SIMPLE RESISTIVE
CIRCUIT ANALYSIS
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3.1 Series Resistive Circuits
3.2 Parallel Resistive Circuits
3.3 Voltage-divider and Current-divider Circuits
3.4 Measuring Current and Voltage
3.5 Measuring Resistance-The Wheatstone Bridge
3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
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3.1 Series Resistive Circuits
Linear resistors (Ohm’s law)
R=ρ
l
A
ρ: resistivity of material
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3.1 Series Resistive Circuits
example of fixed resistors
n wirewound type
n carbon film type
A resistor with zero resistance is called
short circuit.
A resistor with infinite resistance is called
open circuit.
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3.1 Series Resistive Circuits
n
Color code
黑 棕 紅 橙 黃 綠 藍 紫 灰 白
0
1
2
3
4
5
6
7
8
9
金
10-1
銀
10-2
,
error 金 ± 5% , 銀 ± 10%
no color ± 20%
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3.1 Series Resistive Circuits
A:
B:
C:
D:
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yellow
violet
red
gold
47 X 102 Ω ± 5%
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3.1 Series Resistive Circuits
Red
Yellow
Green
blue
Violet
red
Yellow
Green
blue
Violet
2
4
5
6
7
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3.1 Series Resistive Circuits
+ v1 - i
a
R
1
vs
i
d
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b
1
+ v3 -
s
i
3
R
+
v2
R
2
-
i
2
c
3
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3.1 Series Resistive Circuits
High school solution
Req = R1 + R2 + R3
vs
Req
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3.1 Series Resistive Circuits
vs
i1 = i2 = i3 =
Req
v1 = R1 i1
v2 = R2 i2
v3 = R3 i3
is = -i1
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3.1 Series Resistive Circuits
+ v1 - i
a
1
i
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+ v3 -
s
i
3
N=4 nodes
+
v2
R
vs
d
b
1
R
R
2
-
i
c
2
b=4 branches
2b method
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Problem: Find 2b unknowns
i1 , i2 , i3 , is
v1 , v2 , v3, vs
( Passive sign convention )
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3.1 Series Resistive Circuits
n
Solution:
Step1. KCL for N-1=3 nodes
node a
is = -i1
node b
i1 = i2
node c
i2 = i3
node d
i3 = -is
(1)
(2)
(3)
(4)
Only N-1 KCL equations are independent !
e.g. Eq (4) can be obtained from the previous three equations
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3.1 Series Resistive Circuits
Step2. KVL equation
v1 + v2 + v3 + (-vs)= 0
(5)
Step3. Component models
v1 = R1 i1
v2 = R2 i2
v3 = R3 i3
vs = given
(6)
(7)
(8)
(9)
There are 2b (=8) equations for 2b unknowns.
Normally , there exists a unique solution.
Unless the modeling is not accurate enough.
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3.2 Parallel Resistive Circuits
is
vs
n
R1 v 1 R2
i1
i2
v 2 R3
i3
v3
High school solution
1
1
1
1
= +
+
Req R1 R2 R3
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3.2 Parallel Resistive Circuits
vs
Req
is
is = −
ik =
vs
= −(i1 + i2 + i3 )
Req
vs
, k=1,2,3
Rk
Problem : Find 2b unknowns
v1 , v2 , v3 , vs
i1 , i2 , i3 , is
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3.2 Parallel Resistive Circuits
Solution :
Step1. KCL for N-1=1 node
node a is+i1+i2+i3 = 0
Step2. KVL equations
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loop A : vs+(-v1) = 0
loop B : vs+(-v2) = 0
loop C : vs+(-v3) = 0
(1)
(2)
(3)
(4)
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3.2 Parallel Resistive Circuits
Step3. Component models
v1 = R1 i1
v2 = R2 i2
v3 = R3 i3
vs = given
(5)
(6)
(7)
(8)
There are 2b = 8 unknowns and there are 8 equations.
One can find the unique solution.
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3.3 Voltage-divider and Current-divider Circuits
n
The voltage-divider
R1
v
R1 + R2 s
R2
v2 =
v
R1 + R2 s
v1 =
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3.3 Voltage-divider and Current-divider Circuits
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The loading effect ( RL )
R2 R L
R2 + R L
Req
vo =
v
R1 + Req s
Req =
=
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R2

R 
R1  1 + 2  + R2
RL 

vs
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3.3 Voltage-divider and Current-divider Circuits
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Generalized case
R eq =
n
∑R
k
1
v j = R ji =
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Rj
v
R eq s
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3.3 Voltage-divider and Current-divider Circuits
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The current divider
1
1
1
=
+
Req R1 R2
i1
is
R1
v k = i s Req , k = 1, 2
i2
R2
v1
R2
=
i
R1 R1 + R2 s
v
R1
i2 = 2 =
i
R2 R1 + R2 s
i1 =
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3.3 Voltage-divider and Current-divider Circuits
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Generalized case
i2
i1
is
R1
v
j
n
∑
1
in
Rn
1
Rk
= is R eq
∴ ij =
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Rj
R2
1
=
R eq
ij
v j
R eq
=
i
R j
R j s
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3.4 Measuring Current and Voltage
An ammeter is an instrument designed to measure
current and must be placed in series with current
being measured.
Example : An analog ammeter based on the
d’Arsonval meter movement.
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Picture from ELECTRIC CIRCUITS
by Nilsson Riedel
8th EDITION
,
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A dc ammeter equivalent circuit
RA
The movable coil is characterized by a voltage
rating and a current rating.
e.g. a commercial meter movement is rated at
50mV and 1mA
This means when the coil current is 1mA at full-scale
position , the voltage drop across the coil is 50mV.
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A dc ammeter equivalent circuit
An analog ammeter consists of a d’Arsonval
movement in parallel with a resister RA
Example1 : A 50mV , 1mA d’Arsonval movement
is to be used in an ammeter of 1A rating.
Determine RA
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A dc ammeter equivalent circuit
Solution : At full scale rating
(1.0 A − 1mA) RA = 50mV
∴ RA =
50
Ω
999
Loading effect due to ammeter
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or
Rm =
50mV
= 0.05 Ω
1A
Rm =
50 × 50 999
= 0.05 Ω
50+ 50 999
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A dc ammeter equivalent circuit
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Analog DC Ammeter
To allow multiple ranges, shunt resistors are connected in parallel
with the movement meter.
Im =
Rn
I fs , current divider
Rn + Rm
∴ Rn =
Im
Rm
I fs - I m
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A dc ammeter equivalent circuit
Example 2 :Design an ammeter for the following
multiple ranges
(A) 0 ~ 1 A
(B) 0 ~ 100 mA
(C) 0 ~ 10 mA
Assume that Rm=50Ω and Im= 1mA for the adopted
d’Arsonval movement meter.
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A dc ammeter equivalent circuit
Ans : current divider
Im =
Rn
I fs
Rn + Rm
∴ Rn =
I m × Rm
I fs - I m
10 -3 × 50
(A) shunt resistance R1 =
; 0.05 Ω
1A - 1mA
1 × 50
50
(B) R2 =
=
= 0.505 Ω
(100 - 1)mA 99
1 × 50 50
(C) R3 =
=
= 5.556Ω
10 - 1 9
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A dc ammeter equivalent circuit
An ideal ammeter has an equivalent resistance of
0 Ω and functions as a short circuit.
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A dc voltmeter equivalent circuit
An analog voltmeter consists of a d’Arsonval
movement in series with a series resister RV .
RV
The added RV determines the full-scale
reading of the voltmeter.
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A dc voltmeter equivalent circuit
Example 3:
A 50mV , 1mA d’Arsonval movement is
used in a voltmeter of 150V rating.
Determine RV .
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A dc voltmeter equivalent circuit
Solution: From voltage divider formula
50
× 150 V
50mV=
Rv +50
∴ Rv =149950 Ω
Loading effect of the meter
Rm =149950+50=150 kΩ
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which is in parallel with the element to
be measured.
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A dc voltmeter equivalent circuit
V fs =I fs (Rn +Rm )
∴ Rn =
V fs
I fs
-Rm
The design is based on the worst case which
occurs when the full-scale current Ifs=Im flows
through the meters.
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A dc voltmeter equivalent circuit
Example 4 : Design a voltmeter for the following
multiple ranges
(A) 0 ~ 1 V
(B) 0 ~ 5 V
(C) 0 ~ 100 V
Assume the d’Arsonval movement meter has
Rm=2kΩ with full scale Ifs=100μA.
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A dc voltmeter equivalent circuit
Ans :
(A) V fs = I fs (Rm + R1 )
∴ R1 =
V fs
1
- Rm =
- 2000 = 8000Ω
I fs
100μ
5
- 2000 = 48KΩ
100μ
100
(C) R3 =
- 2000 = 998KΩ
100μ
(B) R2 =
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3.5 Measuring Resistance – The Wheatstone Bridge
The Wheatstone bridge circuit is used to measure
resistance of mediumvalues :1Ω ~ 1MΩ , with an
accuracy of about ± 0.1 %
n The Wheatstone bridge circuit
n
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3.5 Measuring Resistance – The Wheatstone Bridge
current detector : a galvanometer
RX : the unknown resister
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3.5 Measuring Resistance – The Wheatstone Bridge
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Measuring RX
Adjust R3 until zero current in the galvanometer.
i1 = i 3
, i2 = i x
V ab = V ac , V bd = V cd
⇒ i1 R1 = i2 R2 , i1 R3 = i2 RX
⇒
R1 R2
=
R3 RX
⇒ RX =
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R2
R3
R1
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3.5 Measuring Resistance – The Wheatstone Bridge
R1 and R2 : 1 , 10 , 100 , 1000 Ω
such that
R2
= 0.001 ~ 1000
R1
in decimal steps
R3 : adjustable in integral values of resistance
from 1 to 11,000 Ω .
n
Lower than 1Ω and higher than 1MΩ resistances are
difficult to measure due to thermal heating effect and
current leakage effect , respectively .
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3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
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Delta (Δ) or Pi (π) circuit
Rb
Ra
Rc
n
Wye (Y) or Tee (T) circuit
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3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
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Δ – Y transformation
Rc
Rb
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R1
Ra
R2
R3
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3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
R ab = R c // (R a + R b ) = R 1 + R 2
R bc = R a // (R b + R c ) = R 2 + R 3
R ca = R b // (R c + R a ) = R 3 + R 1
⇒ R1 =
RbRc
Ra + Rb + Rc
R2 =
RcRa
Ra + Rb + Rc
R3 =
Ra Rb
Ra + Rb + Rc
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3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
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Y to Δ transformation
Similarly , given R1 , R2 , R3 ; one can find Ra , Rb , Rc
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Ra =
R1 R 2 + R 2 R3 + R3 R1
R1
Rb =
R1 R 2 + R 2 R 3 + R3 R1
R2
Rc =
R1 R 2 + R 2 R3 + R 3 R1
R3
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3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
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Special case , if R1=R2=R3=R , then Ra=Rb=Rc=3R
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3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
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Example : Use Y – Δ transformation to find V in
the following circuit.
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3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
Solution :
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3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
Solution :
20×10 +10×5 + 5×20 35
=
Ω
20
2
350
Rb =
= 35Ω
10
350
Rc =
= 70Ω
5
Ra =
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3.6 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
70 / / 28 = 20 Ω
⇒
+
V
35 Ω
-
35
/ / 105 = 15 Ω
2
⇒
∴
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V = 2A ×
35Ω
2
= 35V
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Summary
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Objective 1 : Review series and parallel resistive
circuit solution approach.
n
Objective 2 : Recognize the 2b method by standard
mathematical formulation.
( KCL + KVL + Ohm’s Law )
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Objective 3 : Know how to use simple voltage-divider
and current- divider concepts to solve
simple circuit.
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Summary
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Objective 4 : Be able to determine the reading of
ammeters and voltmeters.
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Objective 5 : Understand how to a Wheatstone bridge
is used to measure resistance.
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Objective 6 : Know when and how to use delta-to-wye
equivalent circuits to solve a simple circuit.
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Assignment : Chapter Problems
Problem : 3.14
3.24
3.30
3.33
3.50
3.64
n Due within one week.
n
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