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1.
CHAPTER 2 Basic LawsCHAPTER 2 Basic Laws
To determine the values of electrical
variables such as current, voltage and
power in a given circuit requires
understanding some fundamental laws for
examples; Ohm’s law and Kirchhoff’s laws
to analyze the circuit. In addition, some
techniques must be used together with
those fundamental laws.
2.
2.2 Ohm’s law
Materials in general have a
characteristic behavior of
resisting the flow of electric
charge which is known as
resistance (R). The resistance of
any material depends on cross –
sectional area A and its lengthl
l
al with resistivity ρ
R = ρ
A
l(Ohm)
ρ – resistivity of materials (ohm – mete
Good conductors low resistivity
insulators high resistivity
See
Tab
l2.
1
3.
The circuit element used to
model the current - resisting
behavior of a material is the
resistor
Resistance of the resistor
Ohm’s law states that the
voltage V across a resistor is
directly proportional to the
current i flowing through the
resistor
iRviv =⇒α
Note : AVΩ /11 =
4.
If current flows from a
higher potential to a lower
potentialiRv =
If current flows from a lower
potential to a higher potential
iRv −=
fect conductor
short - circuit
∠∞∠R0
open - circuit
5.
sistor
fix
e
dvariable
Wire wound
compositionlarge resistance
ar resistor obey ohm’s law
inear resistor does not obey
6.
The reciprocal of resistance
R, known as conductance and
denoted by G
G =I
R = i
v
mho
Or Siemens (S)
G
i
GvpGvi
2
2
, ===
7.
Example 2.2 In the circuit shown in Fig.2.8, calculate the
current i , the conductance G, and the power p.
Solution: The Voltage across the resistor is the same as
the source voltage (30 V) because the resistor and the
voltage source are connected to the same pair of terminals.
Hence, the current is
8.
mA
xR
v
i 6
105
30
3
===
mS
xR
G 2.0
105
11
3
===
mWxvip 180)106(30 3
=== −
mWxxxRip 180105)106( 3232
=== −
mWxxGvp 180102.0)30( 322
=== −
The conductance is
9.
tπsin20
tmA
x
t
R
v
i π
π
sin4
105
sin20
3
===
tmWvip π2
sin80==
Example 2.3 A voltage source of
Hence,
V is connected
ss a 5-kΩ resistor.Find the current through the
r and the power dissipated.
olution:
10.
2.3 Nodes, Branches and Loops
A branch represents a single
element such as a voltage source
or a resistorA Node is the point of
connection between two or more
branches
A Loop is any closed path in
the circuit formed by starting at
a node, passing through a set of
nodes, and returning to starting
node without passing through
11.
Two or more elements are in
series if they exclusively share a
single node and consequent by
carry the same currentTwo or more elements are in
parallel if they exclusively
connected to the same two
nodes and consequent by have
12.
Example 2.4 Determine the number
of branches and node in the circuit
show in Fig.2.12. Identify which
elements are in series and which are
in parallel.
Solution: Since there are four
elements in the circuit, the circuit
has four branches: 10 V, 5 Ω, 6 Ω,
and 2 A. The circuit has three nodes
as identified in Fig. 2.13. The 5 Ω
resistor is in series with the 10-V
voltage source because the same
current would flow in both. The 6-Ω
resistor is in parallel with the 2-A
current source because both are
connected to the same nodes 2 and
13.
2.4 Kirchhoff’s laws
Kirchhoff’s current law (KCL)
states that the algebraic sum of
currents entering a node is zero0
1
=∑
=
N
n
ni
0)()( 54321 =−+++−+ iiiii
52431 iiiii +=++
enteringleaving
14.
A simple application of KCL is
combining current sources in
parallel. The combined or equivalent
current source can be found by
applying KCL to node aIT = I1-I2+I3
IS = I1-I2+I3
IT
b
I1
I2 I3
IT
a
b
15.
Kirchhoff’s voltage law (KVL) states
that the algebraic sum of all voltages
around a loop is zero
-v1+v2+v3-v4+v5 = 0
v2+v3+v5 = v1+v4
When voltage sources are connected
in series, KVL can be applied to
obtain the total voltage
16.
5 For the circuit in Fig.2.21 (a), find voltag
Solution: To find v1 and v2, we apply
Ohm’s law and Kirchhoff’s voltage
law. Assume that current i flow
through the loop as shown in Fig.
2.21(b). From Ohm’s law,v1 =2i , v2=-3i (2.5.1)
17.
Applying KVL around the loop gives
-20+v1-v2=0 (2.5.2)
stituting Eq. (2.5.1) into Eq. (2.5.2), we ob
-20+2i+3i=0 or 5i=20
i =4 A
Substituting i in Eq. (2.5.1) finally gives
v1= 8 V , v2=-12 V
18.
6 Determine v0 and i in the circuit shown in
Solution: We apply KVL around the loop as shown in
Fig.2.23(b). The result is
-12+4i+2v0-4+6i = 0 (2.6.1)
Applying Ohm’s law to the 6-Ω resistor gives
v0 = -6i (2.6.2)
19.
bstituting Eq.(2.6.2) into Eq.(2.6.1) yields
-16+10i-12i = 0
i = -8 A
and v0 = 48 V.
20.
Example 2.8 Find the current and
voltage in the circuit show in Fig.
2.27(a)
e apply Ohm’s law and Kirchhoff’s law. By
v1 = 8i1, v2=3i2, v3 = 6i3 (2
21.
resistor are related by Ohm’s law as
show, we are really looking for three
thing: (v1, v2, v3) or (i1, i2, i3). At
node a, KCL gives
i1- i2- i3 = 0
(2.8.2)
Applying KVL to loop as in Fig.
2.27(b),
-30 + v1 + v2 = 0
ess this in terms of i1 and i2 as in Fig. (2.81
-30 + 8i1 + 3i2 = 0
22.
( )
8
i330
i 2
1
−
= (2.8.3)
Applying KVL to loop 2,
-v2 + v3 = 0
v3 = v2
(2.8.4)
as expected since the two resistors
are in parallel. We express v1 and v2 in
term of i1 and i2 as in Eq. (2.8.1).Eq.
(2.8.4) becomes
23.
6 i3 = 3 i2
2
2
3
i
i =
tuting Eqs. (2.8.3) and (2.8.5) into (2.8.2)
0
28
330 2
2
2
=−−
− i
i
i
or i2 = 2 A. From the value of i2 ,
we now use Eqs. (2.8.1) to (2.8.5) to
obtain
i1 = 3 A, i3 = 1 A, v1 = 24 V, v2 =
6 V, v3 = 6 V.
24.
2.5 Series Resistors and
Voltage divisionv1 =
i
R
1
v2 =
i
R
2
-v+v1+v2
= 0v = v1+v2 = i
(R1+R2)v = i
R
eq
Req = R1 +
R2
...
N resistors in series then,
Req = R1 + R2+
…RN =
Rn Σ
N
n=1
25.
ine the voltage across each resistor
v1
v
iR
1
i
(R1
+R2
)
= v1
R1+R2
R1v=
,
v2 = R1+R2
R2v
Vn= Rn
R1+R2+…+Rn
V
26.
2.6 Parallel resistors and
current divisionv = i1R1 = i2R2
i1
=
R
1
v i2
=
R
2
v,
at node a ;i = i1+ i2
1
Req
= R
1
1
R
2
1+
R1
R1
R2
R2
+
=1
Req
i
R
1
v
R
2
v+ v
R1
1
R2
1 +== v
Req
=
R1+R2
R1R2
=Req.
.
.1 = R2 then Req = R1
/2
สำำหรับ R 2 ตัวต่อ ขนำน
For N resistors1
R
e
=
R
1
1 + R
2
1
R
N
1+…+
If R = R1 = R2 = … = RN Req =R
N
27.
It is more often to use
conductance rather than
resistance when dealing with
resistors in parallel.
Geq = G1 + G2 + G3 + … + GN
Given the total current i enter
node a, how do use obtain
current i1 and i2v = i Req = iR1R2
R1+R2
i1= R
1
v
R
1
=
R1R2
R1+R2
i iR2
R1+R2
=
=i2 iR1
R1+R2
Current divider
er current flow through the smaller resista
28.
e 2.9 Find Req for the circuit shown in Fig.
Solution: To get Req , we combine
resistors in series and in parallel. The
6-Ω and 3-Ω resistors are in parallel, so
equivalent resistance is
Ω=
+
=ΩΩ 2
36
36
3//6
x
29.
is
1Ω+5Ω = 6Ω
Thus the circuit in Fig. 2.34 is
reduced to that in Fig. 2.35(a). In Fig.
2.35(a), we notice that the two 2-Ω
resistors are in series, so the equivalent
resistance is
2Ω + 2Ω = 4Ω
This 4-Ω resistor is now in parallel
with the 6-Ω resistor in Fig. 2.35(a);
30.
Ω=
+
=ΩΩ 4.2
64
64
6//4
x
The circuit in Fig. 2.35(a) is now
replaced with that in Fig. 2.35(b). In Fig.
2.35(b), the three resistors are in series.
Hence, the equivalent resistance for the
circuit is Ω=Ω+Ω+Ω= 4.1484.24Req
31.
Example 2.10 Calculate the
equivalent resistance Rab in the circuit
in Fig. 2.37.
Solution: The 3-Ω and 6-Ω resistors are
in parallel because they are connected
to the same two nodes c and b. Their
combined resistance is
Ω=
+
=ΩΩ 2
63
63
6//3
x
32.
Similarly, the 12-Ω and 4- Ω resistors are
in parallel since they are connected to
the same two nodes d and b. Hence
Ω=
+
=ΩΩ 3
412
412
4//12
x
Also the 1-Ω and 5-Ω resistors are in
series; hence, their equivalent
resistance is Ω=Ω+Ω 651With these three combinations, we can
replace the circuit in Fig. 2.37 with
that in Fig. 2.38(a). In Fig.2.38(a),3-Ω
in parallel with 6-Ω gives 2-Ω, as
calculated in Eq. (2.10.1). This 2-Ω
33.
equivalent resistance is now in series
with the 1-Ω resistance to give a
combined resistance of 1Ω + 2Ω
=3Ω . Thus, we replace the circuit in
Fig. 2.38(a) with that in Fig. 2.38(b).
In Fig. 2.38(b), we combine the 2-Ω
and 3-Ω resistors in parallel to getΩ=
+
=ΩΩ 2.1
32
32
3//2
x
Ω resistor is in series with the 10-Ω resistor
Rab = 10 + 1.2 = 11.2 Ω
34.
Example 2.12 Find iO and vO in the
circuit shown in Fig. 2.42(a). Calculate
the power dissipated in the 3-Ω resistor.
Solution: The 6-Ω and 3-Ω resistors
are parallel, so their combined
resistance isΩ=
+
=ΩΩ 2
36
36
3//6
xThus our circuit reduces to that
shown in Fig. 2.42(b). Notice that vO is
not affected by the combination of the
resistors because the resistors are in
parallel and therefore have the same
voltage vO. From Fig. 2.42(b), we can
obtain vO in two ways. One way is to
35.
Ai 2
24
12
=
+
=
and hence, vO = 2i = 2x2 = 4 V. Another
way is to apply voltage division, since
the 12 V in Fig. 2.42(b) is divided
between the 4-Ω and 2-Ω resistors.
Hence, VVvO 4)12(
42
2
=
+
=
Similarly, iO can be obtained in two way.
One approach is to apply Ohm’s law to
the 3-Ω resistor in Fig. 2.42(a) now that
we know vO; thus,
36.
43 == OO iv
AiO
3
4
=
Another approach is to apply current
division to the circuit in Fig. 2.42(a)
now that we know I, by writing
AAiiO
3
4
)2(
3
2
36
6
==
+
=
ower dissipated in the 3-Ω resistor is
Wivp OOO 333.5
3
4
4 =
==
37.
Example 2.13 For the circuit shown in
Fig. 2.44(a), determine: (a) the voltage
vO, (b) the power supplied by the
current source, (c) the power absorbed
by each resistor.
Solution: (a) The 6-kΩ and 12-kΩ
resistors are in series so that their
combined value is 6 + 12 = 18 kΩ .
Thus the circuit in Fig. 2.44(a)
reduces to that shown in Fig. 2.44(b).
We now apply the current division
38.
mAmAi 20)30(
000,18000,9
000,18
1 =
+
=
mAmAi 10)30(
000,18000,9
000,9
2 =
+
=
Notice that the voltage across the
9-kΩ and 18-kΩ resistors are the
same, and vO = 9,000i1 = 18,000i2 =
180 V, as expected.
(b) Power supplied by the
source is WmWivp OOO 4.5)30(180 ===
39.
Power absorbed by the 12-kΩ resistor is
WxRiRiiivp 2.1)000,12()1010()( 232
222 ===== −
wer absorbed by the 6-kΩ resistor is
WxRip 6.0)000,6()1010( 232
2 === −
wer absorbed by the 9-kΩ resistor is
W
R
v
p O
6.3
000,9
)180( 22
===
WmWivp O 6.3)20(1801 ===
Notices that the power supplied (5.4W)
equals the power absorbed (1.2 + 0.6 +
3.6) = 5.4 W). This is one way of
40.
2.7 Wye – Delta
transformationsWhen the resistors are neither in
parallel nor in series. For example,
the bridge circuit, this circuit can
be simplified by using three –
terminal equivalent networks, wye
(y) and delta ( ) as will be shown
in Ex.2.15.
41.
Delta to Wye
conversion
Wye to delta
= Ra =
R1R2 + R2R3 + R3R1
R1
Rc =R1R2 + R2R3 + R3R1
R3
Rb =R1R2 + R2R3 + R3R1
R2
Y and balance
wh
e
n
R1 = R2 = R3 = Ry
Ra = Rb = Rc = R
Ry = R or R = 3Ry
3
Rb Rc
Ra+Rb+Rc
Rc Ra
Ra+Rb+Rc
Ra Rb
Ra+Rb+Rc
2 =
3 =
42.
Example 2.14 Convert the ∆ network in Fig. 2.50(a) to an
valent Y network.
ution: Using Eqs. (2.49) to (2.51), we obta
Ω==
++
=
++
= 50
50
250
151025
1025
1
x
RRR
RR
R
cba
cb
43.
Ω==
++
= 5.7
50
1525
2
x
RRR
RR
R
cba
ac
Ω==
++
= 3
50
1015
3
x
RRR
RR
R
cba
ba
equivalent Y network is shown in Fig. 2.50
44.
Example 2.15 Obtain the equivalent
resistance Rab for the circuit in Fig.
2.52 and use it to find current i.
45.
Solution: In this circuit, there are two Y network and one ∆
network. Transforming just one of these will simplify the
circuit. If we convert the Y network comprising the 5-Ω, and
20-Ω resistors, we may select
Ω= 101R Ω= 202R Ω= 53R
s from Eqs. (2.53) to (2.55) we have
10
1055202010
1
133221 xxx
R
RRRRRR
Ra
++
=
++
=
Ω== 35
10
350
46.
Ω==
++
= 5.17
20
350
2
133221
R
RRRRRR
Rb
Ω==
++
= 70
5
350
3
133221
R
RRRRRR
Rc
pairs of resistors in parallel, we obtain
Ω=
+
= 21
3070
30x70
30//70
Ω=
+
= 2917.7
5.175.12
5.175.12
5.17//5.12
x
Ω=
+
= 5.10
3515
3515
35//15
x
47.
so that the equivalent circuit is shown
in Fig. 2.53(b). Hence, we find
Ω=
+
=+= 632.9
21292.17
21292.17
21//)5.10292.7(
x
Rab
A
R
v
i
ab
s
458.12
632.9
120
===
48.
2.8 Application
Lighting systems, such as in a
house, often consist of N lamps
connected either in parallel or in
series
2.8.1 Lighting
systems
49.
Assuming that all the lamps are
identical and v0 is the power – line
voltage, the voltage across each
lamp is v0 for parallel connection
and v0/N for series connection.
The series connection is easy to
manufacture but is seldom used
in practice for two reasons. First,
it is less reliable. Second, it is
harder to maintain.
50.
Potentiometer is a three – terminal
device that operates on the principle
of voltage division. It is essentially an
adjustable voltage divider. As a
voltage regulator, it is used as a
volume or level control on radios, TVs
and other devices.
2.8.2 Design of DC
Meters
Potentiometer
controlling
potential levels
Vout = Vbc = Vin
Rbc
Rac
Where Rac = Rab+Rbc
Thus, vout decreases
or increases as the
sliding contact of the
pot moves toward c
51.
Another application where resistors
are used to control current flow is the
analog dc meters, ammeter, voltmeter
and ohmmeter. Each of these meter
employs the d’Arsonval meter
movement.
The movement
consists of a movable
iron – core coil mounted
on a pivot between the
poles of a permanent
magnet. When current
flows through the coil, it
creates torque which
52.
Voltmete
rIt measures the voltage across a load
and is connected in parallel with the
element. It consists of a d’Arsonval
movement in series with a resistor whose
resistance Rm is deliberately made very
large to minimize the current drawn from
the circuit.
( )
m
fs
fs
n
nmfsfs
R
I
V
R
RRIV
−=
+=
53.
Ammete
rIt measures the current through the
load and is connected in series with it.
It consists of a d’ Arsonaval movement
in parallel with a resistor whose
resistance Rm is deliberately made very
small to minimize the voltage drop
across it.
m
mfs
m
n
fs
mn
n
m
R
II
I
R
I
RR
R
I
−
=
+
=
54.
Ohmm
eterIt consists of d’Arsonval
movement, a variable resistor and a
battery
55.
( )
( ) )1.....(....................m
m
x
mxm
RR
I
E
R
IRRRE
+−=
++=
The resistor R is selected such
that the meter gives a full-scale
deflection when Rx = 0
fsm II =
( ) )2........(....................fsm IRRE +=
Substitute eq. (2) in (1)
( )m
m
fs
x RR
I
I
R +
−= 1
56.
connected to a 9-V battery as shown in
Fig. 2.56(a). Calculate: (a) the total
current supplied by the battery, (b) the
current through each bulb, (c) the
resistance of each bulb.
Solution: (a) The total power supplied
by battery is equal to the total power
absorbed by the bulbs, that is,
p = 15 + 10 + 20 = 45 W
Since p = VI, then the total current
57.
A
V
p
I 5
9
45
===(b) The bulbs can be modeled as
resistors as shown in Fig. 2.56(b).
Since R1 (20-W bulb) is in parallel with
the battery as the series combination
of R2 and R3,
V1 = V2 + V3 = 9 V
The current through R1 isA
V
p
I 222.2
9
20
1
1
1 ===
58.
current through the series combination of
AIII 778.2222.2512 =−=−=
(c) Since p = I2
R,
Ω=== 05.4
222.2
20
22
1
1
1
I
p
R
Ω=== 945.1
777.2
15
22
2
2
2
I
p
R
Ω=== 297.1
777.2
10
22
3
3
3
I
p
R
59.
setup of Fig. 2.60, design a voltmeter
for the following multiple ranges:
(a) 0-1 V (b) 0-5 V (c) 0-50 V
(d) 0-100 V
Assume that the internal resistance
Rm = 2 kΩ and the full scale current
.100 AI fs µ=
Solution: We apply Eq. (2.60) and
assume that R1, R2, R3, and R4
correspond with ranges 0-1 V, 0-5 V,
0-50 V, and 0-100 V, respectively.
For range 0-1 V,
Ω=−=−= −
k
x
R 82000000,102000
10100
1
61
60.
or range 0-5 V, Ω=−=−= −
k
x
R 482000000,502000
10100
5
62
or range 0-50 V, Ω=−=−= −
k
x
R 4982000000,5002000
10100
50
63
or range 0-100 V, Ω=−=−= −
k
x
R 9982000000,000,12000
10100
100
64
61.
.9 Summary1. A resistor is a passive element
in which the voltage v across it
is proportional to the current i
through itiRv =;R is the resistance of the resistor
en circuit is a resistor with infinite resistan
2. Short circuit is a resistor with
zero resistance(R = 0)
3. The conductor G of a resistor is
the reciprocal of its resistance
R
G
1
=
62.
4. A branch is a single two-terminal
element in an electric circuit
A node is the point of connection
between two or more branches
A loop is a closed path in a
circuit5. KCL states that the sum of the
currents entering a node equals
the sum of currents leaving the
node6. KVL states that the voltage
around a closed path
algebraically sum to zero
63.
Elements are in parallel if
they have the same voltage
across them8. When two resistors are in
series
7. Elements are in series if the
same current flows through
them
21 RRReq +=
21
21
GG
GG
Geq
+
=
9. The voltage division principle
for two resistors in series is
v
RR
R
vv
RR
R
v
21
2
2
21
1
1 ,
+
=
+
=
64.
10.When two resistors R1 and R2
are in parallel
21
21
21 , GGG
RR
RR
R eqeq +=
+
=
11.The current division principle
for two resistors in parallel is
i
RR
R
ii
RR
R
i
21
1
2
21
2
1 ,
+
=
+
=
12.Delta – to – Wye conversion
cba
ba
cba
ac
cba
cb
RRR
RR
R
RRR
RR
R
RRR
RR
R
++
=
++
=
++
= 321 ,,
65.
13.Wye – to – Delta conversion
3
133221
2
133221
1
133221
R
RRRRRR
R
R
RRRRRR
R
R
RRRRRR
R
c
b
a
++
=
++
=
++
=
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