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ELECTRICAL TECHNOLOGY
EET 103/4
 Explain and analyze series and
parallel circuits
 Explain, derive and analyze Ohm’s
Law, Kirchhoff Current Law,
Kirchhoff Voltage Law, Source
Transformation, Thevenin theorem.
1
PARALLEL DC CIRCUIT
(CHAPTER 6)
2
6.2 – Parallel Resistors
Two or more elements, branches, or circuits
are in parallel if they have two points in
common as in the figure below
3
6.2 – Parallel Resistors
R1 and R2 are in parallel
4
6.2 – Parallel Resistors
R1 and R2 are in parallel, and this parallel
combination is in series with R3:
5
6.2 – Parallel Resistors
R3 is in parallel with the
series combination of
R1 and R2:
6
6.2 – Parallel Resistors
 For resistors in parallel, the total
resistance is determined from:
1
1
1
1
1
 

 ..... 
RT R1 R2 R3
RN
7
6.2 – Parallel Resistors
Note that the equation is for the
reciprocal of RT rather than for RT.
 Once the right side of the equation has
been determined, it is necessary to divide
the result into 1 to determine the total
resistance
8
6.2 – Parallel Resistors
 For parallel elements, the total
conductance is the sum of the individual
conductance values.
GT  G1  G2  G3  .....  GN
As the number of resistors in parallel
increases, the input current level will increase
for the same applied voltage.
 This is the opposite effect of increasing the
number of resistors in a series circuit.
9
6.2 – Parallel Resistors
The total resistance of any number of
parallel resistors can be determined using
1
RT 
1
1
1
1


 ... 
R1 R2 R3
RN
The total resistance of parallel resistors is
always less than the value of the smallest
resistor.
10
6.2 – Parallel Resistors
Example 6.2
Find the total resistance RT in the following figure:
11
6.2 – Parallel Resistors
Example 6.2 - solution
1 1
G1 
  0 .5 S
R1 2
1
1
G2 

 0.005 S
R2 200
1
1
G3 

 0.001 S
R3 1000
GT  G1  G2  G3
1
1
RT 

 1.976 
GT 0.506
 0.506 S
12
6.2 – Parallel Resistors
Example 6.3
Find the total resistance RT in the following figure:
13
6.2 – Parallel Resistors
Example 6.3 - solution
1
G1 
R1
The network redrawn
1
  1S
1
1
G2 
R2
1
  0.25 S
4
1 1
G3 
  0.2 S
R3 5
14
6.2 – Parallel Resistors
Example 6.3 – solution (cont’d)
GT  G1  G2  G3
The network redrawn
 1  0.25  0.2
 1.45 S
1
RT 
GT
1

 0.69 
1.45
15
6.2 – Parallel Resistors
For equal resistors in parallel:
R
RT 
N
Where N = the number of parallel resistors.
16
6.2 – Parallel Resistors
Example 6.6
Find the total
resistance RT in
the following
figure:
17
6.2 – Parallel Resistors
Example 6.6 – solution
R
RT 
N
2

4
The network redrawn
 0.5 
18
6.2 – Parallel Resistors
 A special case: The total resistance of
two resistors is the product of the two
divided by their sum.
R1 R2
RT 
R1  R2
The equation was developed to reduce the
effects of the inverse relationship when
determining RT
19
6.2 – Parallel Resistors
Example 6.8
Find the total resistance RT in the following figure:
20
6.2 – Parallel Resistors
Example 6.8 – solution
R 'T R3
RT 
R 'T  R3
0.8  5

0.8  5
 0.69 
21
6.2 – Parallel Resistors
Example 6.10
Determine the value of R2 to establish a
total resistance of 9 k
22
6.2 – Parallel Resistors
Example 6.10 – solution
R1 R2
RT 
R1  R2
RT R1  R2   R1R2
RT R1
R2 
R1  RT 
9 12

 36 k
12  9
23
6.2 – Parallel Resistors
Example 6.11
Determine R1, R2 and R3 in the following figure if;
R2  2R1;
R3  2R2 ;
and the total resistance is 16 k
24
6.2 – Parallel Resistors
Example 6.11 – solution
R2  2R1
R3  2 R2  22 R1 
1
1
1
1
 

RT R1 R2 R3
1
1
1
7
 


R1 2R1  4R1  4 R1
7 RT
R1 
4
25
6.2 – Parallel Resistors
Example 6.11 – solution (cont’d)
7 RT
R1 
4
7  16

 28 k
4
R2  2R1  2  28  56 k
R3  4R1  4  28  112 k
26
6.3 – Parallel Circuits
 Voltage is always the same across
parallel elements.
27
6.3 – Parallel Circuits
The voltage across resistor 1 equals the voltage
across resistor 2, and both equal the voltage
supplies by the source
V1  V2  E
28
6.3 – Parallel Circuits
 For single-source parallel networks, the
source current (I ) is equal to the sum of
the individual branch currents.
s
I s  I1  I 2
29
6.3 – Parallel Circuits
E
Is 
RT
30
6.3 – Parallel Circuits
V1 E
I1 

R1 R1
I2 
V2
E

R2 R2
E
Is 
RT
I s  I1  I 2
E
E E
 
RT R1 R2
1
1
1
 
RT R1 R2
31
6.3 – Parallel Circuits
 For a parallel circuit, source current equals
the sum of the branch currents. For a series
circuit, the applied voltage equals the sum of
the voltage drops.
32
6.3 – Parallel Circuits
Example 6.12
(a) Find the total resistance
(b) Calculate the source current
(c) Determine the current through each branch
33
6.3 – Parallel Circuits
Example 6.12 – solution
(a)
R1 R2
RT 
R1  R2
9 18

6
9  18
(b)
Is 
E 27

 4.5 A
RT
6
34
6.3 – Parallel Circuits
Example 6.12 – solution (cont’d)
(c)
E
I1 
R1
27

3A
9
E 27
I2 

 1.5 A
R2 18
35
6.3 – Parallel Circuits
Example 6.13
(a) Find the total resistance
(b) Calculate the source current
(c) Determine the current through each branch
36
6.3 – Parallel Circuits
Example 6.13 – solution
(a)
1
RT 
1
1
1


R1 R2 R3

1
1
1
1


10 220 1200
 9.49 
37
6.3 – Parallel Circuits
Example 6.13 – solution (cont’d)
(b)
E
Is 
RT
24

 2.53 A
9.49
38
6.3 – Parallel Circuits
Example 6.13 – solution (cont’d)
(c)
V1 E
I1 

R1 R1
24

 2 .4 A
10
V2
E
24
I2 


 0.11 A
R2 R2 220
V3 E
24
I3 


 0.02 A
R3 R3 1200
39
6.3 – Parallel Circuits
Example 6.14
(a)
(b)
(c)
(d)
Determine R3.
Find the applied voltage E.
Find the source current Is.
Find I2.
40
6.3 – Parallel Circuits
Example 6.14 – solution
(a)
1
1
1
1
 

RT R1 R2 R3
1 1 1
1
  
4 10 20 R3
1 1 1 1
1
  

R3 4 10 20 10
R3  10 
41
6.3 – Parallel Circuits
Example 6.14 – solution (cont’d)
(b)
E  V1  I1R1
 4 10  40 V
(c)
E 40
Is 

 4
RT 10
(d)
V2
E 40
I2 


2A
R2 R2 20
42
6.4 – Power Distribution
in
Parallel Circuits
 For any resistive circuit, the power applied by the
battery will equal that dissipated by the resistive
elements.
PE  PR1  PR2  PR3  ...  PRN
 The power relationship for parallel resistive circuits is
identical to that for series resistive circuits.
43
6.4 – Power Distribution
in
Parallel Circuits
Power delivered by the source;
PE  EI s
44
6.4 – Power Distribution
in
Parallel Circuits
2
P1  V1 I1 
V1
2
 I1 R1
R1
45
6.4 – Power Distribution
in
Parallel Circuits
2
P2  V2 I 2 
V2
2
 I 2 R2
R2
46
6.4 – Power Distribution
in
Parallel Circuits
2
V3
2
P3  V3 I 3 
 I 3 R3
R3
47
6.4 – Power Distribution
in
Parallel
Circuits
Example 6.15
(a) Determine the total resistance RT.
(b) Find the source current and the current through each
resistor.
(c) Calculate the power delivered by the source.
48
6.4 – Power Distribution
in
Parallel
Circuits
Example 6.15 (cont’d)
(d) Determine the power absorbed by each resistor.
(e) Verify:
PE  PR1  PR2  PR3
49
6.4 – Power Distribution
in
Parallel
Circuits
Example 6.15 – solution
(a)
1
1
1
1
 

RT R1 R2 R3
1
1
1



 6.929 10  4 S
1.6 k 20 k 56 k
1
RT 
 1.44 k
4
6.929 10
50
6.4 – Power Distribution
in
Parallel
Example
6.15 –Circuits
solution (cont’d)
(b)
E
28
Is 

RT 1.44 k
 19.44 mA
E
28
I1 

 17.5 mA
R1 1.6 k
E
28
I2 

 1.4 mA
R2 20 k
E
28
I3 

 0.5 mA
R3 56 k
51
6.4 – Power Distribution
in
Parallel
Circuits
Example 6.15 – solution (cont’d)
(c)
PE  EI s
 2819.44 mA
 543.2 mW
(d)
PR1  I1 R1  17.5 mA  1.6 k  490 mW
2
2
PR2  I 2 R2  1.4 mA   20 k  39.2 mW
2
2
52
6.4 – Power Distribution
in
Parallel
Circuits
Example 6.15 – solution (cont’d)
(d) – cont’d
PR3  I 3 R3  0.5 mA   56 k  14 mW
2
2
PR1  PR 2  PR3  490  39.2  14  543.2 mW  PE
53
6.5 – Kirchhoff’s Current Law
 Kirchhoff’s voltage law provides an important
relationship among voltage levels around any closed
loop of a network.
Kirchhoff’s current law (KCL) states that the
algebraic sum of the currents entering and leaving
an area, system, or junction is zero.
 The sum of the current entering an area, system or
junction must equal the sum of the current leaving
the area, system, or junction.
I
in
  I out
54
6.5 – Kirchhoff’s Current Law
55
6.5 – Kirchhoff’s Current Law
56
6.5 – Kirchhoff’s Current Law
 Most common application of the law will
be at the junction of two or more paths of
current.
 Determining whether a current is
entering or leaving a junction is
sometimes the most difficult task.
If the current arrow points toward the
junction, the current is entering the junction.
 If the current arrow points away from the
junction, the current is leaving the junction.
57
6.5 – Kirchhoff’s Current Law
Example 6.16
Determine I3 and I4:
58
6.5 – Kirchhoff’s Current Law
Example 6.16 – solution
I
in
  I out
At node a;
I1  I 2  I 3
I3  2  3  5 A
59
6.5 – Kirchhoff’s Current Law
Example 6.16 – solution (cont’d)
At node b;
I3  I 4  I5
I 4  I3  I5
I 4  5 1  4 A
60
6.5 – Kirchhoff’s Current Law
Example 6.17
Determine I1, I3, I4 and I5:
61
6.5 – Kirchhoff’s Current Law
Example 6.17 – solution
I
in
  I out
At node a;
I  I1  I 2
I1  I  I 2  5  4  1 A
62
6.5 – Kirchhoff’s Current Law
Example 6.17 – solution (cont’d)
At node b; I1  I3  1 A
At node c; I 2  I4  4 A
63
6.5 – Kirchhoff’s Current Law
Example 6.17 – solution (cont’d)
At node d;
I 3  I 4  I5
1  4  I5  5 A
64
6.5 – Kirchhoff’s Current Law
Example 6.18
Determine I3 and I5:
65
6.5 – Kirchhoff’s Current Law
Example 6.18 – solution
I
in
  I out
At node a;
I1  I 2  I3
4  3  I3  7 A
At node b;
I 3  I 4  I5
I5  I 3  I 4  7  1  6 A
66
6.5 – Kirchhoff’s Current Law
Example 6.19
(a) Determine Is.
(b) Find E.
(c) Determine R3.
(d) Calculate RT.
67
6.5 – Kirchhoff’s Current Law
Example 6.19 – solution
(a)
I
(b)
E  V1  I1R1  8 103  2 103  16 V
in
  I out
I s  8  10  2  20 mA
68
6.5 – Kirchhoff’s Current Law
Example 6.19 – solution (cont’d)
(c)
V3 E
16
R3   
 8 k
3
I 3 I 3 2 10
69
6.5 – Kirchhoff’s Current Law
Example 6.19 – solution (cont’d)
(d)
E
16
RT  
 800   0.8 k
3
I s 20 10
70
6.5 – Kirchhoff’s Current Law
Example 6.19 – solution (cont’d)
The circuit redrawn:
71
6.5 – Kirchhoff’s Current Law
Example 6.20
Determine I1.
72
6.5 – Kirchhoff’s Current Law
Example 6.20 – solution
Assume I1 flows into the IC;
I
in
 I1  10  4  8
 I1  22 mA
I
out
 5 4 26
 17 mA
73
6.5 – Kirchhoff’s Current Law
Example 6.20 – solution (cont’d)
I
in
  I out
I1  22  17 mA
I1  22  17  22  5 mA
The negative sign shows
that our assumption is
incorrect.
74
6.6 – Current Divider Rule
The current divider rule (CDR) is
used to find the current through a
resistor in a parallel circuit
75
6.6 – Current Divider Rule
General points:
 For two parallel elements of equal value, the
current will divide equally.
For parallel elements with different values,
the smaller the resistance, the greater the
share of input current.
 For parallel elements of different values, the
current will split with a ratio equal to the
inverse of their resistor values.
76
6.6 – Current Divider Rule
V
IT 
RT
V  I1R1  I 2 R2  I 3 R3  .......  I N RN
77
6.6 – Current Divider Rule
Substituting for V;
I N RN
I1 R1 I 2 R2 I 3 R3
IT 


 ....... 
RT
RT
RT
RT
78
6.6 – Current Divider Rule
 RT
I1  I T 
 R1

;

 RT 
I 3  IT  ;
 R3 
 RT 
I 2  IT  ;
 R2 
 RT
I N  IT 
 RN



79
6.6 – Current Divider Rule
Example 6.22
Determine I1, I2 and I3
I1
I2
I3
80
6.6 – Current Divider Rule
Example 6.22 – solution
1
1
RT 

1
1
1
1
1
1




R1 R2 R3 1 k 10 k 22 k
 873 
I1
I2
I3
81
6.6 – Current Divider Rule
Example 6.22 – solution (cont’d)
RT
873
I1  I T
 12 
 10.48 mA
R1
1000
RT
I 2  IT
R2
873
 12
10000
I1
I2
 1.05 mA
I3
82
6.6 – Current Divider Rule
Example 6.22 – solution (cont’d)
RT
873
I 3  IT
 12 
 0.48 mA
R3
22000
I1
I2
I3
83
6.6 – Current Divider Rule
Example 6.23
Determine I2
84
6.6 – Current Divider Rule
Example 6.23 – solution
R1 R2
RT 
R1  R2
RT
I2  Is
R2
 R1R2 
 R1 
 Is 
  Is 

 R2 R1  R2 
 R1  R2 
 4 
 6
2A

4  8
85
6.6 – Current Divider Rule
Example 6.24
Determine R1
86
6.6 – Current Divider Rule
Example 6.24 – solution
 R2 
I1  I 

 R1  R2 
IR2
R1 
 R2
I1
27  7
R1 
7  2
21
87
6.6 – Current Divider Rule
Example 6.24 – solution
Alternatively;
I 2  I  I1
 27  21  6 mA
V1  V2  I 2 R2  6  7  42 mV
V1 42
R1  
 2
I1 21
88