Transcript Example 2

Circuit Theory I
Part 3. Ohm’s Law, KCL, & KVL
3.1
3.2
3.3
3.4
3.5
Ohm’s Law & Resistance
Passive & Active Conventions
Conductance
Kirchoff’s Current Law
Kirchhoff’s Voltage Law w/o
Ohm’s Law
3.6 KVL with Ohm’s Law
71
3.1 Ohm’s Law and Resistance
+
v
–
For some circuit elements the
voltage-current relationship
is particularly simple—it is
linear:
i
voltage, v
current, i
slope = v/ i
= constant
= R (Ohms
=W )
Circuit elements like the above are called
resistors. The voltage-current relationship is
v = Ri
Resistance causes an opposition to the flow
of current in a circuit.
72
Resistor Relationships
.
.
v = Ri
.
i
+
v
p
R
Memorize
all of me!!
= vi
= R i2
– ..
= v2 / R
.
The resistor never supplies power since the
energy (or power) absorbed is never negative:
E = 
t2
p dt
t1
t2
=  v2 / R dt
t1
>= 0
(never negative  passive)
The resistor is a passive circuit element. It always absorbs
power and never supplies power. It is dissipative,
converting electrical energy into heat. This heat energy is
lost. (It cannot be recovered and converted back to
electrical form.)
73
Example: Power Loss in the Resistance of an
Electrical Cable
The electrical cable in residences is typically 12gauge copper wire. Its resistance is about 0.005 W /
m. If one of the circuits in your residence has 10 m
of this cable, what percentage power loss does it
represent when the load in that circuit is drawing
15A?
Solution:
For 10 m of this cable, the total resistance is 0.05
W The circuit model we use consists of: a series
connection of a voltage source, a 0.05 W resistor
representing the electrical cable, and a circuit
element representing the load. The same current
(15 A) flows through all three of these (series)
15 A
circuit elements.
Model:
+
110 V -
15 A
.05 W
15 A
15 A
load
15 A
Part 3: Ohms Law, etc
74
Example (cont.)
15 A
15 A
110 V
+
–
.05 W
15 A
15 A
load
15 A
The total power supplied by the 110 V source is:
power supplied = v i
= 110 x 15
= 1.65 kW
The power dissipated in the cable is:
power dissipated = R i2
= .05 x 152
= 11.25 W
As a percentage, the power dissipated is:
11.25 W / 1.65 kW  .7 %
Part 3: Ohms Law, etc
75
Resistor Color Code
To determine the value of a given resistor, position
the gold tolerance band (5%) or silver tolerance
band (10%) to the right. Next, match the first two
colors to their corresponding digits. Next, match
the 3rd color band with the corresponding
multiplier. The resistance is then the product of the
first two numbers and the multiplier.
For example, red black yellow
gold
200 K resistor:
2 0 x10,000
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Gray
White
0
1
2
3
4
5
6
7
8
9
5%
gold: 5%
1 = 100
(silver: 10%)
10 = 101
100 = 102
1 K = 103
10 K = 104
100 K = 105
1 M = 106
etc…
silver:divide by 100 = 10-2
gold: divide by 10 = 10-1
Part 3: Ohms Law, etc
76
Resistor Color Code (cont.)
To remember the color codes memorize:
Big Brown Rabbits Often Yield
Great Big Vocal Groans When
Gingerly Slapped.
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Gray
White
0
1
2
3
4
5
6
7
8
9
1W
10 W
100 W
1 KW
10 KW
100 KW
1 MW
silver:divide by 100
gold: divide by 10
Part 3: Ohms Law, etc
77
3.2 Passive & Active Conventions
Ohm’s Law when the Passive Convention (for
associating the voltage reference polarities and the
current reference direction) is being used:
i
+
v
R
v
–
i
–
R
+
v = Ri
The direction of the
reference current is the
same as the direction of
the voltage drop within
the resistor.
Ohm’s Law When the Active Convention Is
Being Used:
i
+
v
–
R
i
–
v
+
R
v =
–Ri
The direction of the
reference current is
opposite to the direction
of the voltage drop
within the resistor. (This
is not a “negative
resistor”!)
Part 3: Ohms Law, etc
78
(cont.)
You must determine the direction of the
reference current within the resistor in relation
to the polarity markings!
i
R
+
–
v
The direction of the reference current within the
resistor is the same as the direction of the voltage
drop, so
v = Ri
i
–
R
v
+
The direction of the reference current is the
opposite of the direction of the voltage drop, so
v =
–Ri
Part 3: Ohms Law, etc
79
Example
Verify that Ohm’s Law holds for the portion of
the circuit shown below. [Note that only part of
a complete circuit is shown. We are not shown
the part of the circuit to the left of the terminal
pair a-b—that’s what the ellipses (… ) are
telling us.]
… a
i=–2A
+
v=–8V 4W
i=–2A
… –b
i=–2A
Solution:
Substitute the values of v, R, and i in Ohm’s
Law:
v = Ri
– 8 = (4) (–2)
– 8 = – 8 checks!
Part 3: Ohms Law, etc
80
Resistivity
Resistivity , measured in W-meters, is an
electrical property of a material that governs
the material’s resistance. For a cylindrical body
of length l and cross-sectional area A,
R=l/A
 (ohm-m)
material
Conductors:
copper
1.7 x 10–8
aluminum
2.7 x10–8
nichrome
1.0 x 10–6
carbon
3.5 x 10–5
Semiconductor: silicon
2.3 x 105
Insulators:
rubber
1012
polystyrene
1015
Part 3: Ohms Law, etc
81
Example 1
For the given circuit, how much energy does the resistor
consume in one year? What is the annual cost of this energy if
purchased from an electrical utility at 15¢ / kilowatt-hour?
…
10 kW
–
10 V
…
+
Solution:
1. Calculate the power absorbed by the resistor:
p = v2 / R = 102 / 10,000 = 0.01 W.
2. Calculate the energy consumed in one year:
(watts) x (hours in one year) = watt-hours / year
= .01 W x 365 days/yr x 24 hr/day
= 87.6 watt-hours / year = 0.0876 kilowatt-hours / yr.
3. Calculate the cost of using 87.6 watt-hours / year:
0.0876 kilowatt-hours / year x $0.15 / kilowatt-hour
= 1.3 cents / year.
Note that the 10 V is a DC voltage, and so it would have to provided by a
power supply that would require more energy to operate than the 87.6 watthours / year.
Part 3: Ohms Law, etc
82
Example 2 (See Special Note on next
page.)
(a) What is the resistance of a 100-W incandescent bulb?
How much current does it draw?
(b) Repeat for a 1500-W portable electric heater.
…
100 watts
–
120 V
…
+
Solution:
(a) We can calculate the resistance of the incandescent bulb
by using the given power rating and the knowledge that the
operating voltage is 120 V:
p = v 2 / R  R = v2 / p
= 1202 / 100
= 144 W
The current is just p / v, or 100 / 120 = 0.833 A.
We note that a typical residence at any one time during the
daylight hours will draw about 10 A although the peak draw
may be much higher over a short period of time (such as when
your toaster, air conditioner, oven, and electric dryer are all on
at the same time. A typical capacity is 200 A.
Part 3: Ohms Law, etc
83
Solution (cont.)
(b) For the 1500-W portable electric heater, the resistance
is
R = v2 / p
= 1202 / 1500
= 9.6 W
The current is p / v, or 1500 / 120 = 12.5 A.
Comment: In many homes, each individual circuit is
protected by a 15-A circuit breaker. If, for example, your
heater and three 100-W bulbs were all being supplied by
the same individual circuit, the circuit breaker would trip.
Special Note:
We are working ahead here, since the quantities in Example 2
are AC quantities, rather than DC quantities. However, the
formulas we used are valid in both the DC case and the AC
case, as we will see later on in the course. To be more precise,
the voltages and currents in the AC case are RMS quantities
and the power in the average power. We will cover AC
circuits in Part 3 of the course.
Part 3: Ohms Law, etc
84
3.3 Conductance
Conductance, G, is the reciprocal of resistance.
Its SI units are siemens, S.
We also use the unit mho, for reciprocal
Ohm (W-1).
i
+
v
G
+
i
v
G
–
–
i = Gv
i =–Gv
p =vi
= G v2
= i2 / G
Part 3: Ohms Law, etc
85
3.4 Kirchoff’s Current Law
KCL: At any instant of time the algebraic sum
of the currents leaving any node is zero.
Example
Find ia.
i = 2A
b
i = ??
ic = – 3 A
a
node
i = –12 A
e
i = 5A
d
Solution:
Set the sum of the currents leaving the node to 0:
Step 1: ia + ib + ic – id – ie = 0
Step 2: Substituting the known numerical values:
ia + 2
– 3 – 5 – (– 12) = 0
Solving for ia gives: ia = – 6 A.
Part 3: Ohms Law, etc
86
Another Formulation of KCL
At any instant of time the algebraic sum of the
currents entering any node is zero.
Same Example
Find ia.
i = 2A
b
i =–3A
c
node
ia = ??
i = –12 A
e
Solution:
i = 5A
d
Set the sum of the currents entering the node to 0:
Step 1: – ia – ib – ic + id + ie
Step 2: – ia – 2 – (– 3) + 5 – 12
= 0
= 0
Solving for ia gives: ia = – 6 A.
Part 3: Ohms Law, etc
87
Still Another Formulation of KCL
The sum of the currents entering any node equals
the sum of the currents leaving the node.
Same Example
Find ia.
i = 2A
b
ia = ??
i =–3A
c
node
i = –12 A
e
Solution:
i = 5A
d
Regard the current in branches d and e as entering, and the
current in branches a, b, and c as leaving (This happens to
be consistent with the assigned reference directions for the
currents, making it less confusing than otherwise.):
Step 1: id + ie
= ia + i b + ic
Step 2: 5 + (– 12) = ia + 2 + (– 3)
Solving for ia gives: ia = – 6 A
Part 3: Ohms Law, etc
88
One Final Formulation of KCL
Designate the unknown current as leaving , and
equate it to the sum of the other (in this case,
known) currents regarded as entering the node.
i = 2A
b
Same Example
i a= ??
Find ia.
i =–3A
c
node
i = –12 A
e
i = 5A
Solution:
d
We regard the current in branch a as leaving and equate that to
the sum of the currents entering, to give:
i a = – i b – i c + id + ie
= – 2 – (– 3) + 5 + (– 12)
= –6A
Part 3: Ohms Law, etc
89
Generalization of KCL
The algebraic sum of the currents entering (or
leaving) any closed surface is zero.
closed surface
Example 1
i = –10 A
1
i5
...
i = ?
2
i4 = 13 A
i=9A
3
Currents entering sum to 0:
i1 – i2 + i3 – i4 = 0
–10 – i2 + 9
– 13 = 0
so that i2 = – 14 A
Part 3: Ohms Law, etc
90
Example 2
Find i1 and i2 for the circuit below.
Solution:
First, we solve the problem by using two closed
surfaces, first finding i1 and then i2. Then we will
solve it by using only one closed surface, without
first finding i1. i
1
1
7A
7A
2
2A
i2
Solution:
@ node 1, i1 = 7 – 2 = 5 A
@ node 2, i2 = i1 – 7 = 5 – 7 = – 2 A
Part 3: Ohms Law, etc
91
Example 2(Cont.)
Since we only need to find i2 (and not i1 ), we can
draw only one closed curve and solve the
problem in one step, without even using i1, as
follows:
i1
7A
7A
2A
i2
The current flowing out through the i2 branch is
equal to the sum of the currents flowing in
through the other branches:
i2 = – 7 + 7 – 2
= –2A
Part 3: Ohms Law, etc
92
3.5 Kirchhoff’s Voltage Law (First, w/o
Ohm’s Law)
The algebraic sum of all the voltage drops* (or voltage rises)
around any closed path is zero.
direction
starting (and ending) point
This must be a closed path (only part of it is being shown here) in
order for the algebraic sum to be zero.
* If the movement around the loop proceeds through an element in
the direction toward the terminal of assumed lower voltage, we say
we have traversed in the direction of a voltage drop of value v.
Example 1
direction
+
12 V
–
starting (and ending) point
+ 8V –
R1
R2
+
v
–
direction
direction
Part 3: Ohms Law, etc
93
Example 2
Find v by applying KVL.
+ 8V –
R1
+
12 V
–
R2
+
v
–
Solution:
Summing voltage rises:
– 12 + v + 8 = 0

v = 4V

v = 4V
Summing voltage drops:
12 – v – 8 = 0
Same answers, of course!
Part 3: Ohms Law, etc
94
Example 2 (Again)
Find v, but use a different starting point
and direction.
+ 8V –
+
12 V
–
+
v
–
Solution:
Summing rises:
12 – 8 – v = 0

v = 4V

v = 4V
Summing drops:
– 12 + 8 + v = 0
Same answers, of course!
Part 3: Ohms Law, etc
95
Alternate Formulation of KVL
The voltage drop (voltage rise) from one node to
another is the same regardless of the path.
Example 2 (Once More)
starting point
+ 8V –
+
12 V
–
+
v
–
ending point
Solution:
Using drops:
v
= – 8 + 12
 v = 4V
Using rises:
– v =
8 – 12
 v = 4V
Same answers, of course!
Part 3: Ohms Law, etc
96
Example 3
Find v for the circuit below.
+
– 35 V
–
b
c
+ v –
a
–
+
25 V
+
50 V
–
d
Solution:
Path abcda, drops: – (v) – (–35) + (50) – (25) = 0
Path adcba, drops: (25) – (50) + (–35) + (v) = 0
 v = 60 V
OR: + (v) + (-35) – (50) + (25) = 0
+ (v) + [
-60
]=0
 v = 60 V
Part 3: Ohms Law, etc
97
3.6 KVL (Now, with Ohm’s Law)
Example. Verify Kirchoff’s Voltage Law.
8W
d
+
24 V
–
a
2A
4W
c
b
Solution:
Note that no voltages are “set up” for the resistors. But
recall the advantage of the passive convention:
d
i
+
v
–
R
v =
Ri
No negative sign needed when the
passive convention is used. So use it
in this example.
a
Part 3: Ohms Law, etc
98
Example 1 (cont.):
+ v8 –
d
+
24 V
–
a
8W
2A
2A
+
4 W v4
2A
–
2A
c
b
Choose convenient set ups for the voltages, and then
sum drops:
v8
+ v4
–
24 = 0
8(2) + 4(2) – 24 =
16
+ 8
0
– 24 = 0 checks!
Part 3: Ohms Law, etc
99
Example 2 (Alexander, p 39)
Find v1 and v2.
4W
+ v1
10 V
–
–
+
+
–
+ v2
Solution:
For the current, choose
the reference direction and
the symbol as indicated.
Choose the closed path,
starting point, and
direction as indicated.
8V
–
2W
4W
10 V
+ v1 –
I
I
I –+ 8 V
I
+
–
2W
+ v2
–
KVL:
v1 – 8 – v2 – 10 = 0
or 4I – 8 – (– 2I) – 10 = 0
or 6I = 18
or I = 3 A
Then
v1 = 4I = 12 V
v2 = – 2I = – 6 V
Part 3: Ohms Law, etc
100
Solution (cont. ):
(Another way of carrying out the solution.)
10 V
+
–
4W
4W
–
I
+ v1
I
+ v2
–
–
+
8V
10 V +
–
2W
–
+
8V
2W
In solving for I, we can also write KVL without reference
to the voltages v1 and v2, as follows:
4I – 8 + 2I – 10 = 0
or
6I – 18 = 0
or
I=3A
At this point v1 and v2 can now be calculated as
previously:
v1 = 4I = 12 V
v2 = – 2I = – 6 V
Part 3: Ohms Law, etc
101
Example 3
Find i and vae.
a
20 V
+ v5 –
b
+ v3 –
5W
i
3W
+
–
10 V
10 W
g – v10 + f
+
–
4W
4V
– +
c
e – v4 +
d
Solution:
To find i, use KVL. Summing voltage drops around the
closed path:
5i + 3i + 10 + 4i + 4 + 10i – 20 = 0
22 i –
6= 0
 i = 3/11 A
Part 3: Ohms Law, etc
102
Solution (cont.)
+ v5 –
a
b + v3 –
5W
20 V
3W
i
+
–
10 V
10 W
g – v10 + f
+
–
4W
4V
– +
c
e – v4 +
d
To find vae, add drops along the path agfe:
vae = 20 – v10 – 4
= 20 – 10 x 3/11 – 4
= 146 / 11 V
Check by finding vae by adding drops along abcde:
vae = vab + vbc + vcd + vde
= 5 x 3/11 + 3 x 3/11 + 10 + 4 x 3/11
= 146 / 11 V Checks!
Part 3: Ohms Law, etc
103
Example 4
Find vag, the voltage drop from a to g.
–6V
a
...
– +
–3A
b
6W
1A
...
c
+ 3V
–
2A
...
d
7A
–5A
10 W
e
...
Solution Strategy:
3A
g
f 5W
–6 A
... h
Step 1. Compute all
currents needed.
Step 2. Apply KVL
along path abcdefg.
Part 3: Ohms Law, etc
104
Solution:
–6V
a
...
– +
–3A
b
6W
1A
...
c
+ 3V
–
2A
...
d
7A
–5A
10 W
ide
e
...
ief
3A
g
f 5W
–6A
... h
Step 1:
@ node d:
ide = – 2 + 5 = 3 A
@ node e:
ief = 3 + 7
Part 3: Ohms Law, etc
= 10 A
105
Solution (cont.):
–6V
a
...
– +
–3A
b
6W
1A
...
c
+ 3V
–
2A
...
d
7A
–5A
10 W
ide= 3A
e
...
Step 2:
ief = 10A
3A
f 5W
g
–6 A
... h
vag = vab
+ vbc
+ vcd + vde
+ vef
+ vfg
= –(– 6) + (6)(–3) + 3 + 10(3) + 5(10) + 0
=
6 –
= 71 V
18
+ 3 + 30
Part 3: Ohms Law, etc
+ 50
+ 0
106
Example 5
Find i0.
10 W
a
i0
120 V
+
–
d
i
b
50 W
1
6A
c
Solution:
b
Step 1. Write KCL @ node b:
– i0 + i1 – 6 = 0 (1 eqns, 2 unknowns)
Step 2. Write KVL around abcda:
– 120 + 10 i0 + 50 i1 = 0
Step 3. Solve the two simultaneous eqns:
… i0 = – 3 A, i1 = 3 A
Part 3: Ohms Law, etc
107
Example 6
Find v, the voltage across the resistor R. Note that
the numerical value of R is not given!
4A
Solution:
1A
6W
3W
18 V +
–
3A
4W
2A
+ v –
R
Strategy:
Step 1. Use KCL to find all the branch
currents.
Step 2. Use KVL to find v.
Part 3: Ohms Law, etc
108
Solution (cont):
4A
1A
6W
3W
18 V +
–
3A
ix
iy
2A
4W
+ v –
R
Step 1. Use KCL to find all the branch currents.
ix = 1 – 3 + 4
= 2A
iy = – 2 + 4 + 2
= 4A
Part 3: Ohms Law, etc
109
Solution (cont):
4A
1A
6W
3W
18 V +
–
2A
3A
4A
4W
2A
+ v –
R
Step 2. Use KVL to find v:
v = – 18 – 6 – 6 + 16
= – 14 V
Notice that we did not need to know the value of R.
Part 3: Ohms Law, etc
110
Example 7 (Controlled source)
Find vab.
Solution:
2 i1
30 V
– 20 V
+
50 V –
+
a
+
–+
–+
i
1
5A
v1
–
+
5v 1
vab
+ 40 V
–
–
–
b
v1
= 30 – 50
= – 20 V
vab
= – 20 – 5 v1 + 40
= – 20 – 5 (– 20) + 40
= 120 V
Part 3: Ohms Law, etc
111
Example 8 (Controlled source) Source unknown.
Find i.
Solution:
3 i1
i
8A
3W
+
i1
v
2W
4A
–
KVL: 3 i – 2 i1 =
inbound
outbound
inbound KCL:
0
i – 2i1 = – 4
Solution to these two
simultaneous linear algebraic
equations:
i = 2A
Part 3: Ohms Law, etc
112
Example 9 (Controlled source)
Find Req , which is the equivalent resistance looking
to the left of x-y. (It will be a function of R0).
b
x
a
4W
i
6i
+
–
Req = ?
R0
y
d
Solution:
c
According to Ohm’s Law, R = V/I. So, insert a current
source to provide a current I, compute the corresponding
voltage V, and then take the ratio V / I to find Req!
a
b
4W
6i
+
–
i
+
R0
V
I
Req =
V/I
–
d
c
Part 3: Ohms Law, etc
113
Solution (cont.):
4W
a
6i
+
–
b
i
+
R0
V
I
–
d
c
KVL:
4(i – I) + R0 i = 6 i
 i = 4 I/ (R0 – 2)
Then V = R0 i
= R0 x 4 I /(R0 – 2)
and Req = V / I
= [ 4 R0 I /(R0 – 2)] / I
= 4 R0 /(R0 – 2)
Note that Req can be negative (if Ro is less than 2)!
Part 3: Ohms Law, etc
114
Comment on Equivalent Resistance:
Note that in the previous circuit we were not able to simply combine
controlled sources using the series-parallel reduction rules developed
for resistors. Instead, we applied a current source and calculated the
corresponding voltage and computed the ratio of the voltage to the
current to find the equivalent resistance. We could also have applied a
voltage source, calculated the current and again computed the ratio to
find the equivalent resistance. Observe that the numerical values of
the voltage and the current are unimportant: only their ratio matters.
It is important to understand that the passive sign convention applies
to the terminals of the equivalent resistance. In terms of the circuit
representation below: The current enters the box at the terminal
associated with the positive terminal of the voltage. (A frequent and
serious mistake is to define the voltage and associated current
contrary to the passive sign convention.)
a
6i
+
–
4W
b
i
+
R0
V
I
–
d
c
Part 3: Ohms Law, etc
115