Download Chapter 3 Simple Resistive Circuits

Chapter 3
Simple Resistive Circuits
3.1
3.2
3.3
3.4
3.5
3.6
3.7
Resistors in Series
Resistors in Parallel
The Voltage-Divider and Current-Divider
Circuits
Voltage Division and Current Division
Measuring Voltage and Current
The Wheatstone Bridge
-to-Y (-to-T) Equivalent Circuits
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Section 3.1
Resistors in Series
2
Definition

Two elements are said to be in series if they
are connected at a single node.

By KCL, all series-connected resistors carry the
same current.
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Equivalent resistor

By (1) KVL, (2) all series resistors share a
common current is, and (3) Ohm’s law,  vs =
isR1 + isR2 +…+ isR7 = is(R1+R2+…+R7) = isReq,
 Req 
k
R
i 1
i
 R1  R2  ...  Rk .
4
Section 3.2
Resistors in Parallel
5
Definition

Two elements are in parallel if they are
connected at a single node pair.

By KVL, all parallel-connected elements have
the same voltage across their terminals.
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Equivalent resistor

By (1) KCL, (2) all parallel resistors share a
common voltage vs, and (3) Ohm’s law, is = vs/R1 +
vs/R2 +…+ vs/R4 = vs(1/R1+1/R2+…+1/R4) = vs/Req,
k
1
1
1
1
1
  
 ...  .

Req i 1 Ri R1 R2
Rk
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Comments about resistors in parallel

Req is always smaller than any resistance in
parallel connection.

The smallest resistance dominates the
equivalent value.
R1R2
Req 
R1  R2
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Section 3.3
The Voltage-Divider &
Current-Divider Circuits
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The voltage-divider circuit
By KVL, vs = iR1 + iR2,
 i = vs /(R1+R2),
R1

v1 iR1  R  R vs ,

1
2

v 2  iR1  R2 vs .

R1  R2

v1 , v2 are fractions of vs depending only on the
ratio of resistance.
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Practical concern: Load resistance
R2 RL
Req 
R2  RL
vo 
Req
R1  Req
vs
R2

vs ,
1  a R1  R2
R2
where a 
.
RL

The effect of R1 is amplified by a factor of a =
R2/RL, reducing the output voltage vo.

Large load resistance RL >> R2 is preferred.
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Example: output voltage vs. load resistance
Let R1  R2  R,  open - circuit output voltage vo  0.5vs .
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Reference: MatlabTM codes
clear
close all
% empty the variables in the working space
% close all existing figures
R2 = 1;
% resistance R2 normalized to resistance R1
RL = linspace(0.1,10,100);
vo_noload = R2/(1+R2);
vo = 1./(2+1./RL);
% load resistance normalized to R1
% output voltage normalized to vs without load
% output voltage normalized to vs with load
plot(RL,vo/vo_noload)
% plot a curve
ylim([0 1])
% y-axis is shown between 0 and 1
xlabel(['normalized load resistance, R_L/R_1'])
% label the x-axis
ylabel(['normalized output voltage, v_o/v"_o'])
% label the y-axis
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Practical concern: Tolerance of resistance
(10%: 22.5-27.5 k)
ideal voltage :
(10%: 90-110 k)
100
100
v o
25  100
 80 V
110
v o ,max 
100  83.02 V  (1  3.78% )v0 ;
22.5  110
90
v o ,min 
100  76.60 V  (1  4.25%)v o .
27.5  90
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Current-divider circuit
By Ohm’s law & resistors in parallel,
v = i1R1 = i2R2 = isReq = is[R1R2/(R1+R2)],
R2
R1
 i1
is , i 2 
is .
R1  R2
R1  R2
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Example 3.3 (1)

Q: Find the power dissipated at the 6-resistor.

Strategy: Find the current i6, then use p = i2R
to calculate the power.
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Example 3.3 (2)
Step 1: Simplifying the circuit with series-parallel
4
reductions.
2.4
io = [16/(16+4)]10 = 8 A,
i6 = [4/(6+4)]8 = 3.2 A,
 p = (3.2)2 6 = 61.44 W.
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Section 3.6
The Wheatstone Bridge
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The Wheatstone bridge

Goal: Measuring a resistor’s value.

Apparatus: Fixed-value resistors 2, variable
resistor 1, current meter 1.
Current meter
Variable
Unknown
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The working principle
Tune R3 until
iab = 0,  vab = 0,
terminals ab
become both
open and short!
Open: i1 = i3, i2 = ix …(1)
Short: i1R1 = i2R2, i3R3 = ixRx …(2)
(1) into (2): i1R3 = i2Rx …(3)
(2) R1 R2


,
(3) R3 Rx
R2
 Rx 
R3
R1
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Section 3.7
-to-Y (-to-T) Equivalent
Circuits
21
Definition of -to-Y (-to-T) transformation

Two circuits of Δ and Y configurations are
equivalent if the terminal behavior of the two
configurations are the same.  Rab, = Rab,Y; Rbc,
= Rbc,Y; Rca, = Rca,Y
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Terminal resistances
+ -
+ -
Rc ( Ra  Rb )
Rab  Rc //( Ra  Rb ) 
 R1  R2  (1)
Ra  Rb  Rc
Rbc  Ra //( Rb  Rb )  R2  R3  ( 2)
Rca  Rb //( Rc  Ra )  R3  R1  (3)
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Transformation formulas

Solving simultaneous equations (1-3),

Rb Rc
 R1  R  R  R ,
a
b
c


Rc Ra
,
 R2 
Ra  Rb  Rc


Ra Rb
.
 R3 
Ra  Rb  Rc


R1R2  R2 R3  R3 R1
,
 Ra 
R
1


R1R2  R2 R3  R3 R1
,
 Rb 
R2


R1R2  R2 R3  R3 R1
.
 Rc 
R3

(-to-Y)
(Y-to-)
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Example 3.7 (1)

Q: Find the source current i.
i=?
100  125
 50 , R2  12.5 , R3  10 .
R1 
25  100  125
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Example 3.7 (2)
i
50 
50 
Req  (5 )  (50 )  (50 // 50 )  80 ,
i  (40 V) (80 )  0.5 A.
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