Transcript Behavior-Based Robotics

EE 1105: Introduction to EE
Freshman Seminar
Lecture 3: Circuit Analysis
Ohm’s Law, Kirkhoff’s Laws
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Voltage and Current
• Voltage is the energy per unit of charge.
dw
v
dq
• Current is the rate of flow of charge.
i
dq
dt
Derivative: slope, Integral: area under curve
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Power and Energy
dw  dw   dq 
p

   vi

dt  dq   dt 
• Power associated with a circuit element
is consumed by that circuit element
when the value of power is positive.
• Conversely, power is generated, or
produced by the element if the value
consumed is negative.
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Independent Voltage Source
• Voltage may be
constant or timedependent
• Delivers nominal
terminal voltage
under all conditions
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Positive Terminal
Vg
Negative Terminal
Independent Current Source
• Current may be
constant or timedependent
• Delivers nominal
terminal current
under all
conditions
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Negative Node
Ig
Positive Node
Passive Sign Convention
i
+
v
-
1
2
• Assign current flow direction according to the
voltage polarities
• Calculate p=vi, p<0 is power supplied by
device, p>0 is power delivered to device
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Ohm’s Law
• Electrical resistance
is the ratio of voltage
drop across a resistor
to current flow
through the resistor.
• Polarities are
governed by the
passive sign
convention.
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
R
i
+
v
v
R
i
-
Power and Energy
dw  dw   dq 
p

   vi

dt  dq   dt 
• Power associated with a circuit element
is consumed by that circuit element
when the value of power is positive.
• Conversely, power is generated, or
produced by the element if the value
consumed is negative.
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Power Consumed by Resistors
• Resistors
consume power.
• v and i are both
positive or both
negative.
+
v
-
p  v i
v  R i
v
i
R
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
R
i
p i R
2
2
v
p
R
Conductance Defined
• Conductance is
the reciprocal of
resistance.
• The units of
conductance are
called siemens (S)
• The circuit symbol
is G
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
1
G
R
i  v G
i
v
G
p  v2  G
2
i
p
G
Calculating Resistance
R
 l
When conductor has uniform
cross-section
A
cu  1.67 10   cm
6
 al  2.70 10   cm
6
Area,
A
l
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Temperature Coefficient of Resistance
Metallic conductors have a linear increase of resistance with
increased temperature.
R(T )  Ro 1  a T  To 
To is the reference temperature (usually 20oC) and Ro is the
resistance at the reference temperature. a is the temperature
coefficient of resistance for the material. At 20oC, some
values for a are:
Material
Aluminum
Copper
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Alpha @ 20oC
0.004308
0.004041
Power and Energy
dw  dw   dq 
p

   vi

dt  dq   dt 
• Power associated with a circuit element
is consumed by that circuit element
when the value of power is positive.
• Conversely, power is generated, or
produced by the element if the value
consumed is negative.
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Power Consumed by Resistors
• Resistors
consume power.
• v and i are both
positive or both
negative.
+
v
-
p  v i
v  R i
v
i
R
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
R
i
p i R
2
2
v
p
R
Conductance Defined
• Conductance is
the reciprocal of
resistance.
• The units of
conductance are
called siemens (S)
• The circuit symbol
is G
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
1
G
R
i  v G
i
v
G
p  v2  G
2
i
p
G
Circuits
• Abstraction describing how (the topology) electrical or
electronic modules are interconnected.
• Closely related to a GRAPH.
• Nomenclature:
–
–
–
–
–
Nodes, Edges(Branches)
Voltage drop computed between nodes
Currents flowing along edges
Paths (collection of edges with no node appearing twice),
Loops (closed paths), meshes (loop containing no other
loop).
– Series connection (elements sharing the same current)
– Parallel connection (elements sharing the same voltage)
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Example of a Circuit Model w/
Series Connection
1000 ft AWG 14
Copper Wire
100 W
Lamp
120 V Battery
0.25 
2.57 
144 
120 V
2.57 
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Kirchhoff’s Voltage Law
• The sum of the voltage drops around a
closed path is zero.
• Example: -120 + V1 + V2 + V3 + V4 = 0
0.25 
+ V1 -
2.57 
+ V2 -
120 V
2.57 
- V4 +
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
+
V3
-
144 
Kirchhoff’s Current Law
• A node is a point where two or more
circuit elements are connected together.
• The sum of the currents leaving a node
is zero.
I1
I4
I1  I 2  I3  I 4  0
I3
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
I2
Apply KCL to Example
I1
Is
0.25  I1
+ V1 -
I2
2.57  I2
120 V
2.57 
Is
I4
+
V3
-
- V4 + I4
I s  I1  I 2  I 3  I 4
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
I3
+ V2 -
144 
I3
Combine KVL & Ohm’s Law
I1
Is
0.25  I1
+ V1 -
I2
2.57  I2
120 V
2.57 
Is
I4
I3
+ V2 +
V3
-
144 
I3
- V4 + I4
120  0.25  I s  2.57  I s  144  I s  2.57  I s
120V
Is 
 0.803 A
149
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Lamp Voltage & Battery Voltage
I1
Is
0.25  I1
+ V1 -
I2
+
2.57  I2
Vb
120 V
-
Is
I4
I3
+ V2 -
2.57 
+
V3
-
144 
I3
- V4 + I4
V3  144  I s  115.67V
Vb  (2.57  2  144)  0.803  119.8V
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Battery Power and Lamp Power
1000 ft AWG 14
Copper Wire
100 W
Lamp
120 V Battery
Pb  119.8V  0.8033 A  96.23W
Pl  115.67V  0.8033 A  92.91W
Loss: Ploss  Pb  Pl  3.32W
Pl 92.91

 96.55%
Efficiency:  
Pb 96.23
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Resistors in Series
I1
Is
R1
+ V1 -
I2
+
R2
Vs
V2
By KCL: Is = I1= I2
Is
Vs
+
Vs
-
Req
By Ohm’s Law: V1 = R1·I1 and V2 = R2·I2
Combine: Vs = R1I1 + R2I2 = (R1 + R2) Is = ReqIs
In General: Req = R1 + R2 +···+ Rn
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Resistors in Parallel (1/2)
Is
+ +
Vs V1
- -
I2
I1
R1
+
V2 R2
-
By KVL: Vs = V1 = V2
By Ohm’s Law:
Combine:
V1
I1 
R1
Is
+
Vs
-
Req
By KCL: Is = I1 + I2
and
V2
I2 
R2
 1 1  Vs
V1 V2
Is  
 Vs    
R1 R2
 R1 R2  Req
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Resistors in Parallel (2/2)
For two resistors:
R1 R2
Req 
R1  R2
For many resistors:
1
1
1
 

Req R1 R2
In terms of conductance:
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
1

Rn
Geq  G1  G2 
 Gn
Voltage Divider Circuit
+ V1 I
Vs
R1
R2
+
V2
-
Measure
V2
Vs
I
Vs  I  R1  R2 
R1  R2
Vs
R2
V2  I  R2 
R2 
Vs
R1  R2
R1  R2
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Loaded Voltage Divider
R1
Vs
+
R2 Vo
-
Vo  Vs
RL
Req
Req  R1
R2 RL
Vo  Vs
R1  R2  RL   R2 RL
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Voltage Divider Equations
R2
Vo  Vs
R1  R2
Unloaded:
Loaded:
Vo  Vs
If RL >> R2:
R2
 R2

R1 
 1  R2
 RL

R2
Vo  Vs
R1  R2
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Using Loops to Write Equations
vb
R2
+ v2 va
a
+
v1
-
R1
b
+
v3
-
R3
c
va  v2  v1  0
KVL @ Loop b: vb  v3  v1  0
KVL @ Loop c: va  v2  vb  v3  0
KVL @Loop a:
Loop c equation same as a & b combined.
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Using Nodes to Write Equations
i2
y
ia
R2
i2
x
+ v2 -
va
ib
vb
ib
z
i3
i1
+
v1
-
+
v3
-
R1
i1
ia
R3
i3
w
i2  i1  ib  0
KCL @ Node y: ia  i2  0
KCL @ Node z: ib  i3  0
KCL @ Node w: ia  i1  i3  0
KCL @ Node x:
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
<== Redundant
Combining the Equations
•
•
•
•
•
•
•
•
•
There are 5 circuit elements in the problem.
va and vb are known.
R1, R2 and R3 are known.
v1, v2 and v3 are unknowns.
ia, ib, i1, i2 and i3 are unknowns.
There are 2 loop (KVL) equations.
There are 3 node (KCL) equations.
There are 3 Ohm’s Law equations.
There are 8 unknowns and 8 equations.
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Example 1 (1/3)
20 A
Ie
50 V
30 A
Ic
Id
+ Va + + Vb - +
25  Vc
Vd 10 
- 50 V -
By KCL:
If
ie  20 A, id  30 A, i f  30 A, ic  10 A
By Ohm’s Law:
Vc  25I c  250V , Vd  10 I d  300V
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Example 1
20 A
Ie
50 V
(2/3)
30 A
Ic
Id
+ Va + + Vb - +
25  Vc
Vd 10 
- 50 V If
By KVL: Va  300V , Vb  600V
Power:
Pa  300V  20 A  6.0 kW
Pb  600V  30 A  18.0 kW
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Example 1
20 A
Ie
50 V
(3/3)
30 A
Ic
Id
+ Va + + Vb - +
25  Vc
Vd 10 
- 50 V If
Pc  250V   10 A  2.5 kW
Pd  300V  30 A  9.0 kW
Pe    50V    20 A  1.0 kW
Pf  50V  30 A  1.5 kW
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Example 2 (1/4)
1
3A
I
12 
84 V
R
4
8
8
12 
Find Source Current, I, and Resistance, R.
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Example 2 (2/4)
1
3A
I
84 V
12 
6A
8
Ohm’s Law: 36 V
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
+
36 V
+4
48 V
12 
KVL: 48 V
R
8
Ohm’s Law: 6 A
Example 2
(3/4)
1
3A
12 
I
84 V
6A
8
KCL: 3 A
+
36 V
3 A -12 V+
+4 +
48 V 60 V
- 12 
Ohm’s Law: 12 V
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
R
8
KVL: 60 V
Example 2 (4/4)
1
3A
I
84 V
12 
6A
8
++
36 V 24 V
3 A -12 V+
+4 +
48 V 60 V
- 12 
6A
R
3A
8
Ohm’s Law: 3 A KCL: 6 A
KVL: 24 V
KCL: I=9 A
Ohm’s Law: R=3 
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
Homework 3 due next class!!
Available online at course website
Acknowledgements: Dr. Bill Dillon
Questions?
Dan O. Popa, Intro to EE, Freshman Seminar Spring 2015
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