Transcript Electrical Engineering 1

DC circuits and methods of
circuits analysis
• Circuits elements:
•
•
•
•
•
Voltage source
Current source
Resistors
Capacitors
Inductors
Voltage source - V [V]
• Ideal source
Constant output voltage, internal
resistance equals to zero
• Real source
Output voltage depends on various
conditions. Dependence may be linear
(battery) on non-linear
Current source - I [A]
• Ideal source
Constant output current, internal
resistance equals to infinity
• Real source
Output current depends on various
conditions. Dependence may be linear on
non-linear (Usually electronic sources)
Resistance - R []
• Coductance G=1/R [S]
• Ideal resistor
linear R = const.
V= I . R
• Real resistor
non-linear
(electric bulb, PN junction)
Resistance (2)
• Resistors in series
R = R1 + R2
• Resistors in parallel
R = R1 // R2 = (R1 . R2) / (R1 + R2)
• Voltage divider
U2 = U . R2 /(R1 + R2)
potential divider (‘pot’)
Passive electronic parts
• Resistors feature electrical
resistivity R
dimensioning according maximal dissipation power (loses) Pmax
• Capacitors feature capacity C
dimensioning according maximal granted voltage Vmax
• Inductors feature inductivity L
dimensioning according maximal granted current Imax
Resistors
• Feature: resistivity
•
r = R = const.
•
nonreversible el. energy transfer to heat
•
•
• Data: R [Ω], P [W]
• Description:
Ω → J, R
4,7 Ω
→ 4R7
•
kΩ → k
68 kΩ → 68k
•
MΩ → M
2.2 MΩ → 2M2
•
0,15 MΩ →
M15
• 47k/0,125W 3R3/ 5W
Resistors
Resistors color codings
color
num
ber
tolerance
Meaning
Blacká
0
Brown
1
±1%
Strip
4 strips
5strips
Red
2
±2%
1
first digit
first digit
Orange
3
2
second digit
second digit
Yelow
4
3
exponent 10x
third digit
Green
5
± 0,5 %
4
tolerance
exponent 10x
Blue
6
± 0,25 %
5
violet
7
± 0,1 %
grey
8
white
9
gold
-1
±5%
silver
-2
± 10 %
no color
± 20 %
tolerance
First strip is near to edge than last
If tolerance is ±20 %, the 4. strip miss
Resistors
Material
• Carbon – non stable, temperature
dependent
• Metalised - stable, precise
• Wired more power dissipation > 5W
Resistors
Potentiometer variable resistor
Potentiometr adjustable by hand
Potentiometer adjustable by tool
Resistors
Capacitors
• Part: Capacitor, condenser
• Feature: capacity
Accumulator of the energy in electrostatic field
symbol
t
dynamic definition
c = C = const.
1
v   i.dt
C0
Capacitors
static definition
q  Cv  i.t
power definition
1
2
W  C.V
2
For calculation should be used SI system only! :
unit:
1 F (Farrad)
dimension: [A.s/V]
Capacitors
Description:
• pF → J, R
• 103 pF → k , n
• 106 pF → M
4,7 pF
→ 4R7
68 000 pF → 68k
3,3 µF
→ 3M3
• 109 pF → G
200 µF
→ 200M
Number code: number, number, exponent in pF
eg. : 474 → 470 000pF → 470k → M47 ±20%
Capacitors
Inductors
Part: Inductor, coil
Feature: inductivity
Accumulator of the energy in electrostatic field
dynamic definition
l = L = konst.
di
uL
dt
Inductors
static definition
power definition
  N .  L.I
1
2
W  L.I
2
For calculation should be used SI system only! :
unit:
1 H (Henry)
dimension: [V.s/A]
Inductors
Details for instalation and ordering
L [H], IMAX [A]
Lower units 1 µH = 10-3 mH = 10-6 H
------------------It use in electronic not very often.
See next semestr
Ohm’s and Kirchhoff’s laws
• Ohm’s law
I=U/R
• 1st Kirchhoff’s law (KCL)  I = 0
At any node of a network, at every instant of
time, the algebraic sum of the currents at the
node is zero
• 2nd Kirchhoff’s law (KVL)  U = 0
The algebraic sum of the voltages across all the
components around any loop of circuits is zero
Nodal analysis
(for most circuits the best way)
• Uses 1st K. law
– Chose reference node
– Label all other voltage nodes
– Eliminate nodes with fixed voltage by source
of emf
– At each node apply 1st K. law
– Solve the equations
Mesh analysis
• Uses 2nd K. law
– Find independent meshs
– Eliminate meshs with fixed current source
– Across each mesh apply 2nd K. law
– Solve the equations
Thevenin equialent circuit
for linear circuit
As far as any load connected across its output
terminals is concerned, a linear circuits
consisting of voltage sources, current sources
and resistances is equivalent to an ideal voltage
source VT in series with a resistance RT. The
value of the voltage source is equal to the open
circuit voltage of the linear circuit. The resistance
which would be measured between the output
terminals if the load were removed and all
sources were replaced by their internal
resistances.
Norton equialent circuit
for linear circuit
As far as any load connected across its output
terminals is concerned, a linear circuits
consisting of voltage sources, current sources
and resistances is equivalent to an ideal current
source IN in parallel with a resistance RN. The
value of the current source is equal to the short
circuit voltage of the linear circuit. The value of
the resistance is equal to the resistance
measured between the output terminals if the
load were removed and all sources were
replaced by their internal resistances.
Principle of superposition
• The principle of superposition is that, in a
linear network, the contribution of each
source to the output voltage or current can
be worked out independently of all other
sources, and the various contribution then
added together to give the net output
voltage or current.
Example
Methods of electrical circuits analysis:
• Node Voltage Method
• Mesh Current Method
Σii = 0 , ΣIi = 0
Σvi = 0 , ΣVi = 0
• Thevenin and Norton Eq. Cirtuits
• Principle of Superposition
• --- and other 15 methods
Topology and Number of Lineary Independent
Equations
No. of elements
No. of nodes
p
u
zv
No. of current sources zi
No. of voltage sources
I2
I1
R
1
R3
3
V1
R
2
I
3
V2
•
•
No of elements p = 5
No. of nodes u = 4
No of voltage sources zv = 2
No of current sources zi = 0
I2
I1
R
1
R3
V3
V1
R
2
I
3
VU
2
•
•
Xi = u – 1 - zu = 4 – 1 - 2 = 1
No of independent meshes Xi = p – u + 1 – zi = 5 – 4 + 1 = 2
No of independent nodes
Node Voltage Analysis Method
1. Select a reference node (usually ground). All
other node voltages will be referenced to this
node.
2. Define remaining n-1 node voltages as the
independent variables.
3. Apply KCL at each of the n-1 nodes,
expressing each current in terms of the
adjacent node voltages
4. Solve the linear system of n-1 equations in n-1
unknowns
R
1
R3
R
2
V3
V1
V2
V3  V1 V3  V2 V3


0
R1
R2
R3
V1 V2

R1 R2
V3 
1
1
1


R1 R2 R3
Mesh Current Analysis Method
1. Define each mesh current consistently. We
shall define each current clockwise, for
convenience
2. Apply KVL around each mesh, expressing
each voltage in terms of one or more mesh
currents
3. Solve the resulting linear system of equations
with mesh currents as the independent
variables
I2
I1
R
1
R3
V3
R
2
I
3
V1
V2
R1I1  R3 I1  I 2   V1  0
R3 I 2  I1   R2 I 2  V2  0
________________
R1  R3 I1  R3 I 2  V1
R3 I1  R2  R3 I 2  V2