CHAPTERS 3
Figure 3.1-1
The circuit being designed provides an
adjustable voltage, v, to the load circuit.
Resistive Circuits
67
1
67
Section 3.3
Kirchhoff’s laws
KCL
R1=10Ω
Kirchhoff’s Current Law (KCL):
„
Node 1
the algebraic sum of the currents into/out
of a node at any instant is zero.
Node 2
_
+
+
+
_
_
I=5A
R3= 5Ω
R2= 20Ω
Kirchhoff’s Voltage Law (KVL):
„
2
the algebraic sum of the voltages around
any closed path in a circuit is zero for all
time.
Node 3
Assume passive sign convention
67
Node 1
I=5A
R1=10Ω
i1
+ v1=50v _
3
+
v2=20v
_
R1=10Ω
Node 2 +i1 - i2 - i3 = 0
Node 3 +i2 + i3 - I = 0
i2 = v2/R2
i3
+
_
LOOP 1
R2= 20Ω
_
Node 3
+V - vR1 - vR2 = 0
Use KCL and
Ohm’s Law
i3 = v3/R3
67
+
V= 5v
R3= 5Ω
v3=20v
_
Start
Node 1 +I - i1 = 0
_
+
+
i2
4
KVL
Node 2
I
R2= 20Ω
67
5
i = V/(R1 + R2)
iV = iR1 = iR2 = i
vR1 = iR1 = VR1 /(R1 + R2)
+V = iR1 + iR2
V = i(R1 + R2)
vR2 = iR2 = VR2/(R1 + R2)
67
6
steps taken
PROBLEM SOLVING METHOD
va
+
Ra
ia
ivs
+
_
vs
vb
_ node2
+
Rb
ib
+
ic R
c
loop1
Apply P.S.C. to passive elements.
Show current direction at voltages
sources.
Show voltage direction at current
sources.
Name nodes and loops.
Name elements and sources.
Name currents and voltages.
node3
_
+
vc
loop2
_
vis
is
_
node4
67
7
WRITE THE
KCL EQUATIONS
v
v
node1
a
+
Ra
ia
ivs
+
_
vs
b
_ node2
+
ic R
c
loop1
node1
Rb
ib
+
vis
is
loop2
vs
_
_ node2
+
ia
ivs
+
_
a
Ra
vc
+
_
loop1
+
ic R
c
node4
ivs − ia = 0
node3:
ib + i s = 0
node2:
ia − i b − i c = 0
node4:
ic − i s − ivs = 0
va
+
Ra
ia
ivs
vs
_ node2
+
+
_
loop1
vb
ic R
c
_
loop2
is
loop2
vis
_
loop2:
+ vc − vb − v is = 0
67
10
CIRCUIT REDUCTION (1)
node3
_
10Ω
is
30Ω
45Ω
iT
+
vc
+
_
+ v s − va − vc = 0
Rb
ib
+
node3
_
vc
loop1:
9
WRITE SUPPLEMENTARY
EQUATIONS
node1
b
Rb
ib
node4
node1:
67
8
WRITE THE
KVL EQUATIONS
v
v
node3
_
67
+
node1
vis
+
_ 5v
_
15Ω
90Ω
5Ω
i1
50Ω
100Ω
node4
ia = va / Ra
ib = vb / Rb
67
ic = vc / Rc
11
67
12
CIRCUIT REDUCTION (2)
10Ω
30Ω
CIRCUIT REDUCTION (3)
45Ω
10Ω
iT
+
_ 5v
15Ω
iT
i1
90Ω
5Ω
50Ω
Begin with loop on far right.
Combine the three resistors that are in
series.
Req = 45+50+100 = 195Ω
67
+
_ 5v
15Ω
90Ω
i1
195Ω
5Ω
100Ω
13
Again using the loop on the far right.
The 90 Ω and 195 Ω resistors are in
parallel.
Req= (90)(195)/(90+195) = 61.58 Ω
67
14
CIRCUIT REDUCTION (5)
CIRCUIT REDUCTION (4)
10Ω
30Ω
30Ω
10Ω
iT
iT
+
_ 5v
15Ω
+
_ 5v
61.58Ω
5Ω
15Ω
91.58Ω
5Ω
Still working with the loop on the far right.
The 30 Ω and the 61.58 Ω resistors are in
series.
Req = 30 + 61.58 = 91.58 Ω
67
Again, the far right loop.
The 15 Ω and 91.58 Ω resistors are in
parallel.
Req = (15)(91.58)/(15+91.58) = 12.9 Ω
15
CIRCUIT REDUCTION (6)
67
16
CIRCUIT REDUCTION (7)
10Ω
10Ω
iT
+
_ 5v
iT
+
_ 5v
12.9Ω
a
0.179A
27.9Ω
+
_ 5v
12.9Ω
5Ω
5Ω
b
Now there is only one loop.
All the resistors are in series.
Req = 10+12.9+5 = 27.9 Ω
67
17
Use Ohm’s Law to determine iT.
iT = 5/27.9 = 0.179A
iT flows in all three resistors, the 12.9 Ω
resistor is the equivalent resistance of
the entire circuit beyond
points a and b18.
67
CIRCUIT REDUCTION (9)
CIRCUIT REDUCTION (8)
10Ω
10Ω
a
ix
0.179A
+_ 5v
30Ω
0.0252A
0.179A
15Ω
+
_ 5v
91.58Ω
5Ω
a
15Ω
61.58Ω
5Ω
b
iT divides at a to flow through the 15 Ω and the
91.58 Ω resistors (the 91.58 Ω is an equivalent
resistance for the rest of the circuit).
Use current divider: ix = (0.179)(15)/(15+91.58) =
0.0252A.
67
19
CIRCUIT REDUCTION (10)
10Ω
67
20
Equivalent Subcircuits
30Ω
a
0.179A
+
_ 5v
No calculations are required at this step because
the 0.0252A is flowing through both resistors in the
right loop.
This circuit must be drawn however, because the
61.58 Ω resistor is an equivalent for the circuit to
the right of a and b.
15Ω
0.0252A
90Ω
i1
Circuit Analysis ->
195Ω
„
„
b
5Ω
Use the current divider equation again to determine
i1.
i1 = (0.0252)(195)/(90+195) = 0.01724A = 17.24mA.
The current through the 195 Ω resistor is 0.0252 0.01724 = 7.96mA
67
Simplify wherever possible
Replace with simpler subcircuits
Subcircuit: Any part of a circuit
„
„
„
+
v
_
21
Terminal Law
i
Two-terminal subcircuit
Terminal current
Terminal voltage
67
22
Examples of Terminal Law
Terminal law describes the behavior of
a two-terminal subcircuit
It is a function described by v = f (i) or i
= g (v), where i, v are terminal variables
Element law for two-terminal element
such as resistors
Resistors -> Ohm’s law v = f (i) = i R
Independent voltage sources
-> by KVL v = vs
Independent current sources
i
i
-> by KCL i = is
+
+
+
_
v
23
v
is
_
_
67
vs
67
24
Example:
Equivalent Subcircuits
More Terminal Law
Two two-terminal subcircuits are said to
be equivalent if they have the same
terminal law
Equivalent subcircuits may be freely
interchanged without altering any
external current or voltage
Find the terminal laws
i
i
+
v
+
2Ω
67
67
27
Example (cont’d)
i
R1
+
v
i
_
v1
+
+
-
v2
R2
v
_
+
v3
+
-
Rs
_
R3
67
28
General Statement
Left figure
For N resistors:
N
Rs = ∑ ri
By KVL on the loop: v = v1 + v2 + v3
By Ohm’s law:
v = (R + R + R ) i
1
Right figure
„ v = R i
s
Same terminal law. Thus
Rs = R1 + R2 + R3
67
26
Example:
Single-Loop Circuit
A single loop circuit is one which has only a
single loop
The same current flows through each
element of the circuit – the elements are in
series
Two adjacent elements are in series if they
share a common node that has no other
currents entering it
Nonadjacent elements are in series if they are
each in series with the same elements
„
_
25
Single-Loop Circuit
„
1Ω
v
_
67
1Ω
2
i =1
3
A chain of series resistors is equivalent
to a single resistor whose resistance is
the sum of the series resistances
Series resistance adds
29
67
30
Section 3.4
Voltage Divider
Example (cont’d)
i1=i2=i3=is
v1+v2+v3+(-vs)=0
From the example:
v1 = R1 i = R1
Similarly:
v
R
= 1v
Rs Rs
i1 =
v2 =
R2
R
and v3 = 3
Rs
Rs
vn =
67
p1=(v1
1
p2=(v2)2/R2
pt=p1+p2
67
R2
R1 + R2
33
67
i
250Ω
250Ω
50 Bulbs
Total
250Ω
The same current i flows through the
source and each light bulb. (Why?)
In terms of i, what is the voltage across
each resistor?
To solve for i, apply KVL around the
loop 250i + 250i + L + 250i = 120
i=
67
34
Solve for i
Example Christmas Lights
+
-
32
The voltage across series resistors
divides up in direct proportion to their
resistances
Voltage Divider: Larger resistances have
larger voltage drops in a voltage divider
V 2
R 1 ×V s 2
R 2 ×V s 2
+
=V s i = s
2
2
(R 1 + R 2 ) (R1 + R 2 )
Rs
67
120V
v s Rn
R1 + R 2 + R 3
Voltage Division
)2/R
P2 = Pt ×
v n = iiR n =
v s Rn
R 1 + R 2 + ... + R n
31
Voltage Divider & Power
Pt =
vs
R1 + R 2 + R 3
35
120
= 9.6 ( mA)
50 ⋅ 250
67
36
Some Comments
Single-Node-Pair Circuit
We can solve for the voltage across
each light bulb:
v = R i = 9.6 mA ⋅ 250 Ω = 2.4V
This circuit has one source and several
resistors. The current is:
(source voltage)/(sum of resistances),
or
series resistances sum
67
Two elements are connected in parallel
if together they form a loop containing
no other elements
By KVL, elements in parallel all have the
same voltage across them
37
67
Example (cont’d)
Example
i = i1 + i2 + i3
Left figure
„
i2
i1
i
38
+
i3
R2
R1
+
„
By KCL at the upper node:
By Ohm’s law: i = (G1 + G2 + G3 )v
Right figure
R3
v
i
Rp
_
v
_
„
Same terminal law. Thus
or
67
Rp =
A set of parallel resistors is equivalent
to a single resistor whose conductance
is the sum of the parallel conductances
Parallel conductance adds
67
67
40
When N = 2:
N
1
1
or R = ∑ R
i =1
p
i
i =1
1
1
1
1
= +
+
R p R1 R2 R3
Special Case
For N resistors:
G p = ∑ Gi
G p = G1 + G2 + G3
39
General Statement
N
i = G pv
41
R1 R2
R1 + R2
Product-by-sum rule
The equivalent resistance of two
resistors in parallel is the product of
their resistance divided by the sum
67
42
Example (cont’d)10/6
Some Comments
Adding parallel resistors can only
increase the equivalent conductance, or
decrease the equivalent resistance
Putting resistors in parallel reduces the
overall resistance below that of any of
them individually
For N equal R Ω resistors in parallel,
Rp = R / N
67
From the example:
i1 = G1v = G1
Similarly:
i2 =
G2
i
Gp
43
i
G
= 1i
Gp Gp
and
i3 =
G3
i
Gp
67
44
Section 3.5
Current Division
is=i1+i2
i1=v/R1 i2=v/R2
Let G=1/R
is=G1v+G2v
i1 =
VS=V1=V2=…=Vn
I1=? I2=?
is與i1 i2 ..之關係為何?
67
G1i s
G1 + G 2
45
i2 =
R 1i s
G2i s
=
G1 + G 2 R 1 + R 2
67
46
Example
in =
Gn is Gn is
= N
Gp
∑G n
R1=1/2 Ω, R2=1/4 Ω, R3=1/8 Ω
Gp=G1+G2+G3
=2+4+8=14
i1=(G1*is)/GP=(2*28)/14=4A
i2=(G2*is)/GP=(4*28)/14=8A
i2=(G3*is)/GP=(8*28)/14=16A
n =1
67
47
67
48
兩電阻並聯
三電阻並聯
(1)兩電阻並聯(如圖4-8)
(2)三電阻並聯(如圖4-9)
67
49
67
50
Current Division
Figure E 3.5-1
The current through parallel resistors
divides up in direct proportion to their
conductance
Current Divider: Smaller resistances
(larger conductances) have larger
current flows in a current divider
(a) A parallel
resistor network.
Courtesy of Dale
Electronics.
(b) The connected
circuit uses four
resistors where R =
1kΩ.
67
51
Special Case
67
52
Some Comments
Current divider is only for two resistors
It tells us how to divide the current
through parallel resistors
We cannot make a simple current
divider equation for three or more
parallel resistors
We have to solve for the voltage and
the solve for the currents of interest
For two resistors in parallel, current divides
inversely with their resistances
Current Divider:
i1 =
R2
i
R1 + R2
i2 =
R1
i
R1 + R2
How?
67
53
67
54
Section 3.6
Example
Sources Connection
Find the current and power dissipated
on each resistor
18 mA
i1
i2
i3
i4
250 Ω
500 Ω
500 Ω
1000 Ω
+
v
_
i
67
55
Section 3.6
Series Voltage Sources
i
+
v
_
(a) A circuit containing voltage sources
connected in series and (b) an equivalent
circuit.
67
56
More Series Voltage Sources
i
vs1
+
_
vs2
+
_
v
+
_
A chain of series voltage sources is
equivalent to a single voltage source
whose source function is the algebraic
sum of the series source functions
Series voltage sources add
+
+
_
vsN
vs
_
v = vs1 + vs 2 + L + vsN
v = vs
67
57
67
58
Series Current Sources
i
+
(a) A circuit containing
parallel current sources
and (b) an equivalent
circuit.
v
_
i
+
is1
is2
isN
_
i = is
is1 = is 2 = L = isN
67
59
is
v
67
60
Parallel Current Sources
i
More Parallel Current Sources
A set of parallel current sources is
equivalent to a single current source
whose source function is the sum of the
parallel source functions
Parallel current sources add
i
+
+
v
v
is1
_
is
isN
is2
_
i = is1 + is 2 + L + isN
i = is
67
61
Example
Since KVL cannot be violated….
A set of voltage sources in parallel must have
equal source functions
In this case, any one of them is equivalent to
the entire set
If the given source functions are not all equal,
there is a mathematical inconsistency and the
circuit cannot be analyzed
13 Ω
+
24 A
62
Parallel Voltage Sources
Find v and i
v
67
4A
6Ω
5Ω
i
_
67
63
Comparison
67
64
67
66
Example
Single-loop circuit:
„
„
„
series interconnection,
voltage division,
KCL
Single-node-pair circuit:
„
„
„
parallel interconnection,
current division,
KVL
67
65