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![Page 1: ECE 1100: Introduction to Electrical and Computer Engineering Notes 16 Resistors Spring 2011 Wanda Wosik Associate Professor, ECE Dept. Notes prepared.](https://reader036.fdocuments.us/reader036/viewer/2022062421/56649f505503460f94c72985/html5/thumbnails/1.jpg)
ECE 1100: Introduction toECE 1100: Introduction toElectrical and Computer EngineeringElectrical and Computer Engineering
Notes 16
Resistors
Spring 2011
Wanda WosikAssociate Professor, ECE Dept.
Notes prepared by Dr. Jackson
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ResistorsResistors
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Resistors (cont.)Resistors (cont.)
Ri
+ -v
Note: passive sign convention is assumed here.
vR
i
v Ri
units of R are Ohms []
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Ohm’s LawOhm’s Law
Ri
+ -v
v Ri i
v
There is a linear relationship between voltage and current. (That is, the value of R does not depend on the current going through it.)
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ExampleExample
10 []2 [A]
+ -V
Find V in each case
10 2 20 [V]V RI
10 []2 [A]
+ -V
10 2 20 [V]V RI
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Example (cont.)Example (cont.)
10 []-2 [A]
+- V
10 2 20 [V]V RI
10 []-2 [A]
+- V
10 2 20 [V]V RI
10 []2 [A]
+- V
10 2 20 [V]V RI
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Power DissipationPower Dissipation
R []i
+ -v
2
absP vi
Ri i
Ri
Note that passive sign convention is used here.
power formula Ohm’s law
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Power Dissipation (cont.)Power Dissipation (cont.)i R []
+ -V
2absP Ri
Also,
2
abs
vP R
R
so2
abs
vP
R
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ExampleExample
2 [k]
-+
12 [V]
Find Pabs by resistor
22
3
120 072
2 10abs
VP .
R
0 072 [W]
(72 [mW])absP .
Note: in the MKS system, we must use [V], [A], [], [W], [J].
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ExampleExample
Find power Pabs (t) absorbed by resistor
22
2120 2cos 2 60
200cos 2 60144abs
tv tP t t
R
R = 144 []+-v (t)
Find average power PabsAVE
absorbed by resistor
120 2cos 2 60 [V]v t t
(60 [Hz] AC line voltage)
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Example (cont.) Example (cont.)
2200cos 2 60 [W]absP t t
Pabs (t)
Tp = T/2 = 0.5 (1/60) [s]t
R = 144 []+-v (t)
T = 1 / f = 1/60 [s]co
s (
t)
t
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Example (cont.) Example (cont.)
0
1 pT
AVEabs abs
p
P P t dtT
+-v (t)
2200cos 2 60 [W]absP t t
2
0
1200cos 2 60
pT
AVEabs
p
P t dtT
Pabs (t)
Tp= 0.5 (1/60) = 1/120 [s] = 0.00833 [s]t
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Example (cont.)Example (cont.)
1 1202
0
1 1202
0
1200cos 2 60
1 120
200 120 cos 2 60
/AVE
abs
/
P t dt/
t dt
Note: cos2 (t) = ½ + ½ cos(2t)
1 120
0
sin 2 2 601200 120
2 2 2 2 60
1 120200 120 100
2
/
AVEabs
* ttP
*
/
Note: the average value of cos2 (t) is ½.
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Example (cont.)Example (cont.)
100 [W]AVEabsP
120 2cos 2 60 [V]v t t
R = 144 []
What is this? A 100 [W] light bulb!
(60 Hz AC line voltage)
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Example (cont.)Example (cont.)
2 2120 120
100 [W]144
AVEabsP
R
120 2cos 2 60 [V]v t t
R = 144 []
2 2
120 2 120 2200 [W]
144AVE
absPR
Observation about average power:
The value 120 [V] is called the effective or RMS (Root Mean Square) voltage.(The meaning of the term “RMS” will become clear in later notes.)
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Example (cont.)Example (cont.)
2RMS
AV
cos [V]v t A t
General Formula
2
RMSAVEabs
VP
R
The RMS voltage is the peak voltage divided by 2.