Слайд 1 - Georgia State University

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Transcript Слайд 1 - Georgia State University

Chapter 31
Fundamentals of Circuits
1
Capacitors in Series
V1



V2

C2
C1


V
V  V1  V2
2
Conductor in Electric Field
no electric field
E
equilibrium
E 0
E
E
3
Conductor in Electric Field
no electric field
E
conducting wire
ELECTRIC CURRENT
E
conducting wire
E
4
Electric Current
 Electric current is the rate
of flow of charge through
some region of space
• The SI unit of current is the
ampere (A), 1 A = 1 C / s
• Assume charges are moving
perpendicular to a surface of
area A
• If Q is the amount of charge
that passes through A in time
t, then the average current is
I av
Q

t
5
Chapter 28
Chapter 31
Ohm’s Law
6
Current Density
Current density is defined as
the current per unit area
I
j
A
This expression is valid only if the current
density is uniform and A is perpendicular to
the direction of the current
j has SI units of A/m2
7
Ohm’s Law
Ohm’s Law:
Current density is proportional
to electric field
j E
E
The constant of proportionality, σ, is called
the conductivity of the conductor.
The conductivity depends only on the material
of conductor.
Simplified model of electron
motion in conductor gives

n q 2

m
- is the material dependent characteristic of conductor.
8
Ohm’s Law
j E
• Ohm’s law states that for many materials, the ratio of the
current density to the electric field is a constant σ that is
independent of the electric field producing the current
– Most metals, but not all, obey Ohm’s law
– Materials that obey Ohm’s law are said to be ohmic
– Materials that do not obey Ohm’s law are said to be
nonohmic
• Ohm’s law is not a fundamental law of nature
• Ohm’s law is an empirical relationship valid only for
certain materials
9
Ohm’s Law
Conductor
B
l
E
Voltage across the conductor (potential
difference between points A and B)
V  VB  VA  El
A
where electric field is the same along
the conductor. Then
I
V
j

E
A
l
j E
V 1
I
E
 j
l

A
Another form of the Ohm’s Law
V 
l
I  RI
A
10
Ohm’s Law: Resistance
Conductor
B
l
E
A
 The voltage applied across the
ends of the conductor is proportional
to the current through the conductor
 The constant of proportionality is
called the resistance of the
conductor
V  RI
resistance
SI units of resistance are ohms (Ω)
1Ω=1V/A
11
Ohm’s Law: Resistance
Conductor
B
l
V  RI
resistance
R
E
A
l
A
Or
R
l
A
where   1 /  is the resistivity –
the inverse of the conductivity
Resistivity has SI units of ohm-meters (Ω m)
12
Resistance: Example
Conductor
l
R
A
l
The same amount of material has
been used to fabricate the wire with
uniform cross-section and length l/3.
What is the resistance of the wire?
l1 A1  lA
l1
R1  
A1
l1  l / 3
lA
A1 
 3A
l1
l1
l/3
l
R
R1  



A1
3A
9A 9
13
Ohm’s Law
j E
V  RI
– Materials that obey Ohm’s law are said to be ohmic
– Materials that do not obey Ohm’s law are said to be
nonohmic
An ohmic device
 The resistance is constant over a
wide range of voltages
 The relationship between current
and voltage is linear
 The slope is related to the
resistance
14
Ohm’s Law
j E
V  RI
– Materials that obey Ohm’s law are said to be ohmic
– Materials that do not obey Ohm’s law are said to be
nonohmic
Nonohmic materials
 The current-voltage relationship
is nonlinear
15
Chapter 31
Electric Power
16
Electrical Power
I
V
R
qE
v f  vi 
t
m
Before the collision
After the collision
17
Electrical Power
 As a charge moves from a to b,
I
V
R
the electric potential energy of the
system increases by QV
 The chemical energy in the
battery must decrease by the
same amount
 As the charge moves through the
resistor (c to d), the system loses
this electric potential energy during
collisions of the electrons with the
atoms of the resistor
 This energy is transformed into
internal energy in the resistor
18
Electrical Power
 The power is the rate at which the
energy is delivered to the resistor
U  QV
V
I
R
- the energy delivered to
the resistor when charge
Q moves from a to b
(or from c to d)
The power:
U Q
P

V  I V
t
t
2

V
P  I V  I 2 R 
R
Units: I is in A, R is in Ω, V is in V, and P is in W
19
Electrical Power
The power:
V 2
P
R(T )
I
2

V
P  I V  I 2 R 
R
V
R
Will increase the
temperature of conductor
Electromagnetic waves (light),
PEMW (T )
T
V 2
P
R(T )
Heat transfer to air
Pair (T )   (T  T0 )
V 2
P
 PEMW (T )   (T  T0 )
R(T )
20
Power: Example
A 1000-W heating coil designed to operate from 110 V is made of
Nichrome wire 0.5 mm in diameter. Assuming that the resistivity of the
Nichrome remains constant at its 20 C value, find the length of wire used.
 N  1.5  106  m
l
R  N
A
d2
A
4
2
U
P  I V  I 2 R 
R
U2
R
P
U2
d2 U2
3.14  0.52  106  1102
lA
A


m  1.58m
6
N
N P
4 N P
4  1.5  10  1000
R
21
Chapter 31
Direct Current
22
Direct Current
• When the current in a circuit has a
constant magnitude and direction,
the current is called direct current
• Because the potential difference
between the terminals of a battery
is constant, the battery produces
direct current
• The battery is known as a source of
emf (electromotive force)
23
Resistors in Series
For a series combination of resistors, the currents are the
same in all the resistors because the amount of charge that
passes through one resistor must also pass through the
other resistors in the same time interval
Ohm’s law:
Vc Vb  IR2
Vb Va  IR1
Vc  Va  Vc  Vb  Vb  Va 
 IR2  IR1  I  R1  R2 
Req  R1  R2
The equivalent resistance has the same effect on the
circuit as the original combination of resistors
24
Resistors in Series
• Req = R1 + R2 + R3 + …
• The equivalent resistance of a series combination of
resistors is the algebraic sum of the individual resistances
and is always greater than any individual resistance
25
Resistors in Parallel
 The potential difference across each resistor is the same
because each is connected directly across the battery terminals
 The current, I, that enters a point must be equal to the total
current leaving that point
I = I1 + I2
- Consequence of Conservation of Charge
Ohm’s law:
Vb Va  V  I1R1
Vb Va  V  I2R2
Conservation of Charge:
I  I1  I2
 1
V V
1  V
I

 V  

R1 R2
 R1 R2  Req
26
Resistors in Parallel
1
1
1


Req R1 R2
 Equivalent Resistance
1
1
1
1




Req R1 R2 R3
– The equivalent is always less than the smallest resistor in
the group
 In parallel, each device operates independently of the others
so that if one is switched off, the others remain on
 In parallel, all of the devices operate on the same voltage
 The current takes all the paths
– The lower resistance will have higher currents
– Even very high resistances will have some currents
27
Example
Req  R1  R2 
 8  4  12
1
1
1 1 1 1


  
Req R1 R2 6 3 2
Req  R1  R2 
 12  2  14
28
Example
Req  ? or
Main question:
R1
I ?
I
R2
V
Req
Req
R1
R2
I
I



V

V

29
Example
Req  ? or
Main question:
I ?
RR
R
in parallel Req ,1  1 1  1
R1  R1 2
in parallel
R1
R2
R1
R2
Req ,2 
R2R2
R
 2
R2  R2
2
I


V
30
Example
Req  ? or
Main question:
Req ,1 
R1
2
Req ,2 
in series
I ?
R2
2
Req  Req ,1  Req ,2 
R1  R2
2
I
V
2V

Req R1  R2
Req
Req ,1
Req ,2
I
I


V


V
31
Example
Main question:
Req  ? or
R1
I ?
I
V
Req
R2
Req
R1
R2
I


I


V
V
32
Example
Main question:
Req  ? or
R1
I ?
R3
Req
R5
I
I
R2

R4

V
To find
Req


V
you need to use Kirchhoff’s rules.
33
Chapter 31
Kirchhoff’s rules
34
Kirchhoff’s rules
 There are two Kirchhoff’s rules
 To formulate the rules you need, at first, to choose the
directions of current through all resistors. If you choose the
wrong direction, then after calculation the corresponding
current will be negative.
R1
I1
I
R3
I5
I2
R2


V
R5
I3
I4
R4
I
35
Junction Rule
 The first Kirchhoff’s rule – Junction Rule:
 The sum of the currents entering any junction must equal
the sum of the currents leaving that junction
- A statement of Conservation of Charge
I I
in
out
I1  I2  I3
In general, the number of times the
junction rule can be used is one fewer than
the number of junction points in the circuit
36
Junction Rule
 The first Kirchhoff’s rule – Junction Rule: Iin Iout
 In general, the number of times the junction rule can be used is one
fewer than the number of junction points in the circuit
 There are 4 junctions: a, b, c, d.
 We can write the Junction Rule for any three of them
R1
I1
b
I5
I2
a
I
R2
R3
R5
c


V
I3
(a) I  I1  I2
(b) I1  I5  I3
I4
(c) I2  I4  I5
d
R4
I
37
Loop Rule
 The second Kirchhoff’s rule – Loop Rule:
 The sum of the potential differences across all the
elements around any closed circuit loop must be zero
- A statement of Conservation of Energy

V  0
closed loop
Traveling around the loop from a to b
38
Loop Rule
 The second Kirchhoff’s rule – Loop Rule:
In general, the number of times the Loop Rule can be used is one
fewer than the number of possible loops in the circuit

V  0
closed loop
39
Loop Rule
 The second Kirchhoff’s rule – Loop Rule:

V  0
closed loop
There are 4 loops.
We need to write the Loop
Rule for 3 loops
Loop 1:
4
R1
I1
I
R3
I1R1  I5R5  I2R2  0
I3
Loop 2:
I2
I5
1
R5
R2
2
R4
3


V
I3R3  I5R5  I4R4  0
I4
Loop 3:
I
V  I2R2  I4R4  0
40
Kirchhoff’s Rules
I I
Junction Rule
in

 Loop Rule
out
V  0
closed loop
R1
I1
I
R3
I5
I2
R2

R5

V
I3
I  I1  I2
I1  I5  I3
I4
R4
I2  I4  I5
I
We have 6 equations and 6 unknown currents.
Req 
V
I
I1R1  I5R5  I2R2  0
I3R3  I5R5  I4R4  0
V  I2R2  I4R4  0
41
Kirchhoff’s Rules
Junction Rule
V2


I1
I
in

 Loop Rule
I  I1  I2
I1  I5  I3
I2  I4  I5
I I
out
V  0
closed loop
I1R1  I5R5  I2R2  V2  0
R1
R3
I5
I2 R
2


R5
I3R3  I5R5  I4R4  0
I3
R4
V1  I2R2  I4R4  0
I4
I
V1
We have 6 equations and 6 unknown currents.
42
Example 1
R3
I3
R2
I2
R1
I


I
V
Req
I

Req  R1 

V
I
I
R2R3
R2  R3
V
Req
43
Example 1
R3
I3
R2
I2
Req
R1
I

I
R2R3
 R1 
R2  R3
V
Req
I

V1
I  I 2  I3
I2R2  I3R3
 R3 
I  I3  1 

R
2 

R3
I 2  I3
R2
IR2
I3 
R2  R3
IR3
I2 
R2  R3
44
Example 1: solution based on Kirchhoff’s Rules
R3
I3
R2
I2
I  I 2  I3
R1
I3R3  I2R2  0
V  I2R2  IR1  0
I

I

V
R3
I 2  I3
R2
I3 
IR2
R2  R3
IR3
V 
R2  IR1  0
R2  R3
IR3
I2 
R2  R3
I
V
V

R3R2
 R1 Req
R2  R3
45
Example 2
V2 

I
R3
I3
R2
I2
R1

I

V1
I  I 2  I3
I3R3  I2R2  0
V1  V2  I2R2  IR1  0
46
Example 3
V
 2
I
R3
I3
R2
I2
R1

I

V1
I  I 2  I3
I3R3  I2R2  0
V1  V2  I2R2  IR1  0
47