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Effect on the current of changing the voltage with the resistance at a constant value.
Ohm’s Law
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
•I = V/R
•V = RI
•R = V/I
Effect on the current of changing the resistance with the voltage at a constant value.
Ohm’s Law
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
•I = V/R
•V = RI
•R = V/I
Graph of current versus voltage
Linear Relationships
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
A graphic aid for the Ohm’s law formulas.
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Determine IT
IT = 50V/1.0k Ohms
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
I = 50mA
Determine RL
RL = 12V/3A
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
RL = 4 Ohms
Determine VS
Vs = (100 Ohms)(5A)
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Vs = 500V
Power in Electric Circuits
P = VI
•1 Watt of Power is 1 Joule of Energy/Second
•Rate of Work Being Done
•Heat being Produced
•Source has to Generate Enough Votage & Current for Circuit Load
P = VI
P = I2 R
P = V2/R
High Power Circuits have:
A. High Voltage
B. High Current
C. A & B
Power dissipation in an electric circuit is seen as heat given off by the resistance.
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
(a) P = (10V)(2A) = 20W
(b) P = (2A)2(47Ohms) = 188W
(c) P = (5V)2/(1Ohms) = 2.5W
Unknown Source Voltage Unknown CurrentUnknown Resistance
Ohm’s Law and Power Wheel
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
All Circuits have a certain amount of Power Loss – Generally Through Heat
Power EfficiencyOften stated as a Percentage
Thomas L. Floyd
Electronics Fundamentals, 6e
Electric Circuit Fundamentals, 6e
Copyright ©2004 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Efficiency = Pout/Pin
49W/50W = 98%
Pout = Pin – Ploss
49W = 50W - 1W
Kilowatt Hour
Kilowatt Hour = How Power Company computes your Bill
Watts over Hours:
Refrigerator: 500W for 18 Hours = 9Kwh
Air Conditioner: 3.8KW for 10 Hours = 38KWh
Ampere Hour Battery Ratings
• How long the battery is rated to last
• 1 Ampere Hour (Ah) = Can Deliver:
– 1 Amp for 1 Hour– 1 Amp for 1 Hour
– 500Ma for 2 Hours
• Generally tested at a Certain Current
– Car Battery – 70 Ah @ 3.5A
– D Battery – 4.5 Ah @ 100mA
– 9V Battery – 400Ah @ 8uA
Troubleshooting – Open Resistor
•Full Source Voltage Across Resistor/Component
•Low Total Current
•Little or No Voltage Across Other Resistors/Components
Troubleshooting – Shorted Resistor
•Low Voltage Across Shorted Resistor/Component
•High Total Current
•Higher Voltages Across Other Resistors/Components
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