Chapter 1 Basic Concepts
Electric circuits --- interconnection of electric circuit elements
that deal with electricity (charges)
• Electric circuit elements
Resistor
Capacitor
Inductor
Voltage sources
Current sources
Operational amplifiers
Diodes
Transistors
• Circuit variables --- being used to quantify electricity
1. Circuit Variables
(1) Charge and Current
Charge --- a physical property associated with certain
fundamental particles and has polarities.
Examples:
e = 1.6 ×10−19 coulomb (C)
electron −e
proton +e
neutron 0
Charges occur in nature in integral multiples of the fundamental charge e.
1
Law of conservation of charge
---- Charges can neither be created nor destructed, only be
transported.
Current --- the time rate of transporting charges across
a cross-section area.
=
unit: ampere (A)
= ( ) • Must have reference direction
positive charges flow in the reference direction, or
negative charges flow in the opposite direction
opposite situations
• A direct current (dc) is constant in
time
• An alternating current (ac) changes
sinusoidally in time
2
(2) Voltage
How to make charges move?
→ under electric force →
electric potential difference
• U --- electric potential at a given point in a circuit
• Voltage = electric potential difference between two points
≡ − • Voltage has polarities
“+” terminal --- (reference) high potential point
“−” terminal --- (reference) low potential point
≡ − = voltage drop from "" point to "" point
= voltage rise from "" point to "" point
• Electric potential energy of ∆q charge at a given point
" = ∆
Electric potential energy difference in moving ∆q charge
between "" and “−” points
∆" = ∆ ( − ) = ∆
"
=
unit: volt (V)
3
*+
*,
= electric energy loss when unit positive charge
(∆ = +1 /) moving from "" to "" points
( = ) − ) =
= electric energy gain when unit positive charge
moving from "" to "" points
Default statements can be rephrased with negative charge
and/or different direction.
(3) Power --- the time rate of supplying or absorbing energy
0=
"
" = 0( ) • In electric circuits, power is associated with moving charges
1=
*+ *+ *,
=
= 34
*2
*, *2
• p>0
power is absorbed by the element
p<0
power is supplied by the element
• Law of conservation of energy (power)
sum of power absorbed = sum of power supplied
or
∑6 06 = 0
(supply and absorb have opposite signs)
4
2. Circuit Elements
(basic, two-terminal elements)
• Passive circuit elements
Resistor
Capacitor
Inductor
• Sources (active circuit elements)
independent source
(value specified)
89 value
specified
dependent source
(dependence specified)
Voltage source
89 = :8;
(89 () specified,
or
9 to be determined)
89 = <;
--------------------------------------------------------------------------------------Current source
9 value
9 = =8;
(9 () specified,
specified
or
89 to be determined)
9 = >;
---------------------------------------------------------------------------------------------------
8; ?@ ; is the voltage or current at other location of the circuit.
5
Examples
70-V is an independent
voltage source
28; is a dependent
voltage source
9-A is an independent
current source
B /4 is a dependent
current source
6
Passive Sign Convention
(Sign used in power equation and device-law equations for any twoterminal element based on the voltage polarity and current direction)
Use "" sign in equation
Use "" sign in equation
Example
Consider 8 > 0 and positive charges entering "" terminal
So the positive charges lose energy.
In 1st case, use "" sign in the power equation
0 = +8
>0
(+ move along current reference direction)
o 0>0
(FGFHFI JKL?@KIM 0?NF@)
In 2nd case, use "" sign in the power equation
0 = −8 < 0
(+ move opposite to current reference direction)
o 0>0
(FGFHFI JKL?@KIM 0?NF@, LJHF @FLQG)
7
Example
Find power for each element.
4 circuit elements in the circuit.
2 elements information given.
Two more pieces information needed.
Let’s make measurements to get two more pieces of information.
RS = 9 U, RV = R; = 5 U
→
dependent voltage source value
0.6 R; = 3 ,
[ = V = 3 \ = 2 + 3 = 5 R[ = 9 − 5 = 4 U
0\ = −5 × 9 U = −45 "
LQ00G^
0S = +2 × 9 U = 18 "
JKL?@K (QIF@ `ℎJ@MIM)
0V = +3 × 5 U = 15 "
JKL?@K
0[ = +3 × 4 U = 12 "
0\ + 0S + 0[ + 0V = 0
8
JKL?@K
`?ILF@8J?I ?b 0?NF@