Transcript EE2003 Circuit Theory

電路學(一)
Chapter 2
Basic Laws
1
Basic Laws - Chapter 2
2.1
2.2
2.3
2.4
2.5
2.6
Ohm’s Law.
Nodes, Branches, and Loops.
Kirchhoff’s Laws.
Series Resistors and Voltage Division.
Parallel Resistors and Current Division.
Wye-Delta Transformations.
2
2.1 Ohms Law (1)
• Ohm’s law states that the voltage across
a resistor is directly proportional to the
current I flowing through the resistor.
• Mathematical expression for Ohm’s Law
is as follows:
v  iR
• Two extreme possible values of R:
0 (zero) and  (infinite) are related
with two basic circuit concepts: short
circuit(短路) and open circuit(開路).
3
2.1 Ohms Law (2)
• Conductance(電阻) is the ability of an element to
conduct electric current; it is the reciprocal
of resistance R and is measured in mhos or
siemens.
1 i
G
R

v
• The power dissipated by a resistor:
2
v
p  vi  i 2 R 
R
4
2.2 Nodes, Branches and
Loops (1)
• A branch(分支) represents a single element such
as a voltage source or a resistor.
• A node(節點) is the point of connection between
two or more branches.
• A loop(迴路) is any closed path in a circuit.
• A network with b branches, n nodes, and l
independent loops will satisfy the fundamental
theorem of network topology:
b  l  n 1
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2.2 Nodes, Branches and
Loops (2)
Example 1
Original circuit
Equivalent circuit
How many branches, nodes and loops are there?
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2.2 Nodes, Branches and
Loops (3)
Example 2
Should we consider it as one
branch or two branches?
How many branches, nodes and loops are there?
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2.3 Kirchhoff’s Laws (1)
• Kirchhoff’s current law (KCL) (克希荷夫電流定律)
states that the algebraic sum of currents
entering a node (or a closed boundary) is zero.
N
Mathematically,
i
n 1
n
0
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2.3 Kirchhoff’s Laws (2)
Example 4
• Determine the current I for the circuit shown in
the figure below.
I + 4-(-3)-2 = 0
I = -5A
We can consider the whole
enclosed area as one “node”.
This indicates that
the actual current
for I is flowing
in the opposite
direction.
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2.3 Kirchhoff’s Laws (3)
• Kirchhoff’s voltage law (KVL) (克希荷夫電壓定律)
states that the algebraic sum of all voltages
around a closed path (or loop) is zero.
Mathematically,
M
v
m 1
n
0
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2.3 Kirchhoff’s Laws (4)
Example 5
• Applying the KVL equation for the circuit of the
figure below.
va − v1 − vb − v2 − v3 = 0
V1 = IR1 v2 = IR2 v3 = IR3
 va − vb = I(R1 + R2 + R3)
v a  vb
I
R1  R2  R3
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2.3 Kirchhoff’s Laws (5)
Example 6
• Determine vo and i in the circuit.
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2.3 Kirchhoff’s Laws (6)
Example 7
• Find current io and voltage vo in the circuit.
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2.3 Kirchhoff’s Laws (7)
Example 8
• Find current and voltage in the circuit.
14
2.4 Series Resistors and Voltage
Division (1)
• Series: Two or more elements are in series if they
are cascaded or connected sequentially
and consequently carry the same current.
• The equivalent resistance (等效電阻) of any
number of resistors connected in a series is the
sum of the individual resistances.
N
Req  R1  R2      RN   Rn
n 1
• The voltage divider can be expressed as
vn 
Rn
v
R1  R2      RN
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2.4 Series Resistors and Voltage
Division (2)
Example 9
10V and 5W
are in series
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2.5 Parallel Resistors and Current
Division (3)
• Parallel: Two or more elements are in parallel if
they are connected to the same two nodes and
consequently have the same voltage across them.
• The equivalent resistance of a circuit with
N resistors in parallel is:
1
1
1
1


  
Req R1 R2
RN
• The total current i is shared by the resistors in
inverse proportion to their resistances. The
current divider can be expressed as:
v iReq
in 

Rn Rn
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2.5 Parallel Resistors and Current
Division (4)
Example 10
2W, 3W and 2A
are in parallel
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2.5 Parallel Resistors and Current
Division (5)
Example 11
• Find Req for the circuit.
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2.5 Parallel Resistors and Current
Division (6)
Example 12
• Calculate the equivalent resistance Rab for the
circuit.
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2.5 Parallel Resistors and Current
Division (7)
Example 13
• Find current io and voltage vo in the circuit.
Calculate the power dissipated in the 3-W resistor.
2.5 Parallel Resistors and Current
Division (8)
Example 14
• Find current io and voltage vo in the circuit.
Calculate the power dissipated in the 3-W resistor.
2.6 Wye-Delta Transformations
(1)
Δ -> Y
Rb Rc
R1 
( Ra  Rb  Rc )
Y -> Δ
R1 R2  R2 R3  R3 R1
Ra 
R1
Rc Ra
R2 
( Ra  Rb  Rc )
Rb 
R1 R2  R2 R3  R3 R1
R2
Ra Rb
( Ra  Rb  Rc )
Rc 
R1 R2  R2 R3  R3 R1
R3
R3 
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2.6 Wye-Delta Transformations
(2)
Example 15
• Obtain the equivalent Rab and the current i.