B , if / 3 flows along B D , then (i2 - / 3) flows along B C , and (i\ - i2 + / 3) flows along D C . N o d e C adds no further information (as before) since (i2 - / 3) and (i\ - i2 + 13) combine to give i\, A total of only three unknown currents has been introduced for the whole circuit and only three independent equations are required for their solution. Using the same loops as before: 250/ + I2OO/3 - 2000(/! - i ) = 0,
(4.8)
95(i2 - i 3) - 1000(/i - i2 + i 3) - 1200/3 = - 0 . 1
(4.9)
2000(/j - / 2) + 1000(/i - i2 + / 3) = 6
(4.10)
2
2
and
which may be rearranged as follows: 2000/j - 2250/ 2 - I2OO/3 = 0,
(4.8a)
1000/, - 1095/2 + 2295/3 = 0 . 1 ,
(4.9a)
3000ii ~ 3000/2 + IOOO/3 = 6
(4.10a)
and subsequently solved for i u i2 and / 3. T h e current in any branch may be found by combining, as appropriate, the values for i u i2 and / 3. 4.3 T H E L O O P - A N D M E S H - C U R R E N T M E T H O D S
Either of these m e t h o d s replaces both of Kirchhoff s Laws by one rule, and is therefore generally simpler to apply in practice. T h e two methods differ only in the choice of closed path in a network round which the total voltage d r o p is equated to the e.m.f. In the mesh-current m e t h o d t h e paths are t h e obvious simple meshes and t h e method is therefore restricted to planar networks. T h e loop-current m e t h o d is
CIRCUIT ANALYSIS
205
m o r e general in that it applies to both planar and non-planar networks, but, as with the loops chosen for Kirchhoff's second Law, care must be taken to ensure a sufficient n u m b e r of independent equations in order to solve for all the u n k n o w n currents. In either m e t h o d a circulating current is associated with each selected closed p a t h . O n e particular direction is arbitrarily chosen as positive and the total voltage d r o p round each path is calculated by taking the algebraic sum of the products of (a) the path circulating current and the total path resistance, and (b) such other circulating currents (from adjoining paths) as impinge upon the chosen path and the resistances through which they flow. T h e result is equal to the total e.m.f. acting r o u n d the p a t h . Referring to Fig. 4.1(c), three possible loops (which in this case are meshes) are chosen, as shown, and the currents / A, h and IC associated with t h e m . R o u n d loop A : / A( 2 5 0 + 1200 4- 2000) - 1 2 0 0 / B- 2 0 0 0 / c = 0 or 3450/A - 1200/B - 2 0 0 0 / c = 0.
(4.11)
Comparing Figs. 4.1 (b) and (c), clearly for the m e t h o d to be valid, IA = h, h = ( ' 2 ~ *3) and IC = / i - Substituting these into eqn. (4.11) gives: 3450/2 ~ 1200/ 2 + I2OO/3 - 2000/, = 0 or 2000/! - 2250/ 2 - I2OO/3 = 0
(4.12)
which is seen to be identical with eqn. (4.8a), and the m e t h o d is therefore justified by Kirchhoff's Laws. R o u n d loop B : 2295/B - 1 2 0 0 / A- 1000/C = - 0 . 1 .
(4.13)
3000/C - 2 0 0 0 / A- 1000/B - 6.
(4.14)
R o u n d loop C:
ELECTRIC CIRCUIT THEORY
206
The three loop equations (4.11), (4.13) and (4.14) are solved to give: 7 I
A = 17.06 m A ,
B = 17.20 m A ,
/ c = 19.11 m A and the branch currents calculated as appropriate. Thus: IAB
= 17.06 m A ,
^BD
= (IA - h)
IAD
= (/c - IA) = 2.05 m A ,
IDC
= (/c -
ICA
= 19.11 m A ,
/B)
=
=
7
BC = 17.20 m A ,
-0.14 mA,
1.91 m A ,
i.e. the same result as for Kirchhoff s Laws. O t h e r loops might have been chosen—for the network of Fig. 4.1 there are seven such possible. They cannot all be independent as there are only three u n k n o w n currents, but in a circuit as simple as that of Fig. 4.1 the most obvious choices are the three mesh currents. 4.4 T H E N O D E - V O L T A G E M E T H O D
O n e principal node in the given network is chosen as a reference node and an unknown p . d . (with respect to the reference node) is assigned to every other principal n o d e . Kirchhoff s first Law is applied to each node in turn, each branch current meeting at a node being written as the product of the branch voltage and the branch conductance. If a current source feeds the node it must, of course, be included in the Kirchhoff Law summation. The method is the exact dual of the loop-current method and is valid for both planar and non-planar networks. P R O B L E M 4.2
Calculate, by the node-voltage m e t h o d , all branch currents and node potential differences for the circuit represented by Fig. 4 . 1 .
CIRCUIT ANALYSIS
207
Solution The circuit diagram is redrawn in Fig. 4.2, and it will be seen that node C has been chosen as the reference n o d e . Since VA (with respect to the reference) is 6 V , as determined by the voltage source, this leaves only two node voltages to be determined. B
FIG. 4.2. Analysing the circuit of Fig.4.1(a) using the node-voltage method (see Problem 4.2).
Designating these VB and VD , we apply Kirchhoff's first Law to n o d e D: Vn-6\
. /VD Vn\ 2000 / \ 1200 /
Vn-0 1000
(4.15)
Each term in t h e equation is a current flowing away from the n o d e —calculated by multiplying a branch conductance by the potential difference between the ends of the branch. Since there is no current source connected to the n o d e , their sum is equal to zero. Similarly, at n o d e B : V B- 6 \ . /VB-VD \ . 250 / V 1200 ) \
/Vu-0.V 95
(4.16)
There is no need to include the secondary node E in the analysis si
ETC - O
ELECTRIC CIRCUIT THEORY 208 _ (VB VE) is uniquely determined by the 0.1 V source, and, as soon as VB is found from (4.15) and (4.16), VE follows. Rearranging (4.15) and (4.16) gives:
(4.15a)
and (4.16a) an arrangement which suggests an alternative, more easily r e m e m b e r e d form of the node-voltage rule. A t each n o d e , calculate the product of the voltage (to the reference node) and the sum of all conductances meeting at the n o d e . For every other node connected to the chosen node through one conductance, calculate the product of its voltage and the conductance and subtract from the first product. T h e result is equal to the total current flowing into the n o d e . Equations (4.15a) and (4.16a) reduce to: 2 3 . 3 3 3 V D- 8.333KB = 30
(4.15b)
- 8 . 3 3 3 V D+ 153.597KB = 250.526
(4.16b)
and
which may be solved to give: VD = 1.905 volts
and
VB = 1.734 volts.
T h e other potential differences follow: yAB = 6 - 1.734 = 4.266 V;
VAD = 6 - 1.905 = 4.095 V,
VBD = 1.734 - 1.905 = - 0 . 1 7 1 V;
VEC = 1.734 - 0.1 - 1.634 V.
A n d the current in each branch calculated by O h m ' s Law:
/AD
CIRCUIT ANALYSIS
^BC ^DC
1.734 - 0.1 95
209
17.20 m A ,
1.905 m A , -0.171 1200
0.143 m A .
T h e network of Fig. 4.2 has four nodes, o n e being chosen as t h e reference. In general, therefore, three node voltages would be u n k n o w n and three equations would be required for their determination. This would have been the case if the 6-V source had been a current source. However, in cases where a voltage source is connected between two principal nodes, one of these nodes is chosen as the reference and this immediately determines the voltage of the other, thus reducing the number of unknown voltages by o n e . If more than o n e pair of principal nodes are connected to voltage sources, a solution by the node-voltage method, directly, is not possible. Source transformation or mesh (or loop) analysis must be used. 4.5 N E T W O R K
TOPOLOGY
A method of determining the correct n u m b e r of independent loops in a network consists of first drawing the network graph. This is a line diagram showing all principal nodes and branches of the original network but omitting the character of each branch. A s an example, the diagram of Fig. 4.3(a) shows the graph of Fig. 4.2. Branches are then removed from the graph so that all nodes remain interconnected but n o loops remain. T h e remaining diagram is called a TREE. For the graph of Fig. 4.3(a) there are several possible trees, three of which are shown by the full lines in 4.3 ( b ) , (c) and (d). T h e full lines are referred to as TREE BRANCHES and the dotted lines (which convert the tree to the original graph) as LINKS.
Clearly, no loop current can flow in a tree, but for each link added (in the process of reforming the graph from the tree) an extra loop current can flow. T h u s , in the tree of Fig. 4.3(b) the reinsertion of link B C would allow the loop A B C D to be formed; link B D would allow the loop A B D and link A C allows the loop A D C F . Each of these loops
ELECTRIC CIRCUIT THEORY
210
B
(d)
(c)
(b)
FIG. 4.3. (b-d) Three possible trees of the graph of Fig. 4.3(a).
is the path of an independent loop current. Obviously, as soon as each link current is known, all branch currents can be calculated from various linear combinations of the link currents. If a network has TV principal nodes, then the tree must have (TV - 1) branches, since the first branch connects two nodes and each subsequent branch adds one n o d e . If there are B branches altogether then L , the number of links, must be: L = B-(N-\)
= B-N+\
(4.17)
and this is the minimum n u m b e r of independent loop equations for solution of the original network. For the network represented by Fig. 4.3(a) B = 6 and /V = 4. H e n c e : L = 6 - 4 + l = 3 so that three equations are required. However, the trees of Fig. 4.3 (b), (c) and (d) show that at least three different sets of three equations are possible. In (b), the three loops containing the link currents are A B D A
CIRCUIT ANALYSIS
211
(which includes the link B D ) , A B C D A (link BC) and A D C F A (link A C ) . In (c) the three loops are A B D A , B C D B and A B D C F A ; and in (d) the three loops correspond to the three meshes chosen for Fig. 4.1(c). Several others are possible. T h e choice of suitable loops can also depend on the type and n u m b e r of energy sources in the network. Thus, for a network containing current sources the tree should be chosen so that, as far as possible, the current sources appear as links, so reducing the n u m b e r of unknown currents to less than that given by eqn. (4.17). P R O B L E M 4.3
Calculate, by both the node-voltage and the loop-current methods, all branch currents and node potential differences for the non-planar network represented in Fig. 4.4(a). Solution (a) Node-voltage method. For the given network N = 4, hence a sufficient n u m b e r of equations to solve for the unknown voltages is three. Choosing node C as the reference, the three equations are set up as follows: For node A : (4.18) ( 1 + i + i) VA~ V -hV =5
B
D
B
5A F i g . 4.4. (a) Diagram of the network for Problem 4.3.
F i g . 4.4. (b) Tree for the network of Fig. 4.4(a).
ELECTRIC CIRCUIT THEORY
212
At node B : ( l + i + i ) V B- V A- ( i + J ) V D= 0.
(4.19)
(i + i + i + i) vD - ivA - (j + i ) vB = o.
(4.20)
At node D :
These m a y b e rearranged as follows: IVA - 4VB - 2 K D= 20,
(4.18a)
- 2 0 V A + 3 9 K B- 9V D= 0,
(4.19a)
-30VA - 27VB + 11VD= 0
(4.20a)
and solved by substitution, determinants o r matrix methods t o give V
A
V and
V
B
= 6 . 3 7 4 1 volts,
(4.21)
= 4 . 1 8 0 1 volts,
(4.22)
D=
(4.23)
3 . 9 4 9 2 volts.
Thus, t h e voltages between nodes a r e : V and
AB =
V
A
-
V
A
-
V
VAD = V
BD — VB —
V
B
= 2 . 1 9 4 0 volts,
(4.24)
2 . 4 2 4 9 volts,
D=
Vo = 0 . 2 3 0 9 volts,
enabling t h e currents to be calculated, by O h m ' s Law, as follows: 2.194 1
he
=^4^
2
2.4249 »AD
2.194 A ,
0.2309 ^BD — '
4
6.3741
= 2.0901 A ,
/ AC=
= 1.2125 A ,
/ BD(through 5 Q )
= 0.0577 A , = 1.5935 A , 0.2309 0.0462 A ,
3 9492 /DC = ^ ^ = 1 . 3 1 6 4 A .
(4.25)
CIRCUIT ANALYSIS
(b) Loop-current N = 4,
213
method. T h e n u m b e r of branches, 5 , is 8 , and since L =
8
- 4 + 1
= 5. H e n c e five loop currents are n e e d e d , requiring five equations. W e begin by drawing the line graph of the network, and proceed by reducing to a tree as shown in Fig. 4.4(b). The links are shown dotted, as before, and, by making the current source one of the links, we can reduce the u n k n o w n currents to four. The loop currents are chosen, as shown, and the equations are set up as follows: R o u n d loop h 8/i + 2 / 2 + 3 / 3 + 3 / 4 - 2 x 5 = 0, ^ R o u n d loop I2
211 + 9 / 2 + 5 / 3 + 2 / 4 - 5 x 5 = 0,
R o u n d loop h
3/i + 5 / 2 + 8 / 3 4- 3 / 4 - 5 x 5 = 0,
(4.26)
R o u n d loop / 4 3/i + 2 / 2 + 3 / 3 + 7 / 4 - 2 x 5 = 0. The unknowns are best found by matrix m e t h o d s , and the result is: /, - 0.04619 A , 12 = 1.5935 A , 13 = 2.0901 A , / 4 = 0.05774 A . By combining these currents in the manners indicated in Fig. 4.4(b), the current in any branch may be calculated. For example, 7 AB = h + h + h = 0.04619 + 2.0901 + 0.05774 = 2.194 A which agrees with / AB in eqns. (4.25). T h e other currents may be found similarly. T h e n o d e potential differences can be found using O h m ' s Law. For example, = 4 x 1.5935 = 6.374 volts.
214
ELECTRIC CIRCUIT THEORY
In general, the m e t h o d chosen to solve a particular network should be that which gives the smaller n u m b e r of unknowns, as calculated from:
and
L = B - N + 1 N - 1
for the loop-current m e t h o d , for the node-voltage m e t h o d .
However, as Problem 4.3 shows, the n u m b e r of unknowns may be reduced by a judicious choice of links (or reference n o d e ) . 4.6 S T E A D Y - S T A T E S I N U S O I D A L
CONDITIONS
T h e application of source voltages and/or currents which are sinusoidal functions of time to a network composed of linear elements will always result in sinusoidal steady-state responses at every point in the network. Hence the m e t h o d s of analysis discussed in the present chapter are applicable to such networks providing complex phasors are substituted for the constant values hitherto used. PROBLEM
4.4
For the circuit of Fig. 4.5 determine the power delivered by the source. 25 a
FIG.
4.5.
Solution T h e r e are two principal nodes and three branches, hence B = 3, N 2. F o r the loop-current m e t h o d , L = 3 — 2 + 1 = 2 (unknowns). F o r the node-voltage m e t h o d , unknowns = N - 1 = 1. H e n c e , choose the latter. All data are frequency-domain quantities, hence the frequency is not required (it is implicit in the reactances). T h e reference n o d e is indicated in the diagram. Let V be the voltage at the other n o d e .
CIRCUIT ANALYSIS
215
Then
v 25
j20
10-J30/
- — •240 Z_0° = 0 25
or V(0.04 - jO.05 + 0.01 + J0.03) - 9.6 zL0° or V
9.6 Z_0°
9.6 Z.0°
- = 178.273 ^ 2 1 ° 4 8 ' 0.05385 ZL-21°48' —
0.05 - jO.02
= (165.522 + J66.209) volts. The current in the 25-Q branch is: 240 - 165.522 - J66.209 25
-(2.979 -j2.648)A.
So that the power delivered by the source, which is: P = 9fce[VP] - 240 x 2.979 = 714.989 W. P R O B L E M 4.5
For the circuit of Fig. 4.6, determine the current in each branch and the power delivered to the circuit by each source.
FIG.
4.6.
Solution In this circuit, B = 5 and N = 3, hence L =5 - 3 + 1 = 3. However, if the three loops (meshes in this case) are chosen as shown, the current h is already known—thus reducing the unknowns to 2. The node-voltage m e t h o d would also produce two unknowns (and two equations) but since we require the branch currents we choose the loop-current m e t h o d .
216
ELECTRIC CIRCUIT THEORY
Let Ii, I 2 and I 3 be the three loop currents, as shown in Fig. 4.6. Then, I 2(23 + j l 5 - j l 2 ) - 30 ^ 2 0 ° - I 3(8 + j l 5 ) = 0
(4.27)
-I 2(8 + J15) + I 3( 8 + j5) = - 6 0 Z45°
(4.28)
and
since
Ii = 3 ^ 2 0 ° . Solving for I 2 by C r a m e r ' s Rule:
I2 =
30^20° - 6 0 zL45°
- ( 8 + J15) (8 + j5)
23 + J3 - ( 8 + J15)
- ( 8 + J15) (8 + J5)
0
30Z.20 - 6 0 Z45°
- 1 7 ^61°56' 9.434 Z.32°
23.195 /J°26' -17
-17^61°56' 9.434 ^ 3 2 °
283.019 /L52° - 1020 ^ 1 0 6 ° 5 6 ' 218.819 ^ 3 9 ° 2 6 ' - 289 ^ 1 2 3 ° 5 2 ' 2.573 Z . - 4 0 ° 5 6 ' A = (1.944 - jl.686) A . Likewise for h (23 + j3) -(8 + J15) (23 + j3) -(8 + J 1 5 )
30 Z.20° -60-/45° I - ( 8 + J15) (8 + j5)
23.195 z!7°26' - 1 7 Z61°56' 23.195 Z.7°26' -17zL61°56'
30 ^ 2 0 ° - 6 0 ^45° -17Z61°56' 9.434 Z32°
CIRCUIT ANALYSIS
217
= 2.841 Z . - 1 2 5 ° 2 3 ' A = (-1.645 - j 2 . 3 1 6 ) A . Current in the 10-Q resistor = Ii — I 2 = 3 Z 2 0 ° - 1.944 + j l . 6 8 6 A = 2.819 + j l . 0 2 6 - 1.944 - j l . 6 8 6 A = (0.875 + J2.712) A . C u r r e n t in the (8 + j l 5 ) - Q impedance = h -
h
= 1 . 9 4 4 - j l . 6 8 6 + 1.645 + J2.316A = 3.589 + J 0 . 6 3 0 A .
0
C u r r e n t in t h e J20-Q impedance = 3 Z.20 A . C u r r e n t in t h e (5 - j 12)-Q impedance = I 2= 2.573 Z - 4 ( ) ° 5 6 ' A . C u r r e n t in the - j l O - Q impedance = I 3= 2.841 Z.—125°23' A.
Power from each source (a) Current source. We first calculate the voltage across the source. Voltage d r o p across the 10-Q resistor = 10(0.875 + J2.712) volts = 8.75 + j27.12 volts. Voltage d r o p across the J20-Q reactor = 3z.20°.20z.90°volts = 60 z. 110° volts = - 2 0 . 5 2 1 + J56.382 volts.
218
ELECTRIC CIRCUIT THEORY
Voltage across the source = (8.75 - 20.521) + j(56.382 + 27.12) volts = - 1 1 . 7 7 1 + j83.502 volts = 84.328 Z.90°l' volts. A n d power delivered
o = 9fte[VI*] = Sfte[84.328 Z98° 1'.3 z l - 2 0 ] W = 2fte[252.984 z.78° 1'] = 52.529 W. (b) Voltage source. T h e current through the source in the same sense as its e.m.f. is ( — 1 3 ) . H e n c e , power = & e [ V r ] = 0 t e [ V ( - I 3) * ] = 2fte[60 zl45° ( - 2 . 8 4 1 ) Z.125°23'] W = 2fte[-170.460 Z.170°23'] - 168.065 W. 4.7
MUTUAL
INDUCTANCE
Mutual inductance is said to exist between two circuits when a changing current in one induces, by electromagnetic induction, an e.m.f. in the
L1
FIG.
L
z
4.7.
other. Since this implies mutual magnetic flux, which is a fraction of the self-flux set up by each circuit, each circuit must also possess selfinductance. The idealised equivalent circuit of a mutual inductor is therefore as shown in Fig. 4.7. L\ and L 2 are the self-inductances of the two circuits and M the mutual inductance between them. The mutual inductance M is defined by the relationships: e2(t)
=
A/12
d/i(0 dt
(4.29)
CIRCUIT ANALYSIS
or
di2(t) *i(0 •M2[ dt
219 (4.30)
where e2 is the e.m.f. in circuit 2 due to i{ in circuit 1, and e{ that in circuit 1 due to i2 in circuit 2. A s in eqn. (1.11), when the units of / and v are amperes and volts, respectively, the units of MX2 and M2{ are henrys. It may be shown* that Ml2 = M2\. 4.7A The dot notation The polarity of the induced e.m.f. due to mutual inductance is identified by placing a dot, on the diagram, adjacent to that end of each equivalent winding which bears the same relationship to the magnetic flux. T h u s , if a current in one winding changes so as to induce an e.m.f. positive at the dot end of that winding, an e.m.f. will be induced in the other winding which is also positive at its dot end. Using this notation, the mesh- or loop-current m e t h o d may be applied to circuits having separate parts coupled by mutual inductance. For steady-state sinusoidal conditions eqns. (4.29) and (4.30) may be transformed, in the m a n n e r used to derive eqn. (3.11), into and
PROBLEM
E 2 - jcvMh
(4.29a)
Ei - ]a)Ml2.
(4.30a)
4.6
A mutual inductor is used to couple a 10-Q resistive load to a 100-V generator, as shown in Fig. 4.8. T h e generator has an internal resistance
FIG.
L,
L
R,
R
4.8.
2 2
* See, for example, E. G. Cullwick, The Fundamentals of Electromagnetism,
C.U.P.
ELECTRIC CIRCUIT THEORY
220
of 5 Q and the parameters of the mutual inductor are: Lx = 0.1 H , # ! = 1 0 Q , L 2= 0 . 2 H , R2=15Q and M = 0 . 1 H . Calculate the generator and load currents for (a) the two windings having the same sense (as shown by the position of the dots in Fig. 4.8) and (b) the windings having the opposite sense. Given co = 100 rad/sec. Solution (a) Referring to the diagram there are two clear mesh currents, as shown. T h e n , for mesh 1: + ](oL ) - ](oM\ = 100 ZL0°
x
2
or 1.(15 + jlO) - j l 0 I 2 = 100 Z0°
(4.31a)
and for mesh 2: -]wMh or
+ h(Ri + 10 + }toL2) = 0
- j l O l ! + I 2(25 + J20) = 0.
(4.32a)
In e q n . (4.31a) the term ]coMl2 is negative in mesh 1, since the flux produced by I 2induces an e.m.f. in L2 away from the d o t , which means that it induces an e.m.f. in Lx also away from the dot—i.e. in the same direction as Ii. This is in the same sense as the 100-V source and is therefore positive on the R . H . S . of eqn. (4.31a) or negative on the L . H . S . Similarly, in e q n . (4.32a) the ]ioM\\ term can be thought of as being the sole e.m.f. in mesh 2 and hence responsible for the current. It is therefore positive on the R . H . S . of e q n . (4.32a) or negative on the L.H.S. Solving eqns. (4.31a) and (4.32a) by Cramer's Rule:
= 1.626 z^26°34' A .
CIRCUIT ANALYSIS
221
And
Ii =
100 0
-J10 (25 + J20)
2500 + J2000
275 + j550
275 + j550
3201.562 Z.38°40' 614.919 Z.63°26' = 5.206 / . - 2 4 ° 4 6 ' A . (b) If the secondary winding (that feeding mesh 2) is reversed, so that the dots a p p e a r at opposite ends of the windings, the two mesh currents can still be applied clockwise. T h e n , eqns. (4.31a) and (4.32a) b e c o m e : 1,(15 + jlO) + j l 0 I 2 = 100
(4.31b)
+ j l 0 I i + I 2(25 + j20) = 0,
(4.32b)
and
giving the solutions: I, = 5.206 L-24°46'
A
and I 2 = 1.626 Z . - 1 5 3 ° 2 6 ' A . I 2 is 180° out of phase with the solution for part (a) since the winding has been reversed. However, I, is unchanged, since I 2 always flows in that direction which opposes the flux set up by I f. H e n c e , the winding direction is immaterial. However, if a mutual inductor forms one of several c o m m o n impedances between two meshes of a circuit, reversal of either winding will affect both primary and secondary currents. PROBLEM
4.7
T h e mutual inductor of Problem 4.6 is connected in the circuit of Fig. 4.9. Calculate the source and load currents for (a) the windings as shown (dots adjacent), (b) one winding reversed (dots at opposite ends). a) = 100 rad/sec.
ELECTRIC CIRCUIT THEORY
222
5 ii
.
M .
10 ii
FIG.
4.9.
Solution (a) Using the two mesh currents, as shown, the mesh equations are: Mesh 1 Ii(20 + J15) - j l O I 2 - I 2(5 + j5) = 100 Z.0°
(4.33a)
1,(20 + j l 5 ) - I 2(5 + j l 5 ) = 100.
(4.33b)
- J 1 0 I , - 1,(5 + j5) + I 2(30 + J25) = 0
(4.34a)
- 1 , ( 5 + j l 5 ) + I2(30 + J25) = 0.
(4.34b)
or
Mesh 2
or
Solving eqns. (4.33b) and (4.34b) by Cramer's Rule:
I, =
100 0
- ( 5 + jl5) (30 + j25)
(20 + J15) -(5 + J15)
- ( 5 + J15) (30 + J25)
= 4.311 Z . - 2 2 ° 1 3 ' A
3000 + J2500 425 + j800
CIRCUIT ANALYSIS
223
and
:
(20 + J15) -(5 + J15)
100 0
500 + J1500 425 + J800
425 + J800 = 1.745
ZL9°33'A.
(b) W h e n one of the windings of the mutual inductor is reversed, the ]coMl term in eqns. (4.33a) and (4.34a) changes sign but not those due to the voltage drop in the (5 + j5)-Q impedance. H e n c e , the two equations become:
o
Ii(20 + j l 5 ) - I 2(5 - j5) - 100 Z 0
(4.35)
- I i ( 5 - j5) + I 2(30 + j25) - 0,
(4.36)
and
whence
o l! = 3 . 8 1 0 z i - 3 7 3 r A
and
o I 2 = 0.690 z i - 1 2 2 1 9 ' A .
4.8 C I R C U I T T H E O R E M S
4.8.1 The superposition theorem T h e t h e o r e m is the most general expression of the linearity of a circuit or network. It states that the current in or voltage across any branch of a linear network containing m o r e than one source is the algebraic sum of the separate currents or voltages (as the case may be) due to each source acting alone, all other sources being temporarily set to zero. The t h e o r e m is useful in comparatively simple circuits, but containing several sources, and in which one branch is of particular interest. It enables the contribution of each source to the current or voltage in the branch to be assessed readily.
ETC - P
224 PROBLEM
ELECTRIC CIRCUIT THEORY 4.8
For the circuit of Fig. 4.10 d e t e r m i n e , using the superposition t h e o r e m , the current I and the component of I due to each source.
FIG. 4 . 1 0 .
Solution Consider the circuit with Ei acting alone, source E 2 being replaced by a short circuit. T h e n Ei acts upon the impedance of branch 1 in series with a parallel combination of branches 2 and 3. T h u s , the current through Z 2 due to Ei is:
Now consider E 2 acting alone with Ei replaced by a short circuit. I acts upon Z 2 in series with a parallel combination of Zi and Z 3. Henc( the current through Z? due to E? is:
CIRCUIT ANALYSIS
225
O r the current through Z 2 due to Ei is m o r e than 7 times greater than that due to E 2. I = (Ii + I 2) = 0.0862 - J4.628 - 0.181 + jO.629 A = 4.000 Z . - 9 1 ° 2 2 ' A . 4.8.2 Thevenin's and Norton's theorems W h e n interest is focused on one particular branch of a circuit or network containing active and passive elements, the theorems of Thevenin and N o r t o n can often provide a much simpler solution than either loop-current or node-voltage analysis. The theorems achieve this by replacing the whole of the rest of the network—viewed from the branch—by a voltage source in series with an impedance (Thevenin's t h e o r e m ) or by a current source in parallel with an impedance (Norton's t h e o r e m ) . T h e e.m.f. of the voltage source is equal to the open circuit p.d. at the terminals of the network to which the branch was connected, and the series impedance is equal to the impedance looking into the opencircuited terminals with all voltage sources in the network replaced by short circuits and all current sources replaced by open circuits. The result is called the Thevenin Equivalent G e n e r a t o r . For N o r t o n ' s t h e o r e m , the current of the current source is equal to the short-circuit current which would flow from the network at the terminals to which the branch was connected and the shunting impedance is defined in exactly the same way as for Thevenin's t h e o r e m . T h e result is called the N o r t o n Equivalent G e n e r a t o r . T h e theorems are therefore duals of each other and either could be deduced from the other by the source transformation process. PROBLEM
4.9
A 600/150-V volt-ratio box having a total resistance of 30 k Q is used in the m e a s u r e m e n t of the voltage of a d.c. source. T h e whole box is connected across the source and a voltmeter is applied to the 150-V tapping. Using Thevenin's t h e o r e m , calculate the lowest resistance which the voltmeter may have so that the m e a s u r e m e n t error shall not exceed 0.1%.
ELECTRIC CIRCUIT THEORY
226
Solution T h e volt-ratio box is a 30,000-Q resistor with a tapping at: 30,000 x ^ = 7500 Q 600
(see Fig. 4.11(a)).
The Thevenin equivalent of the circuit seen from the voltmeter terminals is as shown in Fig. 4.11(b), where R' is given by: R' =
7500 x 22,500 30,000
5625 Q.
R'=5625 G A
Voltmeter Voltmeter
(b)
(a) FIG. 4 . 1 1 .
If R were infinite, n o error would result, but for finite R a current is drawn through R' and the voltage at terminals A B falls. H e n c e , for 0 . 1 % error: R
(VIA)
R + 5625
- (1/1000)
-(VIA)
(VIA)
5625
1
R
999
1
999
1000
1000
giving R = 5,619,375 Q.
P R O B L E M 4.10
A simple d.c. slide-wire potentiometer is used to measure the voltage of a source, the circuit arrangement being as shown in Fig. 4.12(a). T h e slide wire current is 50 m A and its 100-cm length, A B C , has a volt-drop across it of 1.5 V. Balance (indicated by zero galvanometer current) is
227
CIRCUIT ANALYSIS
(a)
(b) FIG. 4 . 1 2 .
achieved when the slider at B is 20 cm from C. Using Thevenin's t h e o r e m calculate the galvanometer current when the slider is moved 1 m m , towards C. Solution A m o v e m e n t of the slider by 1 m m , to B ' , represents an out-of-balance voltage of 1.5 m V (since 100 cm represents 1.5 V ) . H e n c e , from the viewpoint of the galvanometer circuit, the e.m.f. of the Thevenin equivalent generator is 1.5 m V and its internal resistance is the resistance of B ' C shunted by 50 Q and A B ' in series. T h e resistance of B ' C is: 19.9
1.5
100 0.05 and that of A B ' is:
_ _ 5.97 Q
L5 0.05
- 5.97 = 24.03 Q
so that R' (see Fig. 4.12(b)) is given by:
Hence
228
ELECTRIC CIRCUIT THEORY
In fact there would have been negligible error involved had it been assumed that the internal resistance of the Thevenin equivalent generator does not change for this small movement of the slider. PROBLEM
4.11
Calculate, using (a) Thevenin's t h e o r e m , (b) Norton's t h e o r e m , the current in the branch A B of the circuit represented in Fig. 4.13(a). 3ii
j 6 a
A
-j 20
a
-o F i g . 4.13. (a) Diagram of the circuit for Problem 4.11.
j 20
a
(240 + j 120) V
Fig. 4.13. (b) The branch AB removed and the current source replaced by its equivalent voltage source.
6 . 6 6 7 - j 13.333 A
j 12 A
F i g . 4.13. (c) The branch AB removed and the voltage source replaced by its equivalent current source.
CIRCUIT ANALYSIS
229
Solution R e m o v e the branch A B , short-circuit the voltage source and opencircuit the current source. T h e internal impedance of the circuit viewed from A B is that of (3 + j6) Q and (10 - j20) Q in parallel, or Z
= (5.342 + J5.753) Q = 7.851 ^ 4 7 ° 7 ' Q. (a) By Thevenin's
theorem
First change the current source shunted by (10 - j20) Q into a voltage source of e.m.f.: E - j l 2 ( 1 0 - j20) = (240 + j l 2 0 ) V in series with (10 - j20) Q. T h e circuit is now as shown in Fig. 4.13(b). The circulating current is: I.- ^ 5
- (7.507
+, 1 7 . 3 . 5 ) A .
A n d the open-circuit voltage at A B is: 100 Z0° - (7.507 + J17.315) (3 + j6) V = (181.370 - J96.986) V. H e n c e , the current through the branch A B , on replacing it, is:
(b) By Norton's
theorem
First change the voltage source in series with (3 + j6) Q to a current source of generated current I Nshunted by (3 + j6) Q , as shown in Fig. 4.13(c), In
100 3 + J6
(6.667 - j ! 3 . 3 3 3 ) A.
230
ELECTRIC CIRCUIT THEORY
If terminals A B are short-circuited, t h e c u r r e n t which flows, which is also t h e g e n e r a t e d c u r r e n t of t h e N o r t o n equivalent s o u r c e , is: (6.667 - j'13.333) - j l 2 = (6.667 - J25.333) A and this is s h u n t e d by t h e i m p e d a n c e looking into A B — w h i c h is: (5.342 + J5.753) Q. H e n c e , t h e c u r r e n t t h r o u g h the b r a n c h A B , w h e n r e p l a c e d , is: ;6.667 - J25.333) A - 15.112 Z - 8 0 ° 2 0 ' A
the s a m e result as was o b t a i n e d using T h e v e n i n ' s t h e o r e m .
PROBLEMS IV
1. A Wheatstone-bridge circuit whose four arms have resistances of 3, 4, 5 and 6 Q (when taken in cyclic order) is fed by a 5-A current source connected between opposite corners. Calculate (a) the voltage across the other opposite corners, (b) the current taken by a 10-Q resistor connected to these corners. 2. For the circuit represented by Fig. P.IV.l, calculate the current through and the power absorbed by the 10-Q resistor. 3. Calculate the voltage appearing between the terminals A and B in the circuit represented by Fig. P.IV.2. Use the node-voltage rule. 4. For the circuit of Fig. P. IV.3, calculate the current in the 6-Q resistor by (a) Kirchhoff s Laws, (b) the superposition theorem. 5. Using Thevenin's theorem, calculate the current through the 10-Q resistor in the circuit of Fig. P.IV.4. 6. Deduce the conditions under which maximum power is delivered to a load of resistance R by a generator of e.m.f. E and internal resistance r. LFor the circuit of Fig. P.IV.5, calculate the resistance R for which maximum power is absorbed and the magnitude of this power. 7. For the circuit of Fig. P.IV.6, containing current-controlled voltage sources, calculate the mesh currents i i and / . u2 3 8. Use the node-voltage rule to obtain the currents in all the branches of the circuit of Fig. P.IV.7. 9. In the circuit represented by Fig. P.IV.8, the switch S is arranged to short-circuit half the 12-Q resistor. When S is closed the current through its contacts is 2 A. Calculate the current through the 12-Q resistor when S is open. 10. Define the terms graph, tree, link-branch in network topology. For the circuit of Fig. P.IV.9, draw the graph and choose a tree which allows all branch currents to be evaluated with just one equation. Calculate the branch currents and the power delivered by each source.
CIRCUIT ANALYSIS
231 3A
2a
FIG. P.IV.l (Problem IV.2). 3A
O A
5ft FIG. P.IV.2 (Problem IV.3). 4ft
10ft
8ft
FIG. P.IV.3 (Problem IV.4). 4ft
6ft
—I
I
f
FIG. P.IV.4 (Problem IV.5).
24 ft
FIG. P.IV.5 (Problem IV.6).
1
232
CIRCUIT ANALYSIS 3&
233 i,
FIG. P.IV.10 (Problem IV.11).
A
^
j 20
I
a io
30 a
is
aI
J
FIG. P.IV.ll (Problem IV.12).
FIG. P.IV.12 (Problem IV.13).
FIG. P.IV.13 (Problem IV.14).
a
234
ELECTRIC CIRCUIT THEORY
11. For the circuit of Fig. P.IV. 10, draw the graph and choose a suitable tree for evaluating all branch currents. Calculate each branch current and hence determine the power delivered by the 6-A source. 12. For the frequency-domain circuit of Fig. P.IV. 11, determine the voltage V (magnitude and phase) such that no current flows through the 18-Q resistor. 13. Determine the voltage—in polar form—across the source and the power delivered by the source in the circuit of Fig. P.IV. 12. Determine the Thevenin equivalent generator at the terminals AB and hence the current in the 10-Q resistor. 14. Determine the Thevenin equivalent generator for the network represented by Fig. P.IV. 13. Then apply a source-transformation to obtain the Norton equivalent generator. 15. In the bridge circuit represented by Fig. P.IV. 14, a voltage source is connected to D and B. Derive expressions for the values of R and L, in terms of r, C, C R and co, which u x cause the points A and C to be at the same potential (i.e. to balance the bridge).
C
A
B
D
C
FIG. P.IV.14 (Problem IV. 15). 16. For the bridged-T network of Fig. P.IV.15, derive the conditions for which the voltage across the terminals A and B is zero.
R
-i—y
FIG. P.IV.15 (Problem IV. 16).
CIRCUIT ANALYSIS
235
17. The circuit represented by Fig. P.IV. 16 is supplied by the voltage source for which V = 100 Z.0 at co = 1000 rad/sec. Calculate the current Ii, in polar form. M=O.OIH
FIG. P.IV.16 (Problem IV.17).