BEF 12403 - Week 6 - Series-Parallel Resistor Circuits.ppt

Week 6
SERIES-PARALLEL RESISTOR
CIRCUITS

2
Resistors in Series
When circuit elements are connected end to end or in tandem, they are said
to be in series. A series circuit may be alternatively defined as a circuit having
only one path for the current.
R1
v2
R2
R3
v1
v3
v
i
Consider the series connected resistors shown in
Figure 1. Let v be the terminal voltage of the box
and let v1, v2, and v3 be the voltage drops across
resistors R1, R2, and R3, respectively. Application
of KVL to the circuit leads to the voltage equation
3
2
1 v
v
v
v 


3
3
2
2
1
1 R
;
R
;
R i
v
i
v
i
v 


Application of Ohm’s law to the resistors leads to
the equations
Figure 1

3
Resistors in Series
T
v
i
R

R1
v2
R2
R3
v1
v3
v
i
Hence, upon substitution we have
)
R
R
R
(
R
R
R 3
2
1
3
2
1 




 i
i
i
i
v
or
3
2
1 R
R
R
v
i



Figure 1
Since R1, R2, and R3 are all constants, we can
rewrite the previous expression as
where
3
2
1
T R
R
R
R 


RT is called the total resistance of the box.

4
Resistors in Series
The preceding expression
indicates that in a series circuit,
the total (or equivalent) resistance
RT equals the sum of the
resistances; hence we have the
equivalent circuit of Figure 2.
RT
v
i
Figure 2
What is the total resistance of the box
shown in Figure 3?
Example
400 Ω
2.5 kΩ
3 kΩ
Figure 3
A
B

5
Resistors in Series
Solution
The resistor are connected in series
since they have the same current
flowing through each of them. Hence,
the total resistance of the box is
3000
2500
400
RT 


RT= 5900 Ω = 5.9 kΩ
giving
The box can now be redrawn as in
Figure 4.
5.9 kΩ
Figure 4
Note that the total resistance RT in this
case is also the equivalent resistance
of the box; that is, it completely models
what is inside the box when viewed
from the input terminals AB.
A
B

6
Resistors in Parallel
3
3
3
2
2
2
1
1
1 R
;
R
;
R i
v
i
v
i
v 


When circuit elements are connected so that they have the same pair
of terminal points, the elements are said to be in parallel.
Consider the parallel connected resistors shown in Figure 5. Measurement of
the voltage drop across each of them will give the same readings; that is, v =
v1 = v2 = v3. Hence, in a parallel circuit the voltage drops across the parallel
elements are the same. Application of KCL leads to the current equation
Application of Ohm’s law to the
resistors leads to the equations
Hence, upon substitution we have Figure 5
3
3
2
2
1
1
R
v
R
v
R
v
i 


R1
R2 R3
v v1 v2 v3
i1 i2
i3
i
i = i1 + i2 + i3
or
3
2
1 R
v
R
v
R
v
i 


since v = v1 = v2 = v3

7
3
2
1
1
1
1
R
R
R
v
i



ce,
1/resistan
of
dimension
the
has
v
i
Since
3
2
1
T R
1
R
1
R
1
R
1



The preceding expression indicates that in a parallel circuit, the reciprocal of
the total (or equivalent) resistance RT equals the sum of the reciprocals of the
resistances; hence we have the equivalent circuit of Figure 6.
We can rewrite the preceding expression as
Figure 6
RT
T
v
i
R
1

we can write it as
Hence

8
Example
What is the total resistance of the box
shown in Figure 7?
Solution
Figure 7
The resistor are connected in parallel
since they have the same voltage drops
across each of them. Hence
25
1
50
1
10
1
R
1
T



RT = 25/4 Ω
giving
The box can now be redrawn as
in Figure 8.
25/4 Ω
Figure 8
10 Ω 50
Ω
25 Ω

9
2
1
2
1
T
R
R
R
R
R


Note
Frequently, a parallel circuit will consist of
only two resistors, as in Figure 9. The total
resistance in this case may be expressed as
Figure 9
Figure 10
R1
R2
The above expression simply states that
the total resistance of two resistances in
parallel is their product over their sum.
Example
Determine the total resistance for
the network shown below.
12 Ω 24
Ω
Solution






 8
12
24
12
24
R
R
R
R
R
2
1
2
1
T

10
Series-Parallel Resistors
Circuits that contain both series and parallel
combinations of resistors are called series-
parallel circuits.
Figure 11
In a series-parallel circuit parts of the circuits
contain series elements in which the same
current flows, whereas other parts of the
circuit contain parallel elements across
which the same voltage occurs.
R1
R2
R3
One way of simplifying series-parallel resistor
circuits is to begin with the innermost parts and
then progressively simplify the circuit to one
equivalent resistance.
Unfortunately, the “starting point” for simplification of series-parallel circuits are
as varied as the circuits themselves. However, the following example provide
some suggestions.

11
Exercise
Find the resistance of the box in Figure 12 looking into the terminals EF from
the left.
Figure 12
E
F
20 Ω
20 Ω
20 Ω
20 Ω
20 Ω
20 Ω
C
D
A
B

12
L
L
AB
R
R
R
R
R
R
R




2
1
1
2 )
(
A
B
R1
R2 RL vo
RAB
If RAB is the resistance of the subcircuit looking into
terminals A and B, then, using parallel resistances
principle, we can write
The circuit can then be drawn in a simplified
form as shown in Figure 13(b). C
D
R3
R4 RAB vAB
RCD
If RCD is the resistance of the subcircuit looking into
terminals C and D, then, using parallel resistances
AB
AB
CD
R
R
R
R
R
R
R




4
3
3
4 )
(
Solution
Figure 13
(a)
(b)

13
The circuit can then be drawn in a simplified form as shown in Figure 14.
E
R5
RCD
REF
F
  









3
40
20
20
20
20
20
20
)
(
2
1
1
2
L
L
AB
R
R
R
R
R
R
R
Solution (continued)
Terminal resistance of the box is thus
REF = R5 + RCD
Substituting the values given for the resistances
into the expressions obtained earlier, we obtain
  









4
50
3
/
40
20
20
3
/
40
20
20
)
(
4
3
3
4
AB
AB
CD
R
R
R
R
R
R
R
REF = R5 +RCD = 20 + 50/4 = 37. 5 Ω
Hence
Figure 14

14
Exercise
For the circuit shown in Figure 15. find the equivalent resistance of the
box.
Figure 15
2 Ω
4.5 Ω
1 Ω
3 Ω
1.5 Ω

15
Delta and Wye Conversions
Two networks that frequently occur in electric circuit analysis are the
wye (Y) and delta (Δ). These two circuits identified in Figure 16 are
sometimes part of a larger circuit and obtain their names from their
configurations.
Figure 16. Networks: (a) delta; (b) wye.
(a) (b)

16
Although the networks of Figure 16 are three-terminal networks they are
sometimes redrawn as four-terminal networks as shown in Figure 17,
where the delta network has been redrawn as a T network and the wye
network redrawn as a pi network.
Note
(a) (b)
Figure 17. Networks: (a) T ; (b) Pi.

17
1
1
3
3
2
2
1
R
R
R
R
R
R
R
RA



2
1
3
3
2
2
1
R
R
R
R
R
R
R
RB



Sometimes it is required to convert one configuration into the other
to simplify the circuit analysis. The following relationships are used
to convert a Y-configuration into a delta.
3
1
3
3
2
2
1
R
R
R
R
R
R
R
RC



R1 R2
R3
a b
c
a b
c
RC
RA
RB
Figure 18

18
C
B
A
C
B
R
R
R
R
R
R



1
C
B
A
A
C
R
R
R
R
R
R



2
The following relationships are used to convert a Δ-configuration into a wye.
C
B
A
B
A
R
R
R
R
R
R



3
R1 R2
R3
a b
c
a b
c
RC
RA
RB
Figure 19

19
The following example illustrates the use of a Δ-to-Y transformation to
simplify the analysis of a circuit.
Example
Find the current and power supplied by the 40 V source in the circuit
shown in Figure 20.
125 Ω
25 Ω
37.5 Ω
40
Ω
100 Ω
5 Ω
40 V
Figure 20

20
Solution
Here we are interested only in the
current and power drain on the power
source, so the problem has been
solved once we obtain the equivalent
resistance across the terminals of the
source. We can find this equivalent
resistance easily after replacing the
upper Δ or the lower with its
equivalent Y. We choose to replace
the upper Δ. We then compute the
three Y resistances, defined in Figure
21, using the Δ-Y transformation
equations.
125 Ω
25 Ω
37.5 Ω
40
Ω
100 Ω
5 Ω
40 V
Figure 21

21








 50
25
125
100
125
100
1
C
B
A
C
B
R
R
R
R
R
R








 5
.
12
25
125
100
25
125
3
C
B
A
A
C
R
R
R
R
R
R
Thus,
Figure 22
100 Ω 125 Ω
25 Ω
A
B C
A
B C
R1
R2 R3








 10
25
125
100
25
100
2
C
B
A
B
A
R
R
R
R
R
R

22
Substituting the Y–connected resistors shown in Figure 22 into the circuit
shown in Figure 21 produces the circuit shown in Figure 23.
12.5 Ω
37.5 Ω
40
Ω
10 Ω
40 V
5 Ω
50 Ω
From Figure 23 we can easily
calculate the resistance across the
terminals of the 40 V source by
series-parallel simplifications:




 80
100
50
50
55
RT
Figure 23

23
A
5
.
0
80
40


i
The final step is to note that the circuit reduces to an 80 Ω resistor across a
40 V source, as shown in Figure 24.
80 Ω
40 V
i
Thus, the current supplied supplied by the source
to the circuit is
and the power supplied by the source to the circuit is
    W
20
80
5
.
0
2
2


 R
i
P Figure 24

24
Worked Example
Find line currents iA, iB, and iC at time t = 0 for the circuit shown in Figure
25.
Figure 25
vAB(t) = 10sin ωt
vBC(t) = 10sin (ωt – 120o)
vCA(t) = 10sin (ωt +120o)
2 Ω
2 Ω
2 Ω
A
B
C
vCA(t) vAB(t)
vBC(t)
iA
iB
iC

25
         






 6
2
2
2
2
2
2
2
1
1
3
3
2
2
1
R
R
R
R
R
R
R
RA
         






 6
2
2
2
2
2
2
2
3
1
3
3
2
2
1
R
R
R
R
R
R
R
RC
Solution
To simplify the circuit, we convert the Y-connected
resistors into its Δ equivalent configuration. We then
compute the three Δ resistances, defined in Figure
26(b), using the Y-Δ transformation equations. Thus,
Figure 26
RC
RB
RA
A
B C
A
B C
2 Ω
2 Ω
2 Ω
         






 6
2
2
2
2
2
2
2
2
1
3
3
2
2
1
R
R
R
R
R
R
R
RB
(a)
(b)

26
1.443A;
6
)
120
sin(
10





o
BC
BC
BC
t
R
v
i

A;
0
6
sin
10



t
R
v
i
AB
AB
AB

A.
3
44
.
1
6
)
120
sin(
10




o
CA
CA
CA
t
R
v
i

The load currents at t = 0 are and the line currents at t = 0 are
;
A
443
.
1
0
443
.
1 





 AB
BC
B i
i
i
A;
443
.
1
443
.
1
0 




 CA
AB
A i
i
i
 
443
.
1
443
.
1 



 BC
CA
C i
i
i
A.
886
.
2

Figure 27
vAB
vBC
vCA
iA
iB
iC
6 Ω 6 Ω
6 Ω
A
B
C
iCA
iBC
iAB

27
Voltage Divider Rule
3
2
1 R
R
R
v
i



Often in analysing a series circuit it becomes necessary to find the voltage
drops across one or more of the resistances. A simple relationship for the
voltage drop may be obtained by referring to Figure 27.
R1
v2
R2
R3
v1
v3
v
i
and the voltage drops are given by
3
2
1
1
1
1
R
R
R
R
R




v
i
v
3
2
1
2
2
2
R
R
R
R
R




v
i
v
3
2
1
3
3
3 R
R
R
R
vR
i
v




Figure 27
The total current i is given by

28
R1
v2
R2
R3
v1
v3
v
i
Figure
28
Notice in these relationships that the voltage drop across a resistor is
proportional to the ratio of that resistance to the total resistance of the
circuit.
A general statement called the voltage divider rule
can be made as follows:
In a series circuit, the voltage drop
across a particular resistor Rn is the line
voltage times the fraction Rn/RT.
v
vn
T
n
R
R

Mathematically,
where
m
m
i
i
T R
......
R
R
R
R
R 3
2
1
1





 

and m is the total number of resistors in the series circuit.

29
Example
Figure 29
Using voltage division, determine
the voltages across each of the
resistors in Figure 29.
60 Ω
85 Ω
55 Ω
Solution
First, label the resistors and their
respective voltage drops. Let
R1= 60 Ω
R2= 85 Ω
R3= 55 Ω
RT = R1+ R2 + R3
= 60 + 85 + 55 = 200 Ω
Therefore, the total series resistance
is
and v1, v2, and v3 be the respective
voltage drops across R1, R2, and R3.
10 V

30
Figure 30
Then
R1 = 60 Ω
R2 = 85 Ω
R3 = 55 Ω
v1
v2
v3
v = 10 V
V
3
200
60
10
R
R
R
R
3
2
1
1
1 





v
v
V
25
.
4
200
85
10
R
R
R
R
3
2
1
2
2 





v
v
V
75
.
2
200
55
10
3
2
1
3
3 





R
R
R
vR
v
A check by Kirchhoff’s voltage law indicates that
v1+ v2 + v3 = 3 + 4.25 + 2.75 = 10 V = v

31
Exercise
Find the output voltage of the box shown in Figure 31.
Figure 31
10 V
20 Ω
20 Ω
20 Ω
20 Ω
20 Ω
20 Ω vo
The voltage divider rule can be used to analyse more complex circuit
configurations as the following example demonstrates.

32
Solution
Let us use voltage division to solve this problem. To do this, first we need to
label the resistors and nodes so that we can solve the problem
symbolically. One circuit labelling scheme is shown in Figure 32.
R5 R1
RL
R3
vo
E = 10 V R4 R2
A
B
C
D
E
F
Figure 32
R1 = R2 = R3 = R4 = R5 = RL = 20 Ω

33
(a)
L
L
AB
o
R
R
R
v
v


1
By voltage divider rule, the voltage drop across
RL can be written in term of the terminal voltage
vAB as
If RAB is the resistance of the subcircuit looking into
terminals A and B, then, using parallel resistances
A
B
R1
R2 RL vo
RAB
L
L
AB
R
R
R
R
R
R
R




2
1
1
2 )
(
form as shown in Figure 32(b).
C
D
R3
R4 RAB vAB
RCD
Figure 32
(b)

34
Solution (continued) C
D
R3
R4 RAB vAB
RCD (a)
By voltage divider rule, the voltage drop across
RAB can be written in term of the terminal
voltage vCD as
AB
AB
CD
AB
R
R
R
v
v


3
If RCD is the resistance of the subcircuit looking into
terminals C and D, then, using parallel resistances
AB
AB
CD
R
R
R
R
R
R
R




4
3
3
4 )
(
form as shown in Figure y.
E
F
R5
RCD vCD
10 V
Figure 33
(b)

35
L
L
AB
o
R
R
R
v
v


1
CD
CD
AB
AB
L
L
R
R
R
R
R
R
R
R
R
R
E







5
2
1
1
Finally, by voltage divider principle, the voltage
drop across RCD can be written in term of the
source voltage E as
E
F
R5
RCD vCD
E = 10 V
Figure 34
AB
AB
L
L
CD
R
R
R
R
R
R
R
v





2
1
1

36
  









3
40
20
20
20
20
20
20
)
(
2
1
1
2
L
L
AB
R
R
R
R
R
R
R
CD
CD
AB
AB
L
L
o
R
R
R
R
R
R
R
R
R
R
E
v







5
2
1
1
E
F
R5
RCD vCD
E = 10 V
Figure 35
Thus,
Substituting values into the terminal resistance
expressions we obtained earlier, we get
  









4
50
3
/
40
20
20
3
/
40
20
20
)
(
4
3
3
4
AB
AB
CD
R
R
R
R
R
R
R

37
CD
CD
AB
AB
L
L
o
R
R
R
R
R
R
R
R
R
R
E
v







5
2
1
1
Hence
4
/
50
20
4
/
50
3
/
40
20
20
3
/
40
20
20
20
10







giving
vo = 100/208 = 0.481 V

38
Current Divider Rule
1
1
R
v
i 
2
1
2
1
R
R
R
R
v
R
v
i
T



The special case of two resistors in parallel such as shown in Figure x lends
itself to a current division rule.
i
v
R1 R2
i1 i2
Figure 36
Consider the circuit shown in Figure 36. Application of KCL to the circuit leads
to the current equation
and application of Ohm’s law to each resistor gives
i = i1 + i2
If RT is the total resistance of the two parallel resistors,
then by Ohm’s law we have
and
2
2
R
v
i 

39
.
2
1
2
1
R
R
R
R
v
R
v
i
T



.
2
1
2
1
1
1
R
R
R
R
R
i
i


2
1
2
1
R
R
R
i
i


With two resistors in parallel, the current in either resistor is the
total current times the ratio of the opposite resistor over the sum
of the two resistors.
Hence,
Using the relation v = i1R1, we can write Hence,
By a similar argument, .
2
1
2
2
R
R
R
i
i


In words, the current divider rule can be stated as follows:

40
Worked Example
Find vo and vg in the circuit in Figure 37.
30 Ω 60 Ω
25 A
25 Ω
50 Ω
12 Ω
30 Ω
vo
vg
Figure 37.

BEF 12403 - Week 6 - Series-Parallel Resistor Circuits.ppt

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BEF 12403 - Week 6 - Series-Parallel Resistor Circuits.ppt