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