Date post: | 01-Nov-2014 |
Category: |
Business |
Upload: | jassada-sarasook |
View: | 383 times |
Download: | 0 times |
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 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
AMaterial with resistivity
R = A l(Ohm) – resistivity of materials (ohm – meters)Good conductors low resistivity insulators high resistivity
SeeTab
le
2.1
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
resistoriRviv
Note : AVΩ /11
If current flows from a higher potential to a lower
potentialiRv
If current flows from a lower potential to a higher
potentialiRv
perfect conductor
short - circuit
R0
open - circuit
resistorfix
edvariable
Wire wound
compositionlarge resistance
Linear resistor obey ohm’s law
non - linear resistor does not obey
The reciprocal of resistance R, known as
conductance and denoted by G
G =IR= i
vmho
Or Siemens (S)
G
iGvpGvi
22,
Example 2.2 In the circuit shown in Fig.2.8, calculate the current i , the conductance G, and the power p.
30V k50+
-
v
i
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
mAxR
vi 6
105
303
mSxR
G 2.0105
113
mWxvip 180)106(30 3
mWxxxRip 180105)106( 3232
mWxxGvp 180102.0)30( 322
The conductance is
tsin20
tmAx
t
R
vi
sin4105
sin203
tmWvip 2sin80
Example 2.3 A voltage source of
Hence,
V is connected across a 5-kΩ resistor.Find the current through the
resistor and the power dissipated.Solution:
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 branchesA 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 any node more than once
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 the same voltage across them
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 3.
DC 2A6
5
DC 2A6
51 2
3
2.4 Kirchhoff’s laws
Kirchhoff’s current law (KCL) states that the algebraic
sum of currents entering a node is zero
01
N
nni
0)()( 54321 iiiii
52431 iiiii
enteringleaving
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 a IT = I1-I2+I3
IS = I1-I2+I3
ITa
b
I1 I2 I3
ITa
b
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
Example 2.5 For the circuit in Fig.2.21 (a), find voltage v1 and v2.
20V
+ -
2
3+
-
1v
2v 20V
+ -
2
3+
-
1v
2vi
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)
Applying KVL around the loop gives
-20+v1-v2=0 (2.5.2)
Substituting Eq. (2.5.1) into Eq. (2.5.2), we obtain
-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
Example 2.6 Determine v0 and i in the circuit shown in Fig 2.23(a)+ -
4
12V 4V
+
6
ov -
ov2i+ -
4
12V 4V
+
6
ov -
ov2i
i
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)
Substituting Eq.(2.6.2) into Eq.(2.6.1) yields
-16+10i-12i = 0
i = -8 A
and v0 = 48 V.
Example 2.8 Find the current and voltage in the circuit show in Fig. 2.27(a)
30V+
-
+ -
8
1v
32v+
-63v
a1i
2i
3i
20V+
-
+ -
8
1v
32v+
-63v
a1i
2i
3i
Loop1 Loop2
Solution: We apply Ohm’s law and Kirchhoff’s law. By Ohm’s law,
v1 = 8i1, v2=3i2, v3 = 6i3 (2.8.1)
Since the voltage and current of each 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 = 0We express this in terms of i1 and i2 as in Fig. (2.81) to obtain
-30 + 8i1 + 3i2 = 0
8
i330i 21
(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
6 i3 = 3 i2
22
3
ii
Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives0
28
330 22
2 i
ii
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.
2.5 Series Resistors and Voltage divisionv1 =
iR1
v2 = iR2
-v+v1
+v2 = 0
v = v1+v2 = i (R1+R2)v = i
Req
Req = R1 + R2
...
For N resistors in series then,
Req = R1 + R2+…RN
=
Rn N
n=1
To determine the voltage across each resistor
v1viR
1i
(R1
+R2)
= v1 R1+R2
R1
v=,v2 = R1+R2
R2
v
Vn= RnR1+R2+…+Rn
V
2.6 Parallel resistors and current divisionv =
i1R
1
= i2R2
i1
=
R1
v i2
=
R2
v,at node a ; i = i1+ i2
1 Req
= R1
1R
2
1+
R1
R1R2
R2+
=1 Req
i
R1
vR
2
v+ v R1
1R2
1 +== v Req
=
R1
+R2
R1R2=Req...Note : R1 = R2 then Req = R1 2/ สำหรบ R 2 ตวตอ ขนน
For N resistors1 R
e
q
= R1
1+ R2
1R
N
1+…+
If R = R1 = R2 = … = RN Req =RN
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 i2 v = i Req = iR1R2R1+R2
i1 = R1
v R1
= R1R2R1+R2i i R2
R1+R2=
=i2 i R1R1+R2
Current divider
Note : larger current flow through the smaller resistance
Example 2.9 Find Req for the circuit shown in Fig. 2.34.4 1
2
368
5
eqR
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
236
363//6
x
The 1-Ω and 5-Ω resistors are in series; hence their equivalent resistance 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); their equivalent resistance is
4.264
646//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 4.1484.24R eq
4
2
28
6eqR
4
8
4.2eqR
Example 2.10 Calculate the equivalent resistance Rab in the circuit in Fig. 2.37.10
3eqR
1
4
1
5
a
b
c d
b b
12
6
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
263
636//3
x
Similarly, the 12-Ω and 4- Ω resistors are in parallel since they are connected to the same two nodes d and b. Hence
3412
4124//12
x
Also the 1-Ω and 5-Ω resistors are in series; hence, their equivalent resistance is 651
With 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
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 get
2.1
32
323//2
x
This 1.2-Ω resistor is in series with the 10-Ω resistor, so that
Rab = 10 + 1.2 = 11.2 Ω
10
2eqR
1
3 6
a
b
c d
b b b
10
2eqR 3
a
b
c
b b
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 is
2
36
363//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 apply Ohm’s law to get
Ai 224
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,
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
The power dissipated in the 3-Ω resistor is Wivp OOO 333.5
3
44
12V
4
6+
-3ov
ai oi
b
12V
4
2+
-ov
ai
b
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.k6
k9 k12+
-ov30mA k9 k18
+
-ov30mA
oi
1i
2i
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 technique to find i1 and i2.
mAmAi 20)30(000,18000,9
000,181
mAmAi 10)30(000,18000,9
000,92
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
(c) Power absorbed by the 12-kΩ resistor isWxRiRiiivp 2.1)000,12()1010()( 232
222
Power absorbed by the 6-kΩ resistor isWxRip 6.0)000,6()1010( 232
2
Power absorbed by the 9-kΩ resistor is W
R
vp 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 checking results.
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.
Delta to Wye conversion
Wye to delta
R1 = Ra = R1R2 + R2R3 + R3R1
R1
Rc = R1R2 + R2R3 + R3R1
R3
Rb = R1R2 + R2R3 + R3R1
R2
Y and balancewh
en
R1 = R2 = R3 = Ry
Ra = Rb = Rc = R
Ry = R or R = 3Ry3
Rb Rc
Ra+Rb+Rc
Rc Ra
Ra+Rb+Rc
Ra Rb
Ra+Rb+Rc
R2 =
R3 =
Example 2.14 Convert the network in Fig. 2.50(a) to an
equivalent Y network.
25
a b
c
10 15aRbR
cRa b
c
51R
5.7
3
2R
3R
Solution: Using Eqs. (2.49) to (2.51), we obtain
50
50
250
151025
10251
x
RRR
RRR
cba
cb
5.750
15252
x
RRR
RRR
cba
ac
350
10153
x
RRR
RRR
cba
ba
The equivalent Y network is shown in Fig. 2.50(b).
Example 2.15 Obtain the equivalent resistance Rab for the circuit in Fig. 2.52 and use it to find current i. a
c
5.12
15
10
20
120V
b
30n
a
b
5
i
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
101R 202R 53R
Thus from Eqs. (2.53) to (2.55) we have
10
1055202010
1
133221 xxx
R
RRRRRRRa
3510
350
5.1720
350
2
133221
R
RRRRRRRb
705
350
3
133221
R
RRRRRRRc
pairs of resistors in parallel, we obtain
213070
30x7030//70
2917.75.175.12
5.175.125.17//5.12
x
5.103515
351535//15
x
so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we find
632.921292.17
21292.1721//)5.10292.7(
xRab
AR
vi
ab
s 458.12632.9
120
a
c
5.12
15
5.17
b
30
a
b
35
70
5.10
21
b
292.7
a
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
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.
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 RbcRac
Where Rac = Rab+Rbc
Thus, vout decreases or increases as the sliding contact of
the pot moves toward c or a, respectively.
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 causes the pointer to
deflect. The amount of current through the coil
determines the deflection of the
pointer.
Voltmeter It 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.
mfs
fsn
nmfsfs
RI
VR
RRIV
AmmeterIt 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
mn
fsmn
nm
RII
IR
IRR
RI
Ohmmeter It consists of d’Arsonval movement, a variable resistor and a battery
)1.....(....................m
mx
mxm
RRI
ER
IRRRE
The resistor R is selected such that the meter gives a full-scale deflection when Rx = 0fsm II
)2........(....................fsm IRRE
Substitute eq. (2) in (1)
mm
fsx RR
I
IR
1
Example 2.16 Three light bulbs are 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.
9V
15W
10W
20W 9V
+
-
+
-
+
-1v
2v
3v1R
2R
3R
I 1I
2I
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 supplied by the battery is
AV
pI 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 isAV
pI 222.2
9
20
1
11
By KCL, the current through the series combination of R2 and R3 is AIII 778.2222.2512
(c) Since p = I2R,
05.4222.2
2022
1
11 I
pR
945.1777.2
1522
2
22 I
pR
297.1777.2
1022
3
33 I
pR
Example 2.17 Following the voltmeter 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
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, k
xR 82000000,102000
10100
161
For range 0-5 V, kx
R 482000000,50200010100
562
For range 0-50 V, kx
R 4982000000,500200010100
5063
For range 0-100 V, kx
R 9982000000,000,1200010100
10064
2.9 Summary 1. 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
Open circuit is a resistor with infinite resistance(R =∞)
2. Short circuit is a resistor with zero resistance(R = 0)
3. The conductor G of a resistor is the reciprocal of its resistance
RG
1
4. A branch is a single two-terminal element in an electric
circuitA node is the point of connection between two or
more branches A loop is a closed path in a circuit 5. 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
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
21GG
GGGeq
9. The voltage division principle for two resistors in series is
vRR
Rvv
RR
Rv
21
22
21
11 ,
10.When two resistors R1 and R2 are in parallel
2121
21 , GGGRR
RRR eqeq
11.The current division principle for two resistors in
parallel isi
RR
Rii
RR
Ri
21
12
21
21 ,
12.Delta – to – Wye conversion
cba
ba
cba
ac
cba
cbRRR
RRR
RRR
RRR
RRR
RRR
321 ,,
13.Wye – to – Delta conversion
3
133221
2
133221
1
133221
R
RRRRRRR
R
RRRRRRR
R
RRRRRRR
c
b
a