Transcript Document

Circuit Theory I
Part 2: Charge, Current, Voltage,
Energy, Power, Sources
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
Major Topics of the Course
Circuits Quantities
Charge
Current
Voltage
Energy and Power
Practical Examples
Independent & Dependent Sources
Enginr 124 WS 2002 Part 2: Charge, etc.
2.2 Fundamental Circuits Quantities
q
(charge: coulombs, C)
i
(current: amperes, A)
v
(voltage: volts, V)
w
(energy: joules, J)
p
(power: watts, W)
S
complex power:
volt-amps, VA,
watts, W
volt-amps-reactive, VAR
f
(frequency: hertz, Hz)
Enginr 124 WS 2002 Part 2: Charge, etc.
2.3 Charge
When you rub a comb with a woolen cloth, a
negative charge is produced on the comb and a
positive charge is produced on the cloth.
(Benjamin Franklin defined the charge on the
comb as negative.)
The comb acquires its negative charge because
some of the electrons on the cloth are rubbed off
onto the comb.
One coulomb is defined as the amount of charge
on identically charged particles, separated by
one meter in a vacuum, that repel each other
with a force of 10–7 c2 N, where c is the velocity
of light (2.997… x 108 m/s).
Recall that a single electron charge is –1.602 x
10–19 C. This is quite small: For example, the
charge needed to plate 1 g of copper is 3,036 C.
Enginr 124 WS 2002 Part 2: Charge, etc.
2.4 Current
Charge in motion results in energy transfer. Of
special interest is the case where the motion is
confined to a definite path. In this case, current
“flows.” Current is the movement of
electrical charge—the flow of electrons through
an electronic circuit.
Current (i) is the rate at which charge “flows,” or
is transferred.
Memorize me!!
i = dq/dt amperes
1 amp = 1 coulomb/second
It is convenient to think of current as the flowing
motion of positive charge even though we know
that current flow in metallic conductors results
from electron motion (that is, the charge carriers
are negative and move in the opposite direction.)
Enginr 124 WS 2002 Part 2: Charge, etc.
Charge/Current Relationship
++ + ++ +
i(t)
+ ++ +
i(t)
The charge and the current flow through the
conductor, passing from one side of the
imaginary surface to the other.
During the time interval from ta to tb the charge
passing through the imaginary surface is
tb
q   idt
Memorize me!!
ta
Enginr 124 WS 2002 Part 2: Charge, etc.
In defining current, both an arrow and a symbol
are required. It takes two!
Example:
Circuit analysis shows that in the given circuit, i1
has the value 0.5 A. Give the corresponding
physical interpretation.
10 
5
+
1V
_
i1 = 0.5 A
10 
+
12 V
_
Solution:
Since i1 is 0.5 A, this means that 0.5 A, or 0.5 C/s
is flowing in the direction of the currentreference arrow. This current flows through the 5
, to the + side of the battery, through the
battery, exiting from the – side of the battery.
Enginr 124 WS 2002 Part 2: Charge, etc.
Example (cont.)
Suppose that the current-reference direction and
symbol are changed as indicated. What is the value
of i2?
10 
5
+
_
1 V i2 = ?
10 
12 V
Solution:
Here, the circuit is the same, but the direction of
the current reference arrow is reversed. 0.5 A is
still flowing through the 5 , to the + side of
the battery, through the battery, exiting from the
– side of the battery. The value of i2 has to be
– 0.50 A.
Enginr 124 WS 2002 Part 2: Charge, etc.
Example (cont.)
For the same circuit show four ways to represent
the 0.5 A flowing through the 5  in the direction
towards the positive terminal of the 1 V battery.
5
10 
+
_
1V
10 
12 V
Solution:
i1 = 0.5 A 5 
i2 = –0.5 A
– i3
5
5
– i3 = – 0.5 A
i3 = 0.5 A
– i4
5
– i4 = 0.5 A
i4 = – 0.5 A
Enginr 124 WS 2002 Part 2: Charge, etc.
Current and Charge-Flow
i1 = 0.5 A 5 
+
_
Since i1 = 0.5 A, 0.5 C of charge must be flowing
through the circuit elements each second.
As an equation,
i1 = dq/dt
= 0.5 A
Current is a through variable.Current flows, but
does not accumulate.
Charge is also a through variable.Charge flows,
and (sometimes) accumulates.
Enginr 124 WS 2002 Part 2: Charge, etc.
Types of Current
DC:
Doesn’t Change!
i
DC Current
t
1
0.8
Exponential Current
0.6
0.4
0.2
0
0
5
10
15
20
1
0.8
Damped Sinusoidal
Current
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
0
5
10
15
20
0
5
10
15
20
1
Sinusoidal or
AC Current
0.5
0
-0.5
-1
5
4
3
Transient Current
2
1
0
-1
-2
-3
0
5
Enginr 124 WS 2002 Part 2: Charge, etc.
10
15
20
How Large Can Currents Get?
Phenomenon
Type
 Value
Lightning bolt
transient
106 A
industrial motor
AC
103 A
household appliance
AC
101 A
ventricular fibrillation transient
10–1 A
IC memory cell
transient
10–7 A
brain cell
transient
10–13 A
Enginr 124 WS 2002 Part 2: Charge, etc.
Example 1: DC current.
Find the charge passing through a point in a
conductor if the current passing through the
conductor has the waveform indicated.
i
DC:
Direct Current.
Does not Change!
Current does not
change with time.
2A
t
Solution:
In any 1-second interval the charge that has
flowed past a given point in the conductor will be:
q
t a 1
ta
2dt
t a 1
 [2t] |t
a
2C
Memorize me!!
tb
q   idt
Enginr 124 WS 2002 Part 2: Charge, etc.
ta
Example 2
For the given current, which is given in
graphical form, find the charge passing through
a given point in the time interval 0-4 s.
i
10 A
0
1
2
3
4
t
Solution:
For the time interval 0-4 s, the charge passing
through a given point on the conductor that carries
the current is:
4
q =  i dt
0
= ½ x 1 x 10 + 1 x 10 + ½ x 1 x 10 + ½ x 1 x 10
= 25 C
Enginr 124 WS 2002 Part 2: Charge, etc.
2.5 Voltage
The separation of charge creates an electric force, which was
recognized by the 18th century Italian physicist Allessandro Guiseppi
Antonio Anastasio Volta. Voltage is the energy per unit charge that is
created by the separation.
i
i
...
a
circuit
element
...
b
The current (and charge) flowing from a to b, through the circuit
element, requires an expenditure of energy. Voltage is the amount of
energy expended on one unit of charge to move it. The voltage of
point b with respect to point a is the work per unit charge required to
move that charge from point a to point b:
vba = wba/q
Memorize me!!
1 V = 1 J/ C
We say that there is a voltage difference between points a and b, with
point b at the assumed higher voltage level. We use the words
assumed higher voltage level because if this turns out to be a negative
number, then a is actually at the higher voltage level. This will be an
important concept throughout our study: Assume a direction for the
voltage and solve the required equations for its value. If this value
turns out to be a negative number, we can simply reverse the direction
of the assumed voltage and remove the negative sign if we choose.
However, it would be equally correct to retain the original assumed
polarity and the negative sign in its value.
Enginr 124 WS 2002 Part 2: Charge, etc.
i
...
+
a
v
b
–
i
circuit
element
Voltage is an “across variable.” We say that an
electrical voltage of potential difference exists
between a and b, or that there is a voltage across
the circuit element. The voltage across the element
is the amount of energy that must be given up to
move a unit positive charge from the + terminal to
the – terminal. Voltage is the electrical force, or
“pressure,” that causes current to flow in a circuit.
A voltage or potential difference can exist between
a pair of electrical terminals whether or not a
current is flowing. (There could be nothing
connected on the left side of a and b above. That
is, they could be dangling in midair.)
Enginr 124 WS 2002 Part 2: Charge, etc.
Implied-Source Convention
When you see the circuit element with an
indicated voltage across it, drawn as standalone, the implication may be that there is a
source present to supply the indicated voltage,
even though that source is not explicitly shown.
This usage is referred to as the implied-source
convention. This convention is used to simplify
the drawings.
a
a
+
+
+
v
v
v
_

–
b
–
Enginr 124 WS 2002 Part 2: Charge, etc.
b
In defining voltage, a plus-minus sign-pair AND
a symbol are needed. It takes two!
a
+
a
b
–
b
v
+
v
–
–
a
v
OK!
b
+
a
v
+
–
OK!
b
OK!
–
v
+
OK!
Enginr 124 WS 2002 Part 2: Charge, etc.
a is v volts above b.
a
+
b is v volts below a.
v
The voltage rise from b to a is v volts.
–
b The voltage rise from a to b is –v volts.
The voltage drop from b to a is –v volts.
The voltage drop from a to b is v volts.
–
b
v
+
a
Different orientation, same story:
a is v volts above b.
b is v volts below a.
The voltage rise from b to a is v volts.
The voltage rise from a to b is –v volts.
The voltage drop from b to a is –v volts.
The voltage drop from a to b is v volts.
Enginr 124 WS 2002 Part 2: Charge, etc.
Voltage and Potential Difference
The voltage across two nodes in the circuit is
the potential difference between the two nodes.
The potential difference may utilize a third node
as a point of reference.
The analogy with elevation difference:
29,000 feet
+
 = 9,000 feet
20,000 feet
–
Mount
Everest
Sea
0 feet (sea level,
the reference )
Enginr 124 WS 2002 Part 2: Charge, etc.
Double-Subscript Notation
a
vab
b
vab is the drop from a to b.
vba is the drop from b to a. (Note
that this voltage is not specifically
shown here.)
The voltage at a is higher than that at b by vab.
The plus-minus sign-pair is not needed here.
For the symbol vab, regard terminal a as the +
terminal and terminal b as the – terminal.
For the symbol vba, regard terminal a as the –
terminal and terminal b as the + terminal.
Enginr 124 WS 2002 Part 2: Charge, etc.
Voltmeters
The voltage across two nodes is measured by a voltmeter.
The voltmeter measures the voltage drop between its +
terminal and its – terminal.
The voltmeters below are measuring branch voltages: the
voltages across the branches containing the 5  resistor
branch and the 1 V voltage source branch.
–
2.5 V +
5
+
_
1V
10 
10 
12 V
voltmeter
voltmeter
1.00
+ _
–2.50
+ _
Enginr 124 WS 2002 Part 2: Charge, etc.
2.6 Energy and Power
a
circuit
element
q
b
In keeping with the principle of conservation of
energy, the energy that is utilized in moving the
charge through the element is conserved (although it
does change form).
Depending on the particular kind of circuit element
present, this energy:
-- comes from the circuit element, or
-- goes to the circuit element and is converted
irreversible into heat, light, acoustic energy, or
some other non-electrical form, or
-- goes to the circuit element and is stored in
some form that is still available as electric
energy.
Enginr 124 WS 2002 Part 2: Charge, etc.
Power in Terms of v and i
w = energy in joules, J
p = power in watts, W
= energy/time
= dw/dt
Recall that v  dw/dq, so that
dw = vdq so that
p = dw/dt
= vdq/dt
Memorize me!!
p = vi
Check using dimensions: J/C x C/s = J/s = W.
Power is the product of the across variable v and
the through variable i. In a translational system,
p = force x velocity = through x across.
Enginr 124 WS 2002 Part 2: Charge, etc.
Example: Energy Calculation (A&S P1.13)
The voltage and current for a device are shown. Find the
total energy absorbed by the device during the interval 0-4 s.
i
50
(mA)
+
v(t)
i(t)
0
–
v
(V)
Solution:
1
2 3 4
t(s)
1
2 3 4
t(s)
10
0
t
t
0
0
w(t)   p(t)dt   v(t)i(t)dt
By referring to the graph we find that the equations for i(t)
and v(t) are:
for 0  t  2
25t
i(t)  
25t  100 for 2  t  4
for 0  t  1
10t

v(t)  10
for 1  t  3

10t  40 for 3  t  4
Enginr 124 WS 2002 Part 2: Charge, etc.
Solution (cont.):
50
i
(mA)
0
for 0  t  2
25t
i(t)  
25t  100 for 2  t  4
1
2 3 4
t(s)
v 10
(V)
0
for 0  t  1
10t

v(t)  10
for 1  t  3

10t  40 for 3  t  4
1
2 3 4
t(s)
1
2
3
4
0
1
2
3
w(4)   25t 10tdt   25t 10dt    25t  100  10dt    25t  100    10t  40  dt
1
2
3
4
1
2
3


 250  t dt  250  tdt 10   25t  100  dt   250t 2  2000t  4000 dt
2
0

t 3
 25 t 3
250 250 2 t  2
t  3  250 3 t  4

t
 10   t 2
 100 t t  2  
t
 1000 t 2
 4000
t 1
t

2
t

3
t

2
3
2
2
3



250
 125
 250
 125  3  10  
 100  
 37  1000  7  4000
3
2
3


 916.7 mJ
Enginr 124 WS 2002 Part 2: Charge, etc.
Passive Sign Convention
i
...
+
v
i
circuit
element
–
Within the circuit element, the current arrow is
pointing in the direction from + to – (in the
direction of the voltage drop v within the element).
When this passive sign convention is being used:
If v i > 0, then the circuit element is absorbing
energy (getting hot!--if it’s a resistor).
If v i < 0, then the circuit element is supplying
energy. (A resistor can’t do this!)
Enginr 124 WS 2002 Part 2: Charge, etc.
When the passive convention is the one that “fits”*:
i
any
…
+
circuit
+
element: i
v 
v
–
–
…
power absorbed by the circuit element:
pabsorbed = v i
power supplied by the circuit element:
psupplied = – v i
The power absorbed by a circuit element and the
power supplied by that same circuit element are
related by
pabsorbed = – psupplied
* We say it this way to make the point that the current and
voltage notations may already be set up for you on the
circuit and you may not have the flexibility to change them.
Enginr 124 WS 2002 Part 2: Charge, etc.
When the active convention is the one that “fits”:
i
any
+
circuit
+
element: i
v 
v
–
–
power absorbed by the circuit element:
pabsorbed = – v i
power supplied by the circuit element:
psupplied = v i
The power absorbed by a circuit element and the
power supplied by that same circuit element are
related by
pabsorbed = – psupplied.
Enginr 124 WS 2002 Part 2: Charge, etc.
Example 1
Determine how much power the circuit element
is either absorbing or delivering.
3A
...
+ circuit
v = 2 V element
– p = ???
3A
Red (absorbing)
Solution:
Remember: When the passive sign convention is
being used: If the numerical value of the product vi is
positive, then power is truly being absorbed by the
circuit element. If vi is negative, then the circuit
element is absorbing negative power, or delivering it
to the external circuit. In this case,
p = vi
= (2) (3)
= 6 W absorbed by the circuit element
Enginr 124 WS 2002 Part 2: Charge, etc.
Example 2
Determine how much power the circuit element
is either absorbing or delivering.
–3 A
...
–
v=–2V
+
Solution:
circuit
element
p = ???
–3 A
Red (absorbing)
Here, you need to recognize that the current
reference direction is from + to – within the circuit
element, so that the passive convention is again
the one being used. Thus,
p = vi
= (– 2) (– 3)
= 6 W absorbed by the circuit element
Enginr 124 WS 2002 Part 2: Charge, etc.
Example 3
Determine how much power the circuit element
is either absorbing or delivering.
–5A
+
circuit
v = 4 V element
–
p = ???
...
–5A
Blue (supplying)
Solution:
Once more, the passive convention is in effect,
since the direction of the reference current is from
+ to – with the circuit element. Thus,
p = vi
= 4 (–5)
= –20 W absorbed , or 20 W delivered
Enginr 124 WS 2002 Part 2: Charge, etc.
Example 4
Determine how much power the circuit element
is either absorbing or delivering.
–6A
+
circuit
v = 10 V element
–
p = ???
...
–6A
Red (absorbing)
Solution:
Here, the active convention is in effect, since the
direction of the reference current is from – to +
within the circuit element. Remember: When the
active sign convention is being used: The power
absorbed is the negative of the product of the
voltage and the current. In this case,
p = –vi
= –(10) (–6)
= 60 W absorbed by the circuit element
Enginr 124 WS 2002 Part 2: Charge, etc.
Source-Load Circuit
A simple model for many circuits consists of
one source and one load. The source produces
the power and the load consumes the power.
There is a voltage rise at the source and a
voltage drop at the load (when proceeding in
the direction of the reference current).
i
+
Source
V
Load
–
The source get its
energy through this
connection.
Voltage rise in
the direction of
the current:
power
supplied.
Voltage drop in
the direction of
the current:
power
consumed.
Enginr 124 WS 2002 Part 2: Charge, etc.
Example 6
Show the proper cable connections for using a car battery
in a one car to “jump-start” another car that has a
discharged battery.
Solution:
Connect the jumper cables as follows.
i

+
i
Normal battery

+
Discharged battery
i
+
+
v
v

Delivering vi watts
i

Absorbing vi watts
Voltage sources CAN absorb power!
Enginr 124 WS 2002 Part 2: Charge, etc.
Source-Load Circuit: Hydraulic
Analog
At the source (pump) there is a pressure rise in
the direction of the flow.
At the turbine (load) there is a pressure drop in
the direction the flow.
High pressure
Pump
The pump get its
energy through this
connection.
Turbine
Low pressure
Enginr 124 WS 2002 Part 2: Charge, etc.
Power Conservation Theorem
The sum of the powers absorbed by all the
elements in a circuit equals zero.
Example
Verify the power conservation theorem for the
given circuit. + 240 V –
0.5 A
2
+
+
+
2 A 1.5 A
300 V 1
3 60 V
4 60 V
2A
–
–
–
Solution:
p1 = – (300) (2) = – 600 W
Blue
p2 =
Red
(240) (2) =
480 W
p3 = (60) (1.5) =
90 W
Red
p4 = (60) (0.5) =
 pi
30 W
= – 600 + 480 + 90 + 30
Red
White
= 0
Enginr 124 WS 2002 Part 2: Charge, etc.
Passive Element
A passive circuit element is one that never
supplies energy.
Example
Show that the given element behaves passively
for the given conditions.
+
V
–
i
i = sin t, v = cos t
Solution:
p = vi = cos t sin t = .5 sin 2t
t
w =  0 pdt
t
=  .5 sin 2t dt
0
= – .25 cos 2t |
t
0
= 2(1 – cos 2t)
>= 0
Passive!
Enginr 124 WS 2002 Part 2: Charge, etc.
2.7 Practical Examples
Example 1.
The energy stored in car batteries is specified in ampere-hours
instead of joules. A typical 12 V car battery may contain 200
ampere-hours of energy.
Calculate this energy in megajoules:
Solution:
p = vxi
w = v x i x t (since vx i is constant)
= 12 V x 1 A x 200 hr
= 12 V x 1 A x 200 x 3600 s
= 8.5 x 106 J
= 8.5 MJ
This is the total energy stored in the fully charged battery.
We note that if the headlights of the vehicle are inadvertently
left on after the engine is shut down and they consume 50 W,
the current drawn by the lights will be 50 W / 12 V = 4.17 A.
Hence, the battery should last 200 A-h / 4.17 A = 48 h--an
unrealistic answer in that the battery would not be able to
supply 4.17 A right up to the point that it is completely dead.
Enginr 124 WS 2002 Part 2: Charge, etc.
Example 1 (cont.)
We found that the battery contains 8.5 MJ of
energy. Suppose the battery sells for $40.
Calculate the cost of the equivalent amount of
energy if purchased from a typical electrical
utility at a cost of 10¢ per kilowatt-hour.
Solution:
1 kWhr = 1000 J/s x 1 hr x 3600 s/hr
= 3.6 M J
( 10¢ worth of energy)
8.5/3.6 x 10 ¢  24¢
Only a little more than 2 kWhr of energy is stored
in the battery. But it is re-chargeable. And there’s no
cord attached!
Enginr 124 WS 2002 Part 2: Charge, etc.
Example 1 (cont.)
Portable batteries (9 V, D cells, C cells, AA
cells, etc.) typically have less than 1 MJ of
energy stored in them. Their cost can be
1,000 times the cost of the equivalent cost of
energy purchased from an electric utility.
Clearly, batteries are used because of their
portability, convenience, etc.--not their low
price.
Enginr 124 WS 2002 Part 2: Charge, etc.
Example 2
A smoke alarm draws 6 A from a 9 V battery.
Assuming that the battery stores 18.3 kJ of energy,
how often should the battery be replaced?
6 A
Solution:
w = pt
9V
smoke
alarm
= vit
t = w/vi = 18.3 x 103 / 9 x 6 x 10–6
= 338.9 x 106 s
338.9 x 106 s x 1 min/60 s x 1 hr/60 min
x 1 day/24 hr x 1 yr/365 day = 10.7 yr
However, the battery’s internal leakage current will
discharge it in about 4 years, and energy is
consumed each time the smoke alarm is tested.
Play it safe. Replace the battery as recommended
by the manufacturer.
Enginr 124 WS 2002 Part 2: Charge, etc.
Noted in passing:
The human brain consumes about 10–17 J per
lowest-level operation (10 W brain with 1016
synapses operating at about 10 nerve pulses per
second).
The fastest microprocessors require about
10 –9 J per switching operation.
Comparing these energies we see that the
human brain is 10,000,000 times more efficient
than the most efficient conventional computers!
Enginr 124 WS 2002 Part 2: Charge, etc.
2.8 Independent and Dependent Sources
Independent Voltage Source
+
v
+
–
V
i
I
–
Lower-case v indicates a time- UPPER-CASE V indicates a
varying voltage.
constant (DC) voltage.
V
v
t
For an independent voltage source, the
voltage is v, no matter what.
The current is whatever.
Enginr 124 WS 2002 Part 2: Charge, etc.
t
Independent Current Source
+
v
+
V
i
–
I
–
Lower-case v indicates a time- Upper-case I indicates a
varying current.
constant (DC) current.
I
i
t
For an independent current
source, the current is i, no matter
what.
The voltage is whatever.
Enginr 124 WS 2002 Part 2: Charge, etc.
t
Dependent Sources
i
v = vx
VCVS: v =  vx
i = whatever
+
–
vx is somewhere (not shown)
i
v = i x
CCVS: v =  ix
i = whatever
+
–
ix is somewhere (not shown)
+
i = v
x
VCCS: i =  vx
v
v = whatever
–
vx is somewhere (not shown)
Enginr 124 WS 2002 Part 2: Charge, etc.
Dependent Sources (cont.)
+
i =  ix
v
–
CCCS: i =  ix
v = whatever
ix is somewhere (not shown)
There are three items that must be specified in order to
completely specify any dependent source:
1. the polarity of the source (+ and  signs or an arrow with
the diamond),
2. the source constant , , , , and
3. the controlling variable, by completely specifying the
controlling branch and the polarity of the controlling
variable on that branch.
Memorize what these icons signify!

independent source
 dependent source
Enginr 124 WS 2002 Part 2: Charge, etc.
Controlled sources are used primarily to model electronic
devices. For example, the junction field-effect transistor
(JFET) and the bipolar junction transistor (BJT) are
modeled as shown below. These devices are used to
construct electronic circuits such as amplifiers and digital
computers. Without dependent sources we would not be
able to model these important electrical components.
D

G
D
G +
gmvGS
vGS
S

JFET
C
S
D
B

B
ro
iB
ro
E
ro
S
BJT
Enginr 124 WS 2002 Part 2: Charge, etc.
Example Using Dependent Sources
Calculate the power absorbed by the dependent
source. Assume that we are given that Ix = –2 mA.
8 mA
Ix
V=4Ix*
+ –
* Normally the
designation is
just 4Ix, not
V=4Ix.
+ 10 V
–
Solution:
For the power absorbed, we want the product of
the voltage across the dependent source and the
current through it in the direction of the voltage
drop.
p = – (4 Ix) (8 x 10–3)
Minus sign needed because the active
convention is present at the
dependent source.
=
– 4 (–2 x 10 –3) ( 8 x 10 –3)
=
64 mW
Enginr 124 WS 2002 Part 2: Charge, etc.
Does it matter?
Absolutely.
Enginr 124 WS 2002 Part 2: Charge, etc.
Non-Permissible Connections
5A
10 V +
–
+ 8V
–
vx = 5 V +
–
10 V +
–
8A
+
–
vs = 3 vx
“Short
Circuit”
What happens if you make the connections, anyway?
Sparks fly,
circuit breakers pop,
nonlinear things occur,
sometimes, expensive damage results!
Enginr 124 WS 2002 Part 2: Charge, etc.