Chapter 1
Basic Electric Circuit Concepts
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BASIC CONCEPTS
LEARNING GOALS
•System of Units: The SI standard system
•Systeme International unit (=International System of Units)
•(法) 國際單位制
•Basic Quantities: Charge, current, voltage, power and energy
•Circuit Elements: Active and Passive
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4
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SI DERIVED BASIC ELECTRICAL UNITS
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CURRENT AND VOLTAGE RANGES
7
Strictly speaking current is a basic quantity and charge is derived. However,
physically the electric current is created by a movement of charged particles.
What is the meaning of a negative value for q(t)?
+
+
+
+
q(t )
PROBLEM SOLVING TIP
IF THE CHARGE IS GIVEN DETERMINE THE CURRENT BY
DIFFERENTIATION
IF THE CURRENT IS KNOWN DETERMINE THE CHARGE BY
INTEGRATION
A PHYSICAL ANALOGY THAT HELPS VISUALIZE ELECTRIC
CURRENTS IS THAT OF WATER FLOW.
CHARGES ARE VISUALIZED AS WATER PARTICLES
8
EXAMPLE
EXAMPLE
⎧ 0 t <0
i (t ) = ⎨ −2 t
⎩e mA t ≥ 0
q (t ) = 4 × 10 −3 sin(120π t )[C ]
+
+
+
+
i (t ) = 4 × 10 −3 × 120π cos(120π t ) [ A]
i ( t ) = 0.480π cos(120π t ) [mA]
FIND THE CHARGE THAT PASSES
DURING IN THE INTERVAL 0<t<1
1
q(t )
q = ∫e
0
−2 x
1
1
1
1
dx = − e −2 x = − e − 2 − (− e 0 )
2
2
2
0
1
q = (1 − e − 2 )
2
Units?
FIND THE CHARGE AS A FUNCTION OF TIME
q(t ) =
t
t
∫ i ( x )dx = ∫ e
−∞
−2 x
dx
−∞
t ≤ 0 ⇒ q(t ) = 0
t
1
t > 0 ⇒ q ( t ) = ∫ e − 2 x dx = (1 − e −2 t )
2
0
And the units for the charge?...
9
DETERMINE THE CURRENT
Here we are given the
charge flow as function
of time.
Charge(pC)
30
20
10
− 10
− 10 × 10−12 − 10 × 10−12 C
−9
m=
=
−
10
×
10
(C / s )
−3
s
2 × 10 − 0
1 2 3 4 5 6
Time(ms)
Current(nA )
To determine current we
must take derivatives.
PAY ATTENTION TO
UNITS
40
30
20
10
− 10
− 20
1 2 3 4 5 6
Time(ms)
10
CONVENTION FOR CURRENTS
THE DOUBLE INDEX NOTATION
IT IS ABSOLUTELY NECESSARY TO INDICATE
THE DIRECTION OF MOVEMENT OF CHARGED
PARTICLES.
THE UNIVERSALLY ACCEPTED CONVENTION IN
ELECTRICAL ENGINEERING IS THAT CURRENT IS
FLOW OF POSITIVE CHARGES.
AND WE INDICATE THE DIRECTION OF FLOW
FOR POSITIVE CHARGES
-THE REFERENCE DIRECTIONA POSITIVE VALUE FOR
THE CURRENT INDICATES
FLOW IN THE DIRECTION
OF THE ARROW (THE
REFERENCE DIRECTION)
A NEGATIVE VALUE FOR
THE CURRENT INDICATES
FLOW IN THE OPPOSITE
DIRECTION THAN THE
REFERENCE DIRECTION
IF THE INITIAL AND TERMINAL NODE ARE
LABELED ONE CAN INDICATE THEM AS
SUBINDICES FOR THE CURRENT NAME
a
5A
b
I ab = 5 A
a 3A b a −3A b
I ab = 3 A
I ab = −3 A
a −3A b a 3A b
I ba = −3 A
POSITIVE CHARGES
FLOW LEFT-RIGHT
I ba = 3 A
POSITIVE CHARGES
FLOW RIGHT-LEFT
I ab = − I ba
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I = −2 A
a
2A
I
b
I cb = 4 A
I ab =
c
3A
This example illustrates the various ways
in which the current notation can be used
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CONVENTIONS FOR VOLTAGES
ONE DEFINITION FOR VOLT
TWO POINTS HAVE A VOLTAGE DIFFERENTIAL OF
ONE VOLT IF ONE COULOMB OF CHARGE GAINS
(OR LOSES) ONE JOULE OF ENERGY WHEN IT
MOVES FROM ONE POINT TO THE OTHER
b IF THE CHARGE GAINS
ENERGY MOVING FROM
a TO b THEN b HAS HIGHER
VOLTAGE THAN a.
IF IT LOSES ENERGY THEN
b HAS LOWER VOLTAGE
THAN a
+ a
1C
DIMENSIONALLY VOLT IS A DERIVED UNIT
VOLT =
N •m
JOULE
=
COULOMB A • s
VOLTAGE IS ALWAYS MEASURED IN A RELATIVE FORM AS THE VOLTAGE DIFFERENCE
BETWEEN TWO POINTS
IT IS ESSENTIAL THAT OUR NOTATION ALLOWS US TO DETERMINE WHICH POINT
HAS THE HIGHER VOLTAGE
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THE + AND - SIGNS
DEFINE THE REFERENCE
POLARITY
V
IF THE NUMBER V IS POSITIVE POINT A HAS V
VOLTS MORE THAN POINT B.
IF THE NUMBER V IS NEGATIVE POINT A HAS
|V| LESS THAN POINT B
POINT A HAS 2V MORE
THAN POINT B
POINT A HAS 5V LESS
THAN POINT B
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THE TWO-INDEX NOTATION FOR VOLTAGES
INSTEAD OF SHOWING THE REFERENCE POLARITY
WE AGREE THAT THE FIRST SUBINDEX DENOTES
THE POINT WITH POSITIVE REFERENCE POLARITY
V AB = 2V
V AB = −5V
VBA = 5V
V AB = −VBA
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ENERGY
VOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…
CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB OR
RELEASE ENERGY – THEY MAY TRANSFER ENERGY FROM ONE POINT TO ANOTHER
BASIC FLASHLIGHT
Converts energy stored in battery
to thermal energy in lamp filament
which turns incandescent and glows
EQUIVALENT CIRCUIT The battery supplies energy to charges.
Lamp absorbs energy from charges.
The net effect is an energy transfer
Charges gain
energy here
Charges supply
Energy here
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ENERGY
VOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…
CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB OR
RELEASE ENERGY
WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROM
POINT B TO POINT A IN THE CIRCUIT?
THE CHARGES MOVE TO A POINT WITH HIGHER
VOLTAGE -THEY GAINED (OR ABSORBED) ENERGY
THE CIRCUIT SUPPLIED ENERGY TO THE CHARGES
V AB = 2V
V=
W
⇒ W = VQ = 240 J
Q
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EXAMPLE
A CAMCODER BATTERY PLATE CLAIMS THAT
THE UNIT STORES 2700mAHr AT 7.2V.
WHAT IS THE TOTAL CHARGE AND ENERGY
STORED?
ENERGY AND POWER
2[C/s] PASS
THROUGH
THE ELEMENT
CHARGE
THE NOTATION 2700mAHr INDICATES THAT
THE UNIT CAN DELIVER 2700mA FOR ONE
FULL HOUR
s
⎡C ⎤
Q = 2700 × 10−3 ⎢ ⎥ × 3600
× 1Hr
Hr
⎣S⎦
= 9.72 × 103[C ]
TOTAL ENERGY STORED
THE CHARGES ARE MOVED THROUGH A 7.2V
VOLTAGE DIFFERENTIAL
⎡J ⎤
W = Q[C ] × V ⎢ ⎥ = 9.72 × 103 × 7.2[ J ]
⎣C ⎦
= 6.998 × 10 [ J ]
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EACH COULOMB OF CHARGE LOSES 3[J]
OR SUPPLIES 3[J] OF ENERGY TO THE
ELEMENT
THE ELEMENT RECEIVES ENERGY AT A
RATE OF 6[J/s]
THE ELECTRIC POWER RECEIVED BY THE
ELEMENT IS 6[W]
IN GENERAL
P = VI
t2
w (t 2 , t1 ) = ∫ p( x )dx
t1
HOW DO WE RECOGNIZE IF AN ELEMENT
SUPPLIES OR RECEIVES POWER?
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PASSIVE SIGN CONVENTION (慣例)
THIS IS THE REFERENCE FOR POLARITY
+
POWER RECEIVED IS POSITIVE WHILE POWER
SUPPLIED IS CONSIDERED NEGATIVE
+ Vab −
a
b
I ab
P = Vab I ab
IF VOLTAGE AND CURRENT
ARE BOTH POSITIVE THE
CHARGES MOVE FROM
HIGH TO LOW VOLTAGE
AND THE COMPONENT
RECEIVES ENERGY --IT IS
A PASSIVE ELEMENT
A CONSEQUENCE OF THIS CONVENTION IS THAT
THE REFERENCE DIRECTIONS FOR CURRENT AND
VOLTAGE ARE NOT INDEPENDENT -- IF WE
ASSUME PASSIVE ELEMENTS
GIVEN THE REFERENCE POLARITY
+ Vab −
a
b
REFERENCE DIRECTION FOR CURRENT
−
a
b
I ab
IF THE REFERENCE DIRECTION FOR CURRENT
IS GIVEN
EXAMPLE
+ Vab −
2A
a
b
I ab
Vab = −10V
THE ELEMENT RECEIVES 20W OF POWER.
WHAT IS THE CURRENT?
SELECT REFERENCE DIRECTION BASED ON
PASSIVE SIGN CONVENTION
20[W ] = Vab I ab = ( −10V ) I ab
I ab = −2[ A]
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UNDERSTANDING PASSIVE SIGN CONVENTION
We must examine the voltage across the component
and the current through it
I
A
A’
+
S1
B
Current A - A'
positive
positive
positive negative
negative positive
negative negative
Voltage(V)
V
−
S1
supplies
receives
receives
supplies
PS1 = V AB I AB
S2
PS 2 = V A'B ' I A'B '
B’
S2
ON S1
ON S2
receives VAB > 0, I AB < 0 VA B > 0, I A B > 0
supplies
ON S2
supplies
V A'B ' < 0, I A'B ' > 0
receives
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'
'
'
'
CHARGES RECEIVE ENERGY.
THIS BATTERY SUPPLIES ENERGY
CHARGES LOSE ENERGY.
THIS BATTERY RECEIVES THE ENERGY
WHAT WOULD HAPPEN IF THE CONNECTIONS ARE REVERSED
IN ONE OF THE BATTERIES?
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Determine whether the elements are supplying or receiving power
and how much.
a
a
Vab = 2V
I ab = 4 A
−2 A
I ab = −2 A
Vab = −2V
P = −8W
SUPPLIES POWER
b
P = −4W
ABSORBS POWER
b
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Determine the amount of power absorbed or supplied by the elements ?
1
1
2
2
V12 = 12V , I12 = −4 A
V12 = 4V , I12 = 2 A
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I = −8[ A]
V AB = −4[V ]
− 20[W ] = V AB × (5 A)
+
+
−
−
SELECT VOLTAGE REFERENCE POLARITY
BASED ON CURRENT REFERENCE DIRECTION
40[W ] = (−5V ) × I
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V1 = −20[V ]
40[W ] = V1 × ( −2 A)
− 2A
I = −5[ A]
SELECT HERE THE CURRENT REFERENCE DIRECTION
− 50[W ] = (10[V ]) × I
BASED ON VOLTAGE REFERENCE POLARITY
WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION
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COMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENT
P1 = (6V )(2 A)
2 A + 6V −
+
1
24V
−
+
+
-
3
2
2A
18V
−
P1 = 12W
P2 = 36W
P3 = -48W
P2 = (18V )(2 A)
P3 = (24V )(−2 A) = (−24V )(2 A)
Tellegen’s theorem: the sum of the powers absorbed by all elements in an electrical
network is zero. Another statement of this theorem is that the power supplied in a
network is exactly equal to the power absorbed.
IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT
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CIRCUIT ELEMENTS
PASSIVE ELEMENTS
VOLTAGE
DEPENDENT
SOURCES
UNITS FOR μ , g , r , β ?
INDEPENDENT SOURCES
CURRENT
DEPENDENT
SOURCES
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EXERCISES WITH DEPENDENT SOURCES
FIND VO
VO = 40[V ]
FIND I O
I O = 50mA
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DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES
40[V ]
P = (40[V ])(−2[ A]) = −80[W ]
TAKE VOLTAGE POLARITY REFERENCE
P = (−10[V ])(4 × 4[ A]) = −160[W ]
TAKE CURRENT REFERENCE DIRECTION
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POWER ABSORBED OR SUPPLIED BY EACH
ELEMENT
P1 = (12V )(4 A) = 48[W ]
P2 = ( 24V )(2 A) = 48[W ]
P3 = (28V )(2 A) = 56[W ]
PDS = (1I x )(−2 A) = (4V )(−2 A) = −8[W ]
P36V = (36V )(−4 A) = −144[W ]
NOTICE THE POWER BALANCE
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USE POWER BALANCE TO COMPUTE Io
− 12W
(12)(−9)
(6)( I O )
(10)(−3)
(4)(−8)
(8 × 2)(11)
POWER BALANCE
I O = 1[ A]
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