### Chapter 3

```Chapter 3
EGR 271 – Circuit Theory I
1
Reading Assignment: Chapter 3 in Electric Circuits, 9th Edition by Nilsson
Chapter 3: Analysis of simple resistive circuits
Series Resistance
Two or more resistors in series can be replaced by a single equivalent
resistance, Req, as shown below.
I
R1
R2
R3
I
...
+
V
_
Use KVL to show that: Req = R1 + R2 +    + R N
RN

+
V
_
Req
Chapter 3
EGR 271 – Circuit Theory I
2
Example: Determine the current I in the circuit shown below.
I
3
5
8V
6V
15V
2
7
20V
4
Chapter 3
EGR 271 – Circuit Theory I
3
Parallel Resistance
Two or more resistors in parallel can be replaced by a single equivalent
resistance, Req, as shown below.
I
I
...
+
V
_
R1
R2
R3
RN

+
V
_
...
1
R

Use KCL to show that: eq
1
1
1

  
R1
R2
RN
Req
Chapter 3
EGR 271 – Circuit Theory I
4
Example: Find the equivalent resistance for the circuit shown below.
48
36
72
Example: Find the voltage V in the circuit shown below.
+
V
_
3A
120
60
7A
80
240
8A
Chapter 3
EGR 271 – Circuit Theory I
5
Special Case - 2 Resistors in Parallel
An easy formula can be used to calculate the equivalent of two resistors in
parallel.
R1
R2
Show that the general formula for parallel resistance can be simplified as
follows:
R eq 
1
1
1
1

  
R1
R2
RN
(For 2 or more resistors)
R eq
R1  R 2

R1  R 2
 product 


 sum 
(For 2 resistors only)
Chapter 3
EGR 271 – Circuit Theory I
6
Example: Find the equivalent resistance in each case using the formula for
two resistors only.
30
70
80
Equal Value Resistors
• Show the result of two equal value resistors in parallel.
• Show the result of N equal value resistors in parallel.
30
60
Chapter 3
EGR 271 – Circuit Theory I
7
Series/Parallel Combination of Resistors
Circuits can sometimes be reduced through repeated series or parallel
combinations of resistors.
Example: Find the current I in the circuit shown below.
I
100V
40
180
960
160
1.2k
240
Chapter 3
EGR 271 – Circuit Theory I
Two useful tools for analyzing resistive circuits
• Voltage Division (or the Voltage Divider Law)
• Current Division (or the Current Divider Law)
Voltage Division
• Applies to series circuits only.
• Voltage divides proportionally between series R’s with the largest R getting
the most voltage.
• Show that
 Ri 
Vi  VS 
 R 
 eq 
8
Chapter 3
EGR 271 – Circuit Theory I
9
Example: Use voltage division to find the voltage V1 in the circuit shown
40
140
below.
+
80
V1
_
240V
60
Example: Use repeated voltage division to find the voltage V2 below.
100
120
+
40V
600
80
V2
_
Chapter 3
EGR 271 – Circuit Theory I
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Example: Find the voltage V1 using repeated voltage division.
40
80
300
+
480V
240
600
900
V1
_
Chapter 3
EGR 271 – Circuit Theory I
Current Division
• Applies to parallel circuits only.
• Current divides between parallel R’s with the smallest R getting the most
current.
 R total 
• Show that:
I i  IS 

R
 i 
(general form)
11
Chapter 3
EGR 271 – Circuit Theory I
12
Example: Find the current I1 using current division.
I1
20A
40
150
600
Chapter 3
EGR 271 – Circuit Theory I
13
Current Division - Special Case: Two Resistors Only
Show that for two resistors only the general form of current division can be
expressed as follows.
R 
Ii  IS  total 
 Ri 
 R opposite 
I i  IS 

R

R
 i
2 
(general form)
(form for 2 resistors only)
Example: Find the current I1 using current division (special form for two
R’s only).
I1
12mA
100
300
Chapter 3
EGR 271 – Circuit Theory I
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Example: Find the current I1 using repeated current division.
40
200V
80
240
300
600
I1
900
Chapter 3
EGR 271 – Circuit Theory I
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Example: Find I1, V2, I3, I4, V5, and V6 in the circuit shown below.
I1
200
+
V2
_
I4
160
300
800
260
+
V6
_
200V
500
I3
2000
+
V5
_
100
420
EGR 271 – Circuit Theory I
Chapter 3
16
Bridge Circuits
One type of resistive circuit that cannot be simplified through series and/or
parallel combinations is the “bridge circuit.” A bridge circuit is shown below
(drawn twice). Study the circuit to verify that there are no series resistors and
no parallel resistors.
R0
R0
R1
R2
R1
R2
R5
R5
Vs
Vs
R3
R4
R3
R4
Chapter 3
EGR 271 – Circuit Theory I
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Delta-to-Wye (-Y) and Wye-to-Delta (Y-) Transformations
Bridge circuits contain resistors that are connected in delta () and wye (Y)
configurations. One way to analyze this circuit is to use a -Y or a Y-
transformation.
Y and  connections of resistors are shown below:
a
b
a
b
R
R1
R
2
R
R
c
R
b
3
c
c
Wye Circuit
Delta Circuit
a
Chapter 3
EGR 271 – Circuit Theory I
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If the wye and delta circuits to be equivalent, then they should provide the same
resistance between each pair of terminals (a-b, b-c, and c-a).
Development: Determine the
resistance seen at each set of terminals
and equate them as follows:
Ra-b (Delta) = Ra-b (Wye)
Rb-c (Delta) = Rb-c (Wye)
Rc-a (Delta) = Rc-a (Wye)
a
b
a
b
R
R1
R
R
R
c
2
R
b
3
c
c
Wye Circuit
Delta Circuit
a
EGR 271 – Circuit Theory I
Chapter 3
19
Solving the equations on the previous page yields the following relationships:
Y- Conversion Equations:
 -Y Conversion Equations:
R  R  R 2  R 3  R 3  R1
Ra  1 2
R1
R1 
R  R  R 2  R 3  R 3  R1
Rb  1 2
R2
R2 
R1  R 2  R 2  R 3  R 3  R1
R3
R3 
Rc 
a
Ra
Rb  Rc
 Rb  Rc
Ra
Rc  Ra
 Rb  Rc
Ra
Ra  Rb
 Rb  Rc
b
a
b
R
R1
R
2
R
R
c
R
b
3
c
c
Wye Circuit
Delta Circuit
a
Note: The equations
above must be used
along with the circuit
diagrams shown. The
labeling of the resistors
and nodes in the
diagrams is critical.
Chapter 3
EGR 271 – Circuit Theory I
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Example: Determine I in the circuit shown below using: a) -Y conversion
I
30
10
100 V
50
+
_
25
20
40
Chapter 3
EGR 271 – Circuit Theory I
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Example: Determine I in the circuit shown below using: b) Y- conversion
I
30
10
100 V
50
+
_
25
20
40
Chapter 3
EGR 271 – Circuit Theory I
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The “balanced bridge”
A bridge circuit is “balanced” when the following relationship exists:
R1R4 = R2R3
When the bridge is balanced, it can be shown that
I5 = 0
R0
R1
R2
R5
Vs
I5
R3
R4
Chapter 3
EGR 271 – Circuit Theory I
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Applications of bridge circuits
An unknown resistance can be determined using a bridge circuit as follows:
• Replace R4 with an unknown resistor
• Place an ammeter in series with R5
• Adjust R1 until the bridge is balanced (i.e., I5 = 0)
• Solve for R4
+
_
R1  R 4 = R 2  R 3
R2
R1
or
I5
R5
R3
Runknown
R unknown
R 2  R3
= R4 =
R1
A bridge circuit with resistive values, sometimes called a Wheatstone bridge,
can be used to determine unknown resistor values.
Other bridge circuits (such as the Scherring bridge) can be used to determine
unknown values of capacitors and inductors in a similar manner.
Chapter 3
EGR 271 – Circuit Theory I
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Impedance Bridge
A bridge circuits used to find unknown values of resistance, inductance, or
capacitance are sometimes called an impedance bridge.
An impedance bridge is a common piece of test equipment and will be used in
lab classes such as EGR 262 and EGR 270.
impedance bridge.
An older style impedance bridge that
involved adjusting knobs until the needle
indicated that the bridge was balanced.
Ref: http://www.testequipmentconnection.com/products/908
http://www.testequipmentconnection.com/products/914
Chapter 3 EGR 271 – Circuit Theory I
Strain gauges
Strain gauges are often attached to
surfaces to measure forces as the
surface moves (such as in the
deflection of an airplane wing or a
beam). The force can be determined
from the amount of stretch in the wire
by measuring the resistance of the
wire with a bridge circuit. As a wire
stretches note that resistance increases
since
R
 l
A
25
Chapter 3
EGR 271 – Circuit Theory I
Strain gauge mounted on a vehicle suspension system component for
fatigue and durability testing
Ref: http://blog.prosig.com/2006/05/17/fatigue-durability-testing/
26
Chapter 3
EGR 271 – Circuit Theory I
Strain gauge mounted on a human bone to study walking and running
impact
Biomechanics lab at Iowa State University
Ref: http://www.kin.hs.iastate.edu/research/labs/biomechanics_lab.htm
27
```