Download Chapter 2

Electrical Sources (1/3)
• Independent Source: Establishes a voltage or
current in a circuit without relying on voltages or
currents elsewhere in the circuit
• Dependent Source: Establishes a voltage or
current whose value depends on the value of a
voltage or current elsewhere in the circuit (also
known as controlled source)
• Active Circuit Element: Device capable of
generating electric energy
• Passive Circuit Element: Device that cannot
generate electric energy
Principles of Electrical
Engineering – I (14:332:221)
Chapter 2 Notes
14:332:221, Spring 2004
Electrical Sources (3/3)
Electrical Sources (2/3)
• Ideal voltage source: A
circuit element that
maintains a prescribed
voltage across its terminals
regardless of the current
flowing in these terminals
• Ideal current source: A
circuit element that
maintains a prescribed
current through its
terminals regardless of
the voltage across these
terminals
vs
Independent
Voltage Source
+
−
Independent
Current Source
+
−
vs = ρ i x
is
i s = α vx
Ideal dependent
voltage-controlled
current source
Amperes/Volt (A/V)
Ideal dependent
current-controlled
voltage source
Volts/Ampere (V/A)
3
is = β ix
Ideal dependent
current-controlled
current source
Dimensionless
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Electrical Resistance (2/3)
Electrical Resistance (1/3)
• Ohm’s Law: v = iR
• Resistor: A passive circuit element that impedes
the flow of electric charge
– v: voltage in volts (V)
– i: current in amperes (A)
– R: resistance in ohms (Ω) ~ Volts per Ampere (V/A)
R
– Interaction of moving electrons composing the electric
current with the atoms composing the conducting
material
– Electric energy is converted to thermal energy and
dissipated in the form of heat
– Different materials offer different degrees of
resistance
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Ideal dependent
voltage-controlled
voltage source
+ vs = µ vx
−
Dimensionless
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2
5
+
v
R
i=
v
R
+
v
−
R
i=−
v
R
−
i
i
• Conductance (G): Reciprocal of resistance
G = 1/R Siemens (S), e.g., R = 8Ω ↔ G = 0.25S
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1
Electrical Resistance (3/3)
Circuit Switch
• Resistive Power Dissipation
– Resistor always absorbs power from the circuit
– p = vi = (iR)i = i2R = v2/R
– Describing power in terms of conductance:
p = i2/G = v2G
p = vi = (iR )i = i 2 R
+
v
+
p = −vi = −(− iR )i = i 2 R
v
R
−
Short Circuit:
R=0
No current resistance in
ON state
Open Circuit:
R=∞
Infinite resistance to current in
OFF state
OFF
Switch:
−
i
i
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14:332:221, Spring 2004
d
Fig. 2.15:
• To “solve” a circuit, need to know:
vs
– Voltage across every element
– Current in every element
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Fig. 2.18: Multiple
elements meeting
at a single node
a
i1
vs
+
b
+
ic
vc
c
R1
Rc
• E.g., Fig. 2.15, 7 unknowns (Assume vs, R1, Rc and Rl are
given)
– is, i1, ic, il, v1, vc, vl → Require 7 independent equations
– Applying Ohm’s Law, 3 equations provided:
• v1 = i1R1 (2.13)
• vc = icRc (2.14)
• vl = ilRl (2.15)
b
+
10 Ω
120 V
Rl
−
−
c
Rc
50 Ω
6A
10
– A reference direction must be assigned to every current
at a node
– E.g., assign a positive sign to current leaving a node
and a negative sign to current entering a node
– Recalling again 2.15 and applying KCL
•
•
•
•
•
Node a: is − i1 = 0 (2.16)
Node b: i1 + ic = 0 (2.17)
Node c: −ic − il = 0 (2.17)
Node d: il − is = 0 (2.18)
Note: (2.15-2.18) provides only 3 independent equations
– n nodes yields n−1 independent equations via KCL
– Need 4 more independent equations
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+
−
+
ic
vc
• Kirchhoff’s Current Law (KCL): The algebraic
sum of all currents at any node in a circuit equals
zero
Rl
−
−
−
i1
vs
Kirchhoff’s Laws (4/6)
+
−
il vl
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Kirchhoff’s dLaws (3/6)
il vl
+
is
R1
9
is
+
−
a
• Kirchhoff’s Laws provide algebraic
relationships for solving circuits
• Node: A point where two or more circuit
elements meet
+
−
8
Kirchhoff’s Laws (2/6)
Kirchhoff’s Laws (1/6)
vs
ON
R
11
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2
Kirchhoff’s Laws (5/6)
Kirchhoff’s Laws (6/6)
• Loop: Starting at any arbitrary node, one traces a
closed path through selected circuit elements and
returns to the original node without passing
through any intermediate node more than once
• Kirchhoff’s Voltage Law (KVL): The algebraic
sum of all the voltages around any loop in a circuit
equal zero
E.g., Fig. 2.15 has a single loop. Choosing Node d as
the starting point and tracing the circuit clockwise yields
the loop d→vs→Rl→Rc→Rl→d
– Must assign an algebraic sign (reference direction) to
each voltage in a loop
– E.g., assign a positive sign to a voltage drop and a
negative sign to a voltage rise (or visa versa)
– Considering again Fig. 2.15 and applying KVL in a
clockwise direction:
• vl − vc + v1 − vs = 0 (2.20)
• Combining (2.13-2.15, 2.16-2.18 and 2.20) yields 7 independent
equations that may be applied to solve for the 7 unknowns
(is,i1,ic,il,v1,vc,vl)
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Reducing the Number of Unknowns
• Number of unknowns in Fig. 2.15 may be reduced,
thus simplifying the circuit
– If know current across a resistor, also know the voltage
and visa versa
• E.g., Fig. 2.15 need only solve for il, ic, and i1 or alternatively vl, vc
and v1.
• Knowing the current (or voltage) allows voltage (or current) to be
derived via Ohm’s Law
– If elements are in series, then the currents through each
of the series elements are equal
• E.g. in Fig. 2.15: is = i1 = −ic = il
• Thus the problem is reduced to solving for a single unknown, is:
vs = v1 − vc + vl = is(R1 − Rc + Rl)
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3