Download Chapter 2-Resistive Circuit

CHAPTER 2
Resistive
Circuit
School of Computer and Communication
Engineering,
UniMAP
Prepared By:
Mr.Muhamad Sani Mustafa
EKT 101 [Electric Circuit I]: V2010/11
1
RESISTIVE CIRCUIT
 Series/parallel resistor
 Voltage divider circuit
 Current divider circuit
 Voltage and current measurement
 Wheatstone bridge
 Delta-wye (Pi-Tee) equivalent
circuit
2
SERIES/PARALLEL
RESISTOR
 Resistors in series:
R1
I
VS
R2
 V1 

RN
 V2 
 VN 

I
VS


Resistance equivalent
Req = R1 + R2 + ……….+ RN
Req
3
Current in Series Circuit
 Current in series circuit is same at all circuit
elements
I  I1  I 2  I N
VOLTAGE IN SERIES CIRCUIT
 Voltage (VT) in series circuit is the total of voltage
for each elements.
VT  V1  V2  ..  VN
4
Resistors in Parallel
IS

V
I1
I2
R1
R2
IN
RN

IS

V
Req

5
Equivalent Resistors in
Parallel:
1
1
1
1


 ............ 
Req R1 R2
RN
Req 
1
1
R1
 1
R2
 ........  1
RN
6
 Two resistors in parallel:
Req  R1 R2
Req 
1
1
R1
 1
R1R2

R1  R2
R2
7
Current in Parallel Circuit
 Currents in parallel circuit is the total of current
for each elements.
I  I1  I 2  ..  I N
VOLTAGE IN PARALLEL CIRCUIT
 Voltage (VT) in parallel circuit is same at all
circuit elements.
VT  V1  V2  VN
8
Example #1
 Find the equivalent resistor (Req) in the
circuit.
9
RESISTIVE CIRCUIT
 Series/parallel resistor
 Voltage divider circuit
 Current divider circuit
 Voltage and current measurement
 Wheatstone bridge
 Delta-wye (Pi-Tee) equivalent
circuit
10
Voltage Divider
22
2
11
 Using Ohm law, we will get:
V2  R2 I
V
I 
R1  R2
 Voltage at resistor R2:
 V 
 R2 
  V 

V2  R2 
 R1  R2 
 R1  R2 
12
RESISTIVE CIRCUIT
•
•
•
•
•
•
Series/parallel resistor
Voltage divider circuit
Current divider circuit
Voltage and current measurement
Wheatstone bridge
Delta-wye (Pi-Tee) equivalent circuit
13
Current Divider
I
V

_
I1
I2
R1
R2
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 From the Ohm’s law,
 R1R2 

V  I1R1  I 2 R2  I 
 R1  R2 
(1)
 R2 
 I
I1  
 R1  R2 
 R1 
 I
I 2  
 R1  R2 
15
RESISTIVE CIRCUIT
 Series/parallel resistor
 Voltage divider circuit
 Current divider circuit
 Voltage and current measurement
 Wheatstone bridge
 Delta-wye (Pi-Tee) equivalent
circuit
16
Voltage and Current
Measurement
 An ammeter is an instrument designed to
measure current.
 It is placed in series with the circuit element
whose current is being measured.
 An ideal ammeter has an equivalent
resistance of 0Ω and functions as a short
circuit in series with the element whose
current is being measured.
17
 A voltmeter is an instrument designed to
measure voltage.
 It is placed in parallel with the element
whose voltage is being measured.
 An ideal voltmeter has an infinite
equivalent resistance and thus functions
as an open circuit in parallel with the
element whose voltage is being measured.
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 The configurations for an ammeter and
voltmeter to measure current and voltage
R1
A
Vs


R2
V
19
RESISTIVE CIRCUIT
 Series/parallel resistor
 Voltage divider circuit
 Current divider circuit
 Voltage and current measurement
 Wheatstone bridge
 Delta-wye (Pi-Tee) equivalent
circuit
20
Wheatstone Bridge
 The Wheatstone bridge circuit is used to
precisely measure resistance of medium
values, that is in the range of 1Ω to 1MΩ.
 The bridge circuit consists of four resistors, a
dc voltage source and a detector.
21
Wheatstone Bridge
 The Wheatstone bridge circuit:
R1
R2
Vs
R3
RX
22
Wheatstone Bridge
 When the bridge is balanced:
i1  i 3
i 3R 3  i x R x
i2  ix
i1R1  i 2 R 2
 Combining these equation, gives
i1R 3  i 2 R x
23
Wheatstone Bridge
 Solving these equation, yields
R3 Rx

R1 R 2
R2
Rx 
R3
R1
24
RESISTIVE CIRCUIT
Series/parallel resistor
Voltage divider circuit
Current divider circuit
Voltage and current measurement
Wheatstone bridge
Delta-wye (Pi-Tee) equivalent circuit
25
Delta-Wye (PI-TEE)
Circuit
 If the galvanometer in Wheatstone bridge
is replace with its equivalent resistance
Rm,
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 The resistor R1, R2 and Rm (or R3, Rm and
Rx) are referred as a delta (∆)
interconnection.
 It is also referred as a pi (π) interconnection
because the ∆ can be shaped into a π
without disturbing the electrical equivalent of
the two configurations.
27
 Delta configuration
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 The resistors R1, Rm dan R3 (or R2, Rm
and Rx) are referred as a wye (Y)
interconnection because it can be
shaped to look like the letter Y.
 The Y configuration also referred as a
tee (T) interconnection.
29
 Wye configuration
30
The ∆ - Y Transformation
31
 Using series and parallel simplifications in
Δ-connected, yield
Rc ( Ra  Rb )
Rab 
 R1  R2
Ra  Rb  Rc
Ra ( Rb  Rc )
Rbc 
 R2  R3
Ra  Rb  Rc
Rb ( Rc  Ra )
Rca 
 R1  R3
Ra  Rb  Rc
32
 Using straightforward
algebraic
manipulation gives,
Rb Rc
R1 
Ra  Rb  Rc
Rc Ra
R2 
Ra  Rb  Rc
Ra Rb
R3 
Ra  Rb  Rc
33
The expression
for the three Δconnected
R
R

R
R

R
R
1
2
2
3
3
1
resistors as
Ra 
functions of three
R1
Y-connected
R1R2  R2 R3  R3 R1
resistors are
Rb 
R2
R1R2  R2 R3  R3 R1
Rc 
R3
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Example #2
 Find the current and power supplied by
the 40 V sources in the circuit shown
below.
35
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,
36
Example #2
100  125
R1 
 50
250
125  25
R2 
 12.5
250
100  25
R1 
 10
250
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 Substituting the Y-resistor into the
circuit,
38
 The equivalent circuit,
39
 Calculate the equivalent resistance,
50  50
Req  55 
 80
100
 Simplify the circuit,
40
 Then, the current and power values are,
40
i
 0.5 A
50
p  40  0.5  20W
41
Example #3
Find no load value of vo.
Find vo when RL = 150 kΩ
How much power is dissipated in the 25 kΩ
resistor if the load terminals are shortcircuited ?
42
Example #3
a)
 75k 
v0  200
  150V
 100k 
b)
75k  150k
Req 
 50k
75k  150k
 50k 
v0  200
  133.33V
 75k 
43
Example #3
c)
V 200
3
I 
 8 10 A
R 25k
2
3
p  I R  (200)(8 10 )
 1.6W
44
Example #4
 Find the power dissipated in the 6 Ω
resistor.
45
Example #4
Solution:
Equivalent resistance
Req  ( 4 6 )  1.6  4
 current io,
 16 
i0  10
  8A
 16  4 
46
Example #4
 Note that io is the current in the 1.6Ω
resistor.
 Use current divider to get current in the 6Ω
resistor,
 4
i6  8   3.2 A
 10 
 Then the power dissipated by the resistor is
p  I R  (3.2) (6)  61.44W
2
2
47
Example #5
 Find the voltage of vo and vg.
48
Example #5
Solution:
Equivalent resistance
60 30  20
 Current in resistor 30Ω
i30
(25)(75)

 15 A
125
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 Voltage v0
v0  (15)( 20)  300V
 Total voltage at the resistor
v0  30i30  300  450
 750V
50
 Voltage vg
vg  12(25)  750
v g  1050V
51
Example #6
 Find the current of ig and io in the circuit.
Solution:
 Equivalent resistance:
5 20  4
4  6  10
10 40  8
52
Example #6
 The current values,
ig
125

 12.5 A
82
i6 
( 40)(12.5)

 10 A
50
 Thus,
(5)(10)
i0 
 2A
25
53
Example #7
 Determine the value of io
54
Example #8
 Find i and Vo
55
Example #9
 Calculate the value of current; I.
56