Download EE2003 Circuit Theory

Chapter 2
Resistive Circuits
1
Overview of Chapter 2
2.1 Series Resistors and Parallel Resistors
2.2 Voltage Divider Circuit
2.3 Current Divider Circuit
2.4 Voltage and Current Measurement
2.5 Wheatstone Bridge
2.6 Wye-Delta Transformations.
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Resistors
3
4
5
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2.1 Series and Parallel Resistors (1)
 Series Resistors
 Two or more elements are in series if they are
cascaded or connected sequentially and consequently
carry the same current.
 The equivalent resistance of any number of resistors
connected in a series is the sum of the individual
resistances.
N
Req  R1  R2      R N   Rn
n 1
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Series Resistors in
Circuit
 The potential difference (voltage) between the
terminals of the battery (V) equals the sum of the
potential differences across the resistors. KVL
V  VR1  VR 2  VR 3
 I ( R1  R2  R3 )
 IReq
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2.1 Series and Parallel Resistors (2)
 Parallel Resistor
 Two or more elements are in parallel if they are connected to
the same two nodes and consequently have the same
voltage across them.
 The equivalent resistance of a circuit with N resistors in
parallel is:
1
1
1
1


  
Req R1 R2
RN
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Parallel Resistors in Circuit
 The supply current (I) equals the sum of the currents
in the branches.  KCL
I  I1  I 2  I 3
V V V
 

R1 R2 R3
1
1
1 
 V  
 
 R1 R2 R3 
 1 

V
R 
 eq 
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2.1 Series and Parallel Resistors (3)
 Example
 Find Req in circuit below.
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2.1 Series and Parallel Resistors (3)
4×6
4. 4||6 →
4+6
2.4Ω
5. 4 + 8 + 2.4 = 14.4Ω
1. 6||3 →
6×3
6+3
2Ω
2. 1 + 5 = 6Ω
3. 2 + 2 = 4Ω
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2.1 Series and Parallel Resistors (4)
 Example
 Find Req in circuit below.
Answer: 11 ohm
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2.2 Voltage Divider Circuit (1)
 The voltage divider for N series resistors can be
expressed as
Rn
vn 
v
R1  R2      RN
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2.2 Voltage Divider Circuit (2)
 To determine the voltage across each resistor.
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2.2 Voltage Divider Circuit (3)
 Apply KVL to the circuit;
 From Ohm’s law,
v  v1  v2  0 @ v  v1  v2
v1  iR1 and v2  iR2
 From KVL, therefore,
v  iR1  iR2
 Solving equation for current,
i
v
R1  R2
R
n
 Apply Ohm’s law to determine voltage across each resistor; vn 
v
R1  R2      RN
R1
R2
v2 
v
v1 
v
R1  R2
R1  R2
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2.3 Current Divider Circuit (1)
 The total current, I is shared by the resistors in
inverse proportion to their resistances. The
current divider can be expressed as:
v iReq
in 

Rn Rn
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2.3 Current Divider Circuit (2)
 To determine the current flow on each resistor.
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2.3 Current Divider Circuit (3)
 1 
v
v

 Apply KCL, i  i1  i2  
 v
R 
R1 R2
 eq 
iR1 R2
 Using ohm’s law on each branch, therefore: v  iReq 
R1  R2
 Current flow through each branch, i 
1
iR2
R1  R2
iR1
i2 
R1  R2
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Example (voltage divider)
 Determine vo and io in the circuit below.
Answer: Vo=4V, io=4/3A
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2.4 Voltage and Current Measurement
• An ammeter is an instrument designed to measure
current and is placed in series with the circuit
element.
• A voltmeter is an instrument designed to measure
voltage and is placed in parallel with the element.
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2.5Wheatstone Bridge
 A Wheatstone bridge circuit is an accurate
device for measuring resistance
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 Under balance condition where no current flow
between BD,
VAD  VAB or I1 R1  I 2 R2 and
VDC  VBC or I 3 R3  I 4 R4
 Current in each resistance arm,
I1  I 3 and I 2  I 4
 Therefore,
R2 R4

R1 R3
R2 R3
R4 
R1
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2.6 Wye-Delta Transformations
• Wye-delta transformation occurs when the resistors are
neither in parallel or in series.
• This circuit can be simplified to a three-terminal equivalent
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Wye-Delta Transformations II
 Two topologies can be interchanged:
 Wye (Y) or tee (T) networks
 Delta (Δ) or pi (Π) networks
 Transforming between these two
topologies often makes the solution
of a circuit easier
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Wye-Delta Transformations III
 The superimposed wye and
delta circuits shown here will
used for reference
 The delta consists of the outer
resistors, labeled a,b, and c
 The wye network are the
inside resistors, labeled 1,2,
and 3
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Y-Δ Transformations
Rc
Rb
Ra
Ra 
R1 R2  R2 R3  R3 R1
R1
Rb 
R1 R2  R2 R3  R3 R1
R2
Rc 
R1 R2  R2 R3  R3 R1
R3
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Δ - Y Transformations
R1 
Rb Rc
( Ra  Rb  Rc )
R2 
Rc Ra
( Ra  Rb  Rc )
Ra Rb
R3 
( Ra  Rb  Rc )
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Example 1
 Transform Y circuit to Δ circuit
Ra = 15
Rb = 10
Rc = 25
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Example #2
 Find the current and power supplied by the 40 V sources
in the circuit shown below.
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Example #2
Solution:
 We can find this equivalent resistance easily after
replacing either the upper Δ (100Ω, 125Ω, 25Ω) or the
lower Δ (40Ω, 25Ω, 37.5Ω) with its equivalent Y.
 We choose to replace the upper Δ. Thus,
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Example #2
100 125
R1 
 50
250
125  25
R2 
 12.5
250
100  25
R3 
 10
250
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 Substituting the Y-resistor into the circuit,
 Equivalent resistor, Req:
50  50
Req  55 
 80
100
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 Simplified circuit:
 Hence, current and power values are:
V 40
i 
 0.5 A
R 80
p  V * I  40  0.5  20W
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Another option:
100
Ra
Rb
Rb
Rc
40
Rc
Another option is transform from Why to Delta..
You can try this as well.
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Example #3
(a) Find no load value of vo.
(b) Find vo when RL = 150 kΩ
(c) How much power is dissipated in the 25 kΩ resistor
if the load terminals are short-circuited ?
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Example #3
a) Find no load value of vo
R
v

 Apply ohm law: R  R v
1
1
1
2
 75k 
v0  200
  150V
 75k  25k 
b) Find vo when RL = 150 kΩ
75k 150k
Req 
 50k
75k  150k
 50k 
v0  200
  133.33V
 50k  25k 
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Example #3
c) How much power is dissipated in the 25 kΩ
resistor if the load terminals are short-circuited ?
V 200
3
I 
 8 10 A
R 25k
3
P  VI  (200)(8 10 )
 1.6W
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Example #4
• Find the power dissipated in the 6 Ω
resistor.
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Example #4
Solution:
Req
 Equivalent resistance
Req  ( 4 6 )  1.6  4
 current io,
iR2
i1 
R1  R2
iR1
i2 
R1  R2
 16 
i0  10
  8A
 16  4 
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Example #4
 Note that io is the current in the 1.6Ω resistor.
 Use current divider to get current in the 6Ω
resistor,
i1 
i0
iR2
R1  R2
i2 
iR1
R1  R2
 4 
i6  8
  3.2 A
46
 Then the power dissipated by the resistor is
P  I R  (3.2) (6)  61.44W
2
2
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Example #5
• Find the voltage of vo and vg.
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Example #5 - Solution
step 1  simplified the circuit
1)60 30  20
2)20  30  50
3)50  25  75
Current in resistor 30Ω (at Vo) is:
i1 
i30 
iR2
R1  R2
i2 
iR1
R1  R2
(25)(75)

 15 A
50  75
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 Voltage v0 where i0 is 15A and R0 is 20
v0  (15)(20)  300V
 Total voltage at the resistor, Vr:
v0  30i30  300  450
 750V
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V12
Vr
 Voltage vg
v g  v12  vr
v g  12(25)  750
v g  1050V
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Example #6
 Find the current of ig and io in the circuit
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Example #6
• Find the current of ig and io in the circuit.
Solution:
• Equivalent resistance:
5 20  4
4  6  10 15  12  12  40
10 40  8
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Example #6
 The current values,
125
ig 
 12.5 A
82
i6  
( 40)(12.5)
 10 A
(15  12  13)  (6  4)
iR2
i1 
R1  R2
 Thus,
iR1
i2 
R1  R2
(5)(10)
i0 
 2A
20  5
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Exercise 1
 Determine the value of io
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Exercise 2
 Find i and Vo
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Exercise 3
 Calculate the value of current; I.
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