Download Sources and Resistors

Topic 3
Engineering 1333: Electrical
Circuits
Sources and Resistance
(Sections 2.1 & 2.2)
12-Sep-07
Sources and Resistance
1
Independent Voltage Sources
i
+
vs
+
v
-
The ideal voltage source produces a fixed voltage
across itself no matter what current passes through it
v
vs
i
An ideal voltage source is a model for a
battery…
…albeit a very simplified one
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Sources and Resistance
2
Independent Current Source
The ideal current source produces a fixed current
through itself no matter what voltage appears across it
i
+
is
v
v
-
is
i
There is no familiar, practical equivalent to a current
source although they do show up in electronics in
transistors
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Sources and Resistance
3
Resistance
A Resistor is a physical element
which impedes the flow of current
It is represented by an idealized
property called Resistance
R
R = Resistance in Ohms
Ohm’s Law
Current defined to be going downhill
+
v
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+
i
R
-
Current defined to be going uphill
v  Ri
The symbol for
resistance is Ω
Sources and Resistance
R
v
-
v  R  i
i
4
Power in Resistors
Since
p  vi
+
For a resistor, by Ohm’s Law
-
Which means that the power is
p  v  i  R  i  i  i2R
By inverting Ohm’s Law to find i in terms of v
…we may also write the power in terms
of the voltage and resistance
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R
v
v  Ri
i
Sources and Resistance
v
i
R
v2
p
R
5
Conductance
1
G
R
Occasionally we will use the inverse of
Resistance which is Conductance
The unit of Conductance is a Sieman and is represented by an S
It is mostly used to allow us to write the
current directly in terms of the voltage
i
v
 vG
R
It is straightforward to compute power in terms of
conductance instead of resistance
v2
p   v 2G
R
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i2
pi R
G
2
Sources and Resistance
6
i
Problem 1.20
v  100e500t V
a)
i  20  20e500t mA
1.
2.
3.

1
-
2
v
The voltage and current at the
terminals of the circuit are 0 for t < 0.
For t ≥ 0
+
Find the maximum value
of the power delivered to
the circuit

p  100e500t 20  20e500t mW  2000e500t  2000e1000t mW
dp
 1000e500t  2000e1000t
dt
2000e
1000t
 1000e
5.
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t
500t
0
4.
2  e500t
1
ln  2   .001386 S  1.386 mS
500
Sources and Resistance
7
Plot to check for maxima
Power
1.38 mS represents a
maxima not a minima
0.6
Power in watts
0.5
0.4
0.3
Power
0.2
0.1
0
0
1
2
3
4
5
Time in mS
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Sources and Resistance
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i
Problem 1.20
v  100e500t V
b)
i  20  20e500t mA


1.

w   pdt  2  e
0

 2 e
500t
0
500t

dt  2  e
0
2

e
500
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e
1000t
 dt
dt
0

2 1
1000t 
1000
0
e
2
Since
e
 xt 
0
 0  1  1
We get
0
500t 
-
Find the total energy
delivered to the element
2.
1000t
1
v
The voltage and current at the
terminals of the circuit are 0 for t < 0.
For t ≥ 0
+
500 Sources and Resistance
2
1
w

J  2 mJ
500 500
9
Assesment 2.3
(a)
ig
If vg=1 kV and ig=5 mA, find R and
the power absorbed by the resistor
+
vg
By Ohm’s Law
So
For power
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R
vr
-
Clearly the two
voltages are the same
+
-
vr  v g
vr  R  ig
vr vg
1000
3
R 


200

10
 200 K 
3
ig ig 5 10
p  vr  ig  vg  ig  1000  5 103  5W
Sources and Resistance
10
Assesment 2.3
(b)
ig
If ig=75 mA and the power delivered
by the voltage source is 3 watts, find
vg, R and the power absorbed by the
resistor.
Power delivered by
the source
Inverting
As before
+
vg
+
R
vr
-
-
p  v g  ig
p
3
vg  
 40V
3
ig 75 10
vr vg
40
R 

 533.3 
3
ig ig 75 10
power absorbed = power delivered = 3 Watts
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Sources and Resistance
11
Assesment 2.3
(c)
ig
If R=300 Ω and the power absorbed
by R is 480 mW, find ig and vg.
+
vg
R
vr
-
In general the power
absorbed is given by
+
-
pi R
2
Solving for ig
ig 
So
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p
480 103

 16 104  40 mA
R
300
vg  R  ig  300  40 103  12 v
Sources and Resistance
12
Dependent Sources
Dependent sources have a voltage or current that
depends on another variable
Vs=µVx
+
-
The s subscript denotes the
dependent source variable
Vs=ρix
+
-
The x subscript denotes an
independent variable
is=αvx
is=βix
The independent x variable is usually
somewhere else in the same circuit
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Sources and Resistance
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For example
Here is a simple model for an amplifier
+
+
vi
+
+
-
vx
-
Vs=µVx
-
vo
-
vx  vi
vo   vx   vi
So long as µ > 1 we have amplification
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Sources and Resistance
14
ib
Assesment Problem 2.1
a) What value of vg is
required in order for
the interconnections
to be valid?
+
+
-
ib/4
8A
vg
-
b) For this value of vg find the power associated with the 8A source.
a)
ib/4 is actually a voltage
Connecting it across the vg source is like
connecting two batteries together
So
b)
ib
vg 
4
ib  8 A
But
Therefore, vg should be -2V
p  vg  8  16 w
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They better be the
same voltage!
delivered
Sources and Resistance
15