e e 1205 circuit analysis
DESCRIPTION
E E 1205 Circuit Analysis. Lecture 2 - Circuit Elements and Essential Laws. Five Fundamental Elements. Ideal Voltage Sources Independent Dependent Ideal Current Sources Independent Dependent Resistors Inductors (to be introduced later) Capacitors (to be introduced later). - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/1.jpg)
E E 1205 Circuit Analysis
Lecture 2 - Circuit Elements and Essential Laws
![Page 2: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/2.jpg)
Five Fundamental Elements• Ideal Voltage Sources
– Independent– Dependent
• Ideal Current Sources– Independent– Dependent
• Resistors• Inductors (to be introduced later)• Capacitors (to be introduced later)
![Page 3: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/3.jpg)
Independent Voltage Source
• Voltage may be constant or time-dependent
• Delivers nominal terminal voltage under all conditions
Vg
Positive Terminal
Negative Terminal
![Page 4: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/4.jpg)
Independent Current Source
• Current may be constant or time-dependent
• Delivers nominal terminal current under all conditions
Ig
Negative Node
Positive Node
![Page 5: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/5.jpg)
Voltage-Controlled Dependent Voltage Source
• Terminal voltage is a function of the voltage drop of a different branch
• Delivers nominal terminal voltage under all conditions
v
Positive Terminal
Negative Terminal
+v-
![Page 6: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/6.jpg)
Current-Controlled Dependent Voltage Source
• Terminal voltage is a function of the current flow in a different branch
• Delivers nominal terminal voltage under all conditions
i
Positive Terminal
Negative Terminal
i
![Page 7: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/7.jpg)
Voltage-Controlled Dependent Current Source
• Current is a function of the voltage drop of a different branch
• Delivers nominal terminal current under all conditions
v
Positive Node
Negative Node
+v-
![Page 8: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/8.jpg)
Current-Controlled Dependent Current Source
• Source current is a function of the current flow in a different branch
• Delivers nominal terminal current under all conditions
i
Positive Node
Negative Node
i
![Page 9: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/9.jpg)
Electrical Resistance (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.
R
+ v -
i
vRi
![Page 10: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/10.jpg)
Power Consumed by Resistors
• Resistors consume power.
• v and i are both positive or both negative.
R
+ v -
i
p v i
v R i 2p i R
viR
2vp
R
![Page 11: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/11.jpg)
Conductance Defined
• Conductance is the reciprocal of resistance.
• The units of conductance are called siemens (S)
• The circuit symbol is G
1GR
i v G ivG
2p v G 2ip
G
![Page 12: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/12.jpg)
Creating a Circuit Model• A circuit model is usually two or more
circuit elements that are connected.• A circuit model may have active elements
(sources) as well as passive elements (such as resistors).
• By the assumption that electric signal propagation is instantaneous in a circuit, our circuit model has lumped parameters.
![Page 13: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/13.jpg)
Example of a Circuit Model1000 ft AWG 14 Copper Wire
100 WLamp
120 V Battery
120 V
0.25 2.57
2.57
144
![Page 14: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/14.jpg)
Kirchhoff’s Voltage Law• The sum of the voltage drops around a
closed path is zero.• Example: -120 + V1 + V2 + V3 + V4 = 0
120 V
0.25 2.57
2.57
144
+ V1 - + V2 -
- V4 +
+V3-
![Page 15: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/15.jpg)
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
I2I3
I4
1 2 3 4 0I I I I
![Page 16: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/16.jpg)
Apply KCL to Example
120 V
0.25 2.57
2.57
144
+ V1 - + V2 -
- V4 +
+V3-
I s
I1 I1 I2 I2
I3
I3
I4I4
I s
1 2 3 4sI I I I I
![Page 17: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/17.jpg)
Combine KVL, KCL & Ohm’s Law
120 V
0.25 2.57
2.57
144
+ V1 - + V2 -
- V4 +
+V3-
I s
I1 I1 I2 I2
I3
I3
I4I4
I s
120 0.25 2.57 144 2.57s s s sI I I I
120 0.803149s
VI A
![Page 18: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/18.jpg)
Lamp Voltage & Battery Voltage
3 144 115.67sV I V
(2.57 2 144) 0.803 119.8bV V
120 V
0.25 2.57
2.57
144
+ V1 - + V2 -
- V4 +
+V3-
I s
I1 I1 I2 I2
I3
I3
I4I4
I s
+
Vb
-
![Page 19: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/19.jpg)
Battery Power and Lamp Power
Loss:
Efficiency:
119.8 0.8033 96.23bP V A W
115.67 0.8033 92.91lP V A W
3.32loss b lP P P W
92.91 96.55%96.23
l
b
PP
1000 ft AWG 14 Copper Wire
100 WLamp
120 V Battery
![Page 20: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/20.jpg)
Using Loops to Write Equations
KVL @Loop a:KVL @ Loop b:KVL @ Loop c:Loop c equation same as a & b combined.
va
R2vb
R1 R3
+ v2 -+v1-
+v3-a b
c
2 1 0av v v
3 1 0bv v v
2 3 0a bv v v v
![Page 21: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/21.jpg)
Using Nodes to Write Equations
KCL @ Node x:
KCL @ Node y:
KCL @ Node z:KCL @ Node w: <== Redundant
va
R2vb
R1 R3
+ v2 -+v1-
+v3-
xy z
w
ia
i2 i2 ib ib
iai3
i1
i1
i3
2 1 0bi i i
2 0ai i
3 0bi i 1 3 0ai i i
![Page 22: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/22.jpg)
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.
![Page 23: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/23.jpg)
Working with Dependent Sources
KVL @ left loop:
KCL @ top right node:
Substitute and solve:
48 V
4
3 i
i+vo-
i
48 4 3 oV i i
4oi i
3i A 36ov V
![Page 24: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/24.jpg)
Example 1 (1/3)
By KCL:
By Ohm’s Law:
50 V
20 A
25
30 A
50 V10
+ Va - + Vb -+ Vc-
+Vd-
Ie
If
Ic Id
20 , 30 , 30 , 10e d f ci A i A i A i A
25 250 , 10 300c c d dV I V V I V
![Page 25: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/25.jpg)
Example 1 (2/3)
By KVL:
Power:
50 V
20 A
25
30 A
50 V10
+ Va - + Vb -+ Vc-
+Vd-
Ie
If
Ic Id
300 , 600a bV V V V
300 20 6.0aP V A kW
600 30 18.0bP V A kW
![Page 26: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/26.jpg)
Example 1 (3/3)
50 V
20 A
25
30 A
50 V10
+ Va - + Vb -+ Vc-
+Vd-
Ie
If
Ic Id
250 10 2.5cP V A kW
300 30 9.0dP V A kW
50 20 1.0eP V A kW 50 30 1.5fP V A kW
![Page 27: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/27.jpg)
Example 2 (1/4)
Find Source Current, I, and Resistance, R.
1
84 V4
12 8
12 R
8
3 A
I
![Page 28: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/28.jpg)
Example 2 (2/4)
Ohm’s Law: 36 V KVL: 48 V Ohm’s Law: 6 A
1
84 V4
12 8
12 R
8
3 A
I+
36 V-
+48 V
-
6 A
![Page 29: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/29.jpg)
Example 2 (3/4)
KCL: 3 A Ohm’s Law: 12 V KVL: 60 V
1
84 V4
12 8
12 R
8
3 A
I+
36 V-
+48 V
-
6 A3 A -12 V+
+ 60 V-
![Page 30: E E 1205 Circuit Analysis](https://reader035.vdocuments.us/reader035/viewer/2022081515/56816940550346895de0bf95/html5/thumbnails/30.jpg)
Example 2 (4/4)
Ohm’s Law: 3 A KCL: 6 AOhm’s Law: R=3 KCL: I=9 A
KVL: 24 V
1
84 V4
12 8
12 R
8
3 A
I+
36 V-
+48 V
-
6 A3 A -12 V+
+ 60 V-
+ 24 V -
3 A
6 A