Basic Laws of Electric Circuits
Equivalent Resistance
Lesson 5
Basic Laws of Circuits
Equivalent Resistance:
We know the following for series resistors:
R1
R2
. . .
Req
RN
. . .
Figure 5.1: Resistors in series.
Req = R1 + R2 + . . . + RN
1
Basic Laws of Circuits
Equivalent Resistance:
We know the following for parallel resistors:
. . .
Req
R1
RN
R2
. . .
Figure 5.2: Resistors in parallel.
1
1
1
1


 . . . 
Req
R1 R2
RN
2
Basic Laws of Circuits
Equivalent Resistance:
For the special case of two resistors in parallel:
Req
R1
R2
Figure 5.3: Two resistors in parallel.
3
R1 R2
Req 
R1  R2
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
By combination we mean we have a mix of series and
Parallel. This is illustrated below.
R1
Req
R3
R2
R4
R5
Figure 5.4: Resistors In Series – Parallel Combination
To find the equivalent resistance we usually start at
the output of the circuit and work back to the input.
4
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
R1
R3
R2
Req
Rx
R4 R5
Rx 
R4  R5
R1
R eq
5
R2
Ry
Figure 5.5: Resistance reduction.
R y  Rx  R3
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
R1
RZ
R eq
R eq
6
RZ 
R2 RY
R2  RY
Req  RZ  R1
Figure 5.6: Resistance reduction, final steps.
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
It is easier to work the previous problem using numbers than to
work out a general expression. This is illustrated below.
Example 5.1: Given the circuit below. Find Req.
10 
Req
8
10 
3
Figure 5.7: Circuit for Example 5.1.
7
6
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.1: Continued . We start at the right hand side
of the circuit and work to the left.
10 
Req
10 
8
10 
2
R eq
Figure 5.8: Reduction steps for Example 5.1.
Ans:
8
Req  15 
5
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.2: Given the circuit shown below. Find Req.
6
12 
c
Req
10 
b
4
d
Figure 5.9: Diagram for Example 5.2.
9
a
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.2: Continued.
6
12 
c
Req
10 
b
a
4
d
c
Fig 5.10: Reduction
steps.
10
Req
4
12 
b
6
10 
d, a
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.2: Continued.
c
Req
4
12 
b
6
10  resistor
shorted out
10 
d, a
Req
Fig 5.11: Reduction
steps.
11
4
6
12 
Basic Laws of Circuits
Equivalent Resistance:
Resistors in combination.
Example 5.2: Continued.
4
Req
Fig 5.12: Reduction
steps.
12
Req
6
4
12 
4
Extension of Resonant Circuits
What can be learned from this example?
 wr does not seem to have much meaning in this problem.
What is wr if R = 3.99 ohms?
 Just because a circuit is operated at the resonant frequency
does not mean it will have a peak in the response at the
frequency.
 For circuits that are fairly complicated and can resonant,
It is probably easier to use a simulation program similar to
Matlab to find out what is going on in the circuit.
14
Introduction To Resonant
Circuits
15
Resonance In Electric Circuits

Any passive electric circuit will resonate if it has an inductor
and capacitor.

Resonance is characterized by the input voltage and current
being in phase. The driving point impedance (or admittance
is completely real when this condition exists.

In this presentation we will consider (a) series resonance, and
(b) parallel resonance.
16
Series Resonance
Consider the series RLC circuit shown below.
V = VM 0
R
L
+
V _
C
I
The input impedance is given by:
1
Z  R  j ( wL 
)
wC
The magnitude of the circuit current is;
I | I |
Vm
R 2  ( wL 
1 2
)
wC
17
Series Resonance
Resonance occurs when,
1
wL 
wC
At resonance we designate w as wo and write;
1
wo 
LC
This is an important equation to remember. It applies to both ser
And parallel resonant circuits.
18
Series Resonance
The magnitude of the current response for the series resonance c
is as shown below.
|I|
Vm
R
Vm
2R
Half power point
w1 wo w2
w
Bandwidth:
BW = wBW = w2 – w1
19
Series Resonance
The peak power delivered to the circuit is;
2
V
P m
R
The so-called half-power is
Vm
I

given when .
2R
We find the frequencies, w1 and w2, at which this half-power
occurs by using;
1 2
2 R  R  ( wL 
)
wC
2
20
Series Resonance
After some insightful algebra one will find two frequencies at which
the previous equation is satisfied, they are:
2
R
1
 R 
w1  
 
 
2L
 2 L  LC
an
d
2
R
1
 R 
w2 
 


2L
 2 L  LC
The two half-power frequencies are related to the resonant frequ
wo  w1w2
21
Series Resonance
The bandwidth of the series resonant circuit is given by;
BW  wb  w2  w1 
R
L
We define the Q (quality factor) of the circuit as;
wo L
1
1 L
Q


 
R
wo RC R  C 
Using Q, we can write the bandwidth as;
wo
BW 
Q
These are all important relationships.
22
Series Resonance
An Observation:
If Q > 10, one can safely use the approximation;
BW
w1  wo 
2
and
BW
w2  wo 
2
These are useful approximations.
23
Series Resonance
An Observation:
By using Q = woL/R in the equations for w1and w2 we have;
2
 1



1
 
 1
w1  wo 

 2Q

 2Q 


and
2
 1



1
w2  wo 
 
 1

2Q 
 2Q




24
Series Resonance
In order to get some feel for how the numerical value of Q influence
the resonant and also get a better appreciation of the s-plane, we co
the following example.
It is easy to show the following for the series RLC circuit.
1
s
1
I ( s)
L


V ( s) Z ( s) s 2  R s  1
L
LC
In the following example, three cases for the about transfer functio
will be considered. We will keep wo the same for all three cases.
The numerator gain,k, will (a) first be set k to 2 for the three cases
(b) the value of k will be set so that each response is 1 at resonance
25
Series Resonance
An Example Illustrating Resonance:
The 3 transfer functions considered are:
Case 1:
ks
s 2  2s  400
Case 2:
ks
s 2  5s  400
Case 3:
ks
s 2  10s  400
26
Series Resonance
An Example Illustrating Resonance:
The poles for the three cases are given below.
Case 1:
s 2  2s  400  ( s  1  j19.97)( s  1  j19.97)
Case 2:
s 2  5s  400  ( s  2.5  j19.84)( s  2.5  j19.84)
Case 3:
s 2  10s  400  ( s  5  j19.36)( s  5  j19.36)
27
Series Resonance
Comments:
Observe the denominator of the CE equation.
R
1
s  s
L
LC
2
Compare to actual characteristic equation for Case 1:
s  2s  400
2
w  20
wo  400
2
R
BW   2
L
rad/sec
rad/sec
wo
Q
 10
BW
28
Series Resonance
Poles and Zeros In the s-plane:
( 3)
(2)
x
x
jw axis
(1)
x
20
s-plane
 axis
-5
Note the location of the poles
for the three cases. Also note
there is a zero at the origin.
x
( 3)
0
-1
-2.5
x
(2)
0
x
-20
(1)
29
Series Resonance
Comments:
The frequency response starts at the origin in the s-plane.
At the origin the transfer function is zero because there is a
zero at the origin.
As you get closer and closer to the complex pole, which
has a j parts in the neighborhood of 20, the response starts
to increase.
The response continues to increase until we reach w = 20.
From there on the response decreases.
We should be able to reason through why the response
has the above characteristics, using a graphical approach.
30
Series Resonance
Matlab Program For The Study:
% name of
% written
%CASE ONE
K = 2;
num1 = [K
den1 = [1
program is freqtest.m
for 202 S2002, wlg
DATA:
grid
H1 = bode(num1,den1,w);
magH1=abs(H1);
0];
2 400];
H2 = bode(num2,den2,w);
magH2=abs(H2);
num2 = [K 0];
den2 = [1 5 400];
H3 = bode(num3,den3,w);
magH3=abs(H3);
num3 = [K 0];
den3 = [1 10 400];
plot(w,magH1, w, magH2, w,magH3)
grid
xlabel('w(rad/sec)')
ylabel('Amplitude')
gtext('Q = 10, 4, 2')
w = .1:.1:60;
31
Series Resonance
Program Output
1
0.9
Q = 10, 4, 2
0.8
Amplitude
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
10
20
30
w(rad/sec)
40
50
60
32
Series Resonance
Comments: cont.
From earlier work:
2
 1



1
w1 , w2  wo 
 
 1

2Q 
 2Q




With Q = 10, this gives;
w1= 19.51 rad/sec,
w2 = 20.51 rad/sec
Compare this to the approximation:
w1 = w0 – BW = 20 – 1 = 19 rad/sec, w2 = 21 rad/sec
So basically we can find all the series resonant parameters
if we are given the numerical form of the CE of the transfer
function.
33
Series Resonance
Next Case: Normalize all responses to 1 at wo
1
0.9
Q = 10, 4, 2
0.8
Amplitude
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
10
20
30
w(rad/sec)
40
50
60
34
Series Resonance
Three dB Calculations:
Now we use the analytical expressions to calculate w1 and w2.
We will then compare these values to what we find from the
Matlab simulation.
Using the following equations with Q = 2,
1

 1 
w ,w w  w 
 1
 2Q 
 2Q

2
1
2
o
o
we find,
w1 = 15.62 rad/sec
w2 = 21.62 rad/sec
35
Series Resonance
Checking w1 and w2
(cut-outs from the simulation)
w1
15.3000
15.4000
15.5000
15.6000
15.7000
15.8000
0.6779
0.6871
0.6964
0.7057
0.7150
0.7244
w2
25.3000
25.4000
25.5000
25.6000
25.7000
25.8000
25.9000
0.7254
0.7195
0.7137
0.7080
0.7023
0.6967
0.6912
This verifies the previous calculations.
Now we shall look at Parallel Resonance.
36
Parallel Resonance
Background
Consider the circuits shown below:
V
I
R
L
1
1 
I  V   jwC 

R
jwL


C
L
R
V
C
I

1 
V  I  R  jwL 

jwC


37
Series Resonance
Duality
1
1 
I  V   jwC 

R
jwL



1 
V  I  R  jwL 

jwC


We notice the above equations are the same provided:
I
V
R
1
R
L
C
If we
we make
makethe
theinner-change,
inner-change,
If
thenone
oneequation
equationbecomes
becomes
then
the same
sameas
asthe
theother.
other.
the
For such
suchcase,
case,we
wesay
saythe
theone
one
For
circuitisisthe
thedual
dualofofthe
theother.
other.
circuit
38
Parallel Resonance
Background
What
isisis
that
for
all
the
equations
we
Whatthis
What
this
thismeans
means
means
that
that
forfor
allall
thethe
equations
equations
wehave
we
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derived
resonant
circuit,
we
use
derivedfor
derived
for
forthe
the
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parallel
parallel
resonant
resonant
circuit,
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wecan
we
cancan
useuse
for
circuit
provided
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make
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for
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make
the
thesubstitutions:
the
substitutions:
substitutions:
R
replaced be
1
R
L replaced by
C
C replaced by
L
39
Parallel Resonance
Parallel Resonance
1
LC
w 
O
Q
Series Resonance
O
wL
R
Q  w RC
O
o
R
BW  ( w  w )  w 
L
2
1
LC
w 
1
1
BW  w 
RC
ww
,w
1
2
BW
BW
 R
1 
 R
w ,w  
   

2
L
2
L
LC




 1
1 
 1 
w ,w  
 
 

 2 RC  LC 
 2 RC
1

 1 
w ,w  w   
 1
 2Q 
 2Q

1

 1 
w ,w  w   
 1
 2Q 
 2Q

2
1
2
2
1
2
o
2
1
2
2
1
2
o
40
Resonance
Example 1:Determine the resonant frequency for the circuit belo
1
jwL( R 
)
( w LRC  jwL)
jwC
Z 

1
(
1

w
LC
)

jwRC
R  jwL 
jwC
2
IN
2
At resonance, the phase angle of Z must be equal to zero.
41
Resonance
( w LRC  jwL)
(1  w LC )  jwRC
Analysis
2
2
For zero phase;
wL
wRC

( w LCR ) (1  w LC
2
This gives;
2
w LC  w R C 1
2
2
2
2
or
1
w
( LC  R C )
o
2
2
42
Parallel Resonance
Example 2:
A series
parallel
RLC
RLC
resonant
resonant
circuit
circuit
has
has
a resonant
a resonant
frequency
frequency
admittanc
admitta
2x10-2 S(mohs). The Q of the circuit is 50, and the resonant freque
10,000 rad/sec. Calculate the values of R, L, and C. Find the halffrequencies and the bandwidth.
First, R = 1/G = 1/(0.02) = 50 ohms.
wL
, we solvefor L, knowing Q, R, and wo to
Second, fromQ 
R
O
find L = 0.25 H.
Q
50

100  F
Third, we can useC 
w R 10,000 x50
O
43
Parallel Resonance
Example 2: (continued)
4
w 1x10

 200 rad / sec
Fourth: We can usew 
Q
50
o
BW
and
Fifth: Use the approximations;
w1 = wo - 0.5wBW = 10,000 – 100 = 9,900 rad/sec
w2 = wo - 0.5wBW = 10,000 + 100 = 10,100 rad/sec
44
Extension of Series Resonance
Peak Voltages and Resonance:
+
VR _
+
VL
_
L
R
+
+
VS
_
I
C
_
VC
We know the following:
1
 When w = wo =LC , VS and I are in phase, the driving point impeda
is purely real and equal to R.
 A plot of |I| shows that it is maximum at w = wo. We know the s
equations for series resonance applies: Q, wBW, etc.
45
Extension of Series Resonance
Reflection:
 A question that arises is what is the nature of VR, VL, and VC? A
reflection shows that VR is a peak value at wo. But we are not sur
about the other two voltages. We know that at resonance they ar
and they have a magnitude of QxVS.
 Irwin shows that the frequency at which the voltage across the ca
is a maximum is given by;
wmax  wo 1 
1
2Q 2
 The above being true, we might ask, what is the frequency at whi
voltage across the inductor is a maximum?
We answer this question by simulation
46
Extension of Series Resonance
Series RLC Transfer Functions:
The following transfer functions apply to the series RLC circuit.
1
VC ( s )
LC

VS ( s ) s 2  R s  1
L
LC
VL ( s )
s2

VS ( s ) s 2  R s  1
L
LC
R
s
VR ( s )
L

VS ( s ) s 2  R s  1
L
LC
47
Extension of Series Resonance
Parameter Selection:
We select values of R, L. and C for this first case so that Q = 2 and
wo = 2000 rad/sec. Appropriate values are; R = 50 ohms, L = .05 H
C = 5F. The transfer functions become as follows:
VC
4 x106
 2
VS s  1000s  4 x106
VL
s2
 2
VS s  1000s  4 x106
VR
1000s
 2
VS s  1000s  4 x106
48
Extension of Series Resonance
Matlab Simulation:
% program is freqcompare.m
% written for 202 S2002, wlg
numC = 4e+6;
denC = [1 1000 4e+6];
numL = [1 0 0];
denL = [1 1000 4e+6];
numR = [1000 0];
denR = [1 1000 4e+6];
grid
HC = bode(numC,denC,w);
magHC = abs(HC);
HL = bode(numL,denL,w);
magHL = abs(HL);
HR = bode(numR,denR,w);
magHR = abs(HR);
plot(w,magHC,'k-', w, magHL,'k--', w, magHR, 'k:')
grid
w = 200:1:4000;
grid
HC = bode(numC,denC,w);
magHC = abs(HC);
xlabel('w(rad/sec)')
ylabel('Amplitude')
title(' Rsesponse for RLC series circuit, Q =2')
gtext('VC')
gtext('VL')
gtext(' VR')
49
Exnsion of Series Resonance
Simulation Results
Rsesponse for RLC series circuit, Q =2
2.5
Q=2
2
Amplitude
VC
VL
1.5
1
VR
0.5
0
0
500
1000
1500
2000
2500
w(rad/sec)
3000
3500
4000
50
Exnsion of Series Resonance
Analysis of the problem:
Given the previous circuit. Find Q, w0, wmax, |Vc| at wo, and |Vc|
+
VR _
+
R=50 
VL
_
L=5 mH
+
+
VS
Solution:
C=5 F
I
_
_
VC
1
1
w 

 2000 rad / sec
50 x10 x5 x10
LC
2
O
6
3
2
w L 2 x10 x5 x10

2
Q
R
50
O
51
Exnsion of Series Resonance
Problem Solution:
w
MAX
1
 w 1
 0.9354w
2Q
O
2
o
| V | at w  Q | V |  2 x1  2 volts ( peak )
R
O
| V | at w
C
MAX
S
Qx | V |
2


 2.066 volts  peak ) 
1
0.968
1
4Q
S
2
Now check the computer printout.
52
Exnsion of Series Resonance
Problem Solution (Simulation):
1.0e+003 *
1.8600000
1.8620000
1.8640000
1.8660000
1.8680000
1.8700000
1.8720000
1.8740000
1.8760000
1.8780000
1.8800000
1.8820000
1.8840000
0.002065141
0.002065292
0.002065411
0.002065501
0.002065560
0.002065588
0.002065585
0.002065552
0.002065487
0.002065392
0.002065265
0.002065107
0.002064917
Maximum
53
Extension of Series Resonance
Simulation Results:
Rsesponse for RLC series circuit, Q =10
12
Q=10
10
Amplitude
8
VC
VL
6
4
2
VR
0
0
500
1000
1500
2000
2500
w(rad/sec)
3000
3500
4000
54
Exnsion of Series Resonance
Observations From The Study:
 The voltage across the capacitor and inductor for a series RLC c
is not at peak values at resonance for small Q (Q <3).
 Even for Q<3, the voltages across the capacitor and inductor are
equal at resonance and their values will be QxVS.
 For Q>10, the voltages across the capacitors are for all practical
purposes at their peak values and will be QxVS.
 Regardless of the value of Q, the voltage across the resistor
reaches its peak value at w = wo.
 For high Q, the equations discussed for series RLC resonance
can be applied to any voltage in the RLC circuit. For Q<3, this
55
is not true.
Extension of Resonant Circuits
Given the following circuit:
R
+
C
I
_
+
V
L
_
 We want to find the frequency, wr, at which the transfer funct
for V/I will resonate.
 The transfer function will exhibit resonance when the phase a
between V and I are zero.
56
Extension of Resonant Circuits
The desired transfer functions is;
V (1/ sC )( R  sL)

I
R  sL  1/ sC
This equation can be simplified to;
V
R  sL

I LCs 2  RCs  1
With s
jw
V
R  jwL

I (1  w2 LC )  jwR
57
Extension of Resonant Circuits
Resonant Condition:
For the previous transfer function to be at a resonant point,
the phase angle of the numerator must be equal to the phase angle
of the denominator.
num  dem
or,
 wL 
 num  tan 

 R ,
1
 wRC 
 den  tan 
 .
2
w
LC

(1
)


1
Therefore;
wL
wRC

R (1  w2 LC )
58
Extension of Resonant Circuits
Resonant Condition Analysis:
Canceling the w’s in the numerator and cross multiplying gives,
L(1  w2 LC )  R2C or
This gives,
w2 L2C  L  R2C
1 R2
wr 
 2
LC L
Notice that if the ratio of R/L is small compared to 1/LC, we hav
wr  wo 
1
LC
59
Extension of Resonant Circuits
Resonant Condition Analysis:
What is the significance of wr and wo in the previous two equations
Clearly wr is a lower frequency of the two. To answer this question
the following example.
Given the following circuit with the indicated parameters. Write a
Matlab program that will determine the frequency response of the
transfer function of the voltage to the current as indicated.
R
+
C
I
_
+
V
L
_
60
Extension of Resonant Circuits
Resonant Condition Analysis: Matlab Simulation:
We consider two cases:
Case 1:
R = 3 ohms
C = 6.25x10-5 F
L = 0.01 H
wr= 2646 rad/sec
Case 2:
R = 1 ohms
C = 6.25x10-5 F
L = 0.01 H
wr= 3873 rad/sec
For both cases,
wo = 4000 rad/sec
61
Extension of Resonant Circuits
Resonant Condition Analysis: Matlab Simulation:
The transfer functions to be simulated are given below.
Case 1:
V
0.001s  3

I 6.25x108 s 2  1.875x107  1
Case 2:
V
0.001s  1

I 6.25x108 s 2  6.25x105  1
62
Extension of Resonant Circuits
Rsesponse for Resistance in series with L then Parallel with C
18
16
14
R=1 ohm
Amplitude
12
10
2646 rad/sec
8
6
R= 3 ohms
4
2
0
0
1000
2000
3000
4000
5000
w(rad/sec)
6000
7000
8000
63
Extension of Resonant Circuits
What can be learned from this example?
 wr does not seem to have much meaning in this problem.
What is wr if R = 3.99 ohms?
 Just because a circuit is operated at the resonant frequency
does not mean it will have a peak in the response at the
frequency.
 For circuits that are fairly complicated and can resonant,
It is probably easier to use a simulation program similar to
Matlab to find out what is going on in the circuit.
64
Basic Electric Circuits
Wye to Delta Transformation:
You are given the following circuit. Determine Req.
I
9
10 
V
+
_
5
10 
Req
8
4
Figure 5.1: Diagram to start wye to delta.
13
Basic Electric Circuits
Wye to Delta Transformation:
You are given the following circuit. Determine Req.
I
9
10 
V
+
_
5
10 
Req
8
4
Figure 5.13: Diagram to start wye to delta.
14
Basic Electric Circuits
Wye to Delta Transformation:
I
9
10 
V
+
_
5
10 
Req
8
4
We cannot use resistors in parallel. We cannot use
resistors in series. If we knew V and I we could solve
15
V
Req =
I
There is another way to solve the problem without solving
for I (given, assume, V) and calculating Req for V/I.
Basic Electric Circuits
Wye to Delta Transformation:
Consider the following:
a

a

Ra
R1
Rc
c 
R2
Rb
 b
c 
 b
R3
(a) wye configuration
(b) delta configuration
Figure 5.14: Wye to delta circuits.
We equate the resistance of Rab, Rac and Rca of (a)
to Rab , Rac and Rca of (b) respectively.
16
Basic Electric Circuits
Wye to Delta Transformation:
Consider the following:
a

a

Ra
R1
Rc
c 
R2
Rb
 b
c 
 b
R3
(a) wye configuration
(b) delta configuration
R2(R1 + R3)
Eq 5.1
Rab = Ra + Rb =
R1 + R2 + R3
R1(R2 + R3)
Rac = Ra + Rc =
Eq 5.2
R1 + R2 + R3
R3(R1 + R2)
17
Rca = Rb + Rc =
R1 + R2 + R3
Eq 5.3
Basic Electric Circuits
Wye to Delta Transformation:
Consider the following:
a

a

Ra
R1
Rc
R2
Rb
c 
 b
c 
 b
R3
(a) wye configuration
18
(b) delta configuration
Ra 
R1 R2
R1  R2  R3
R1 
Ra Rb  Rb Rc  Rc Ra
Rb
Eq 5.4
Rb 
R2 R3
R1  R2  R3
R2 
Ra Rb  Rb Rc  Rc Ra
Rc
Eq 5.5
Rc 
R1 R3
R1  R2  R3
R3 
Ra Rb  Rb Rc  Rc Ra
Ra
Eq 5.6
Basic Electric Circuits
Wye to Delta Transformation:
Observe the following:
Go to wye
Go to delta
Ra Rb  Rb Rc  Rc Ra
Rb
Eq 5.4
R2 R3
Rb 
R1  R2  R3
R R  Rb Rc  Rc Ra
R2  a b
Rc
Eq 5.5
R1 R3
Rc 
R1  R2  R3
R R  Rb Rc  Rc Ra
R3  a b
Ra
Eq 5.6
Ra 
R1 R2
R1  R2  R3
R1 
We note that the denominator for Ra, Rb, Rc is the same.
We note that the numerator for R1, R2, R3 is the same.
We could say “Y” below: “D”
19
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.3: Return to the circuit of Figure 5.13 and find Req.
I
9
a
10 
V
+
_
Req
5
10 
c
8
b
4
Convert the delta around a – b – c to a wye.
20
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.3: continued
9
2
Req
4
2
8
4
Figure 5.15: Example 5.3 diagram.
It is easy to see that Req = 15 
21
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.4: Using wye to delta. The circuit of 5.13
may be redrawn as shown in 5.16.
9
a
10 
Req
c
5
10 
8
4
b
Figure 5.16: “Stretching” (rearranging) the circuit.
Convert the wye of a – b – c to a delta.
22
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.4: continued
9
9
a
10 
7.33 
27.5 
Req
c
8
a
11 
Req
c
22 
11 
5.87 
b
(a)
b
(b)
Figure 5.17: Circuit reduction of Example 5.4.
23
Basic Electric Circuits
Wye to Delta Transformation:
Example 5.4: continued
9
Req
13.2 
11 
Figure 5.18: Reduction of Figure 5.17.
Req = 15 
This answer checks with the delta to wye solution earlier.
24
Basic Laws of Circuits
circuits
End of Lesson 5
Equivalent Resistance
Self Inductance
Solenoid Flux
• Coils of wire carrying current
generate a magnetic field.
NI,
l
B
– Strong field inside solenoid
F
• If the current increases then the
magnetic field increases.
– Increased magnetic flux
B
• The magnetic flux through all
coils depends on the coil area
and the number of turns.
0 NI
l
F  NAB 
0 N 2 A
l
I
Inductance
• The ratio of magneticFflux to current is the
L
inductance.
I
• Inductance is measured in henrys.
– 1 H = 1 T m2 / A0 N 2 A 0 N 2r 2
L

l = 1 Vl / A / s
– More common, 1 H
Coil Length
• An inductor is made by
wrapping a single layer of wire
around a 4.0-mm diameter
cylinder. The wire is 0.30 mm
in diameter.
• What coil length is needed to
have an inductance of 10 H?
• The formula is based on the
2
0 N 2of
rthe
radius and length
coil.
L
l
– Radius r = 2.0 x 10-3 m
– Turns N = l / d
2 in
2
• Substitute
0 (l for
/ d )N
r 2 theformula.
0lr
L

l
d2
d 2L
l
 0.057 m
0r 2
Electric Inertia
• A changing current will create a
changing flux.
Increasing I,
F
B
• Faraday’s law states that the
changing flux will create an
emf.
– Direction from Lenz’s law
• The emf acts to oppose the
change in flux.



Decreasing I,
F
– Inertial response
B



Back-Emf
• Motors have internal coils.
• The self-inductance will oppose
a change in current by creating
an emf.
• This back-emf is responsible for
excess power draw when a
motor starts.
Induced EMF
• Faraday’s law gives the
magnitude of the induced emf.
– Depends on rate of change
• The definition of inductance
gives a relationship between
voltage and current.
– More useful in circuits
• Inductive elements in a circuit
act like batteries.
– Stabilizes current
F M
 
t
I
  L
t
Mutual Inductance
• The definition of inductance
applies to transformers.
VA
R
NA
– Mutual inductance vs selfinductance
F M
I
M
  NB
t
t
NB
F M
VB   N B
t
• Mutual inductance applies to
both windings.
Stored Energy
• Electrical power is voltage times current.
– True for emf from inductance
– Average current isapproximately
1 I one half
I
I av  L
I
Pav  I av  L
maximum
t
2 t
– Use one half to get average power
1 2
U  Pavt  LI
2
• Magnetic energy is stored in a magnetic
Energy Density
• The energy density in a
solenoid is based on its volume.
1
U
u B  2  0 n 2 I 2
r l 2
• The energy density can be
expressed in terms of the
magnetic field.
B  0 nI
1
u B  0 B 2
2
• The energy density in a
capacitor is based on its
volume.
2
1
CV
1 K 0 2

uE 
V
2
Ad
2 d
2
• The energy density can be
expressed in terms of the
electric
V field.
1
E
d
uE 
2
K 0 E 2
next