1. Lecture 2 Resistance Ohm’s Law Electric Energy Electric Power Efficiency http://dramermejbel.weebly.com

2. The common units: Volts, Amps, Ohms, Coulombs are all named after people VoltaAmpereOhmCoulomb

3. Resistance The resistance of a conductor is due to the collisions between free electrons and atoms. When the same voltage is applied across different conductors, different currents flow. Some conductors offer more opposition or resistance to the passage of current than others.

4. Factors Affecting Resistance 1. The length L of the material. Longer materials have greater resistance. 1 1 L 2 2 2L If the wire doubles in length, it doubles in resistance

5. 2. The cross-sectional area A of the material. Larger areas offer LESS resistance. 2. The cross-sectional area A of the material. Larger areas offer LESS resistance.. Notice that the electrons seem to be moving at the same speed in each one but there are many more electrons in the larger wire. This results in a larger current which leads us to say that the resistance is less in a wire with a larger cross sectional area. It can be shown that R  1/A.

6. For metals and a lot of insulators, when temperature is raised, the lattice ions vibrate more vigorously, increasing the frequency of collision between electrons and the lattice. The resistance therefore increases. 3. The higher temperatures result in higher resistances.

7. 4. The kind of material. Iron has more electrical resistance than a geometrically similar copper conductor. Ig R steel > R copper

8. Introductory Circuit Analysis, 12/e Boylestad Copyright ©2011 by Pearson Education, Inc. publishing as Pearson [imprint] TYPES OF RESISTORS (Fixed and variable) Fixed Resistors FIG. 3.13 Fixed-composition resistors: (a) construction; (b) appearance.

9. Calculating Resistance It’s possible to calculate resistance of a resistor using the color bands on it –AB represent a 2 digit number –C represents the magnitude –Resistance = AB * 10 C + D

10. Example: Calculating Resistance The first two bands correspond to 4 and 7. The third band tells you the number of zeros following. 47*10 3 = 47,000 Ω + 10%

11. Larger devices have a larger surface area. A physically larger device is able to dissipate more heat and handle more power. 11 Since a device loses heat through its surface, the larger the surface area, the more heat a device dissipates. Heat dissipation is related to device surface area. Copyright © Texas Education Agency, 2013. All rights reserved.

12. Variable Resistors -- Potentiometers is a three-terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. R ac = R ab + R bc

13. Construction

14. Setting the Value of Resistance EGR 10114

15. Rheostats A variable resistance used to control current Usually a two-terminal variable resistor Potentiometer can be wired as a rheostat

16. Introductory Circuit Analysis, 12/e Boylestad Copyright ©2011 by Pearson Education, Inc. publishing as Pearson [imprint] TEMPERATURE EFFECTS Demonstrating the effect of a positive temperature coefficient( a-for conductors) and a negative temperature coefficient (b-for semiconductors )on the resistance

17. Temperature Coefficient For most materials, the resistance R changes in proportion to the initial resistance and to the proportion to the initial resistance R o and to the change in temperature  t. Change in resistance: The temperature coefficient of resistance,  is the change in resistance per unit resistance per unit degree change of temperature.

18. Example : The resistance of a copper wire is 4 m  at 20 0 C. What will be its resistance if heated to 80 0 C? Assume that  = 0.004 /C o. R o = 4 m  t = 80 o C – 20 o C = 60 C o  R = 1.03 m  R = R o +  R R = 4 mm R = 4 m  + 1.03 m  R = 5.03 m 

19. Resistivity of a material is a measure of how strongly a material opposes the flow of electric current  is called the resistivity of the material. The unit of  is .m Conductivity of material, 

20. Resistivity of various materials Material Class  ( .m ) CopperGood conductor 1.72  10 -8 AluminumGood conductor 2.65  10 -8 GermaniumSemiconductor0.6 SiliconSemiconductor2300 Quartzinsulator 5  10 16

21. Ohm’s Law  Every conversion of energy from one form to another can be related to this equation.  In electric circuits the effect we are trying to establish is the flow of charge, or current.  The potential difference, or voltage between two points is the cause (“pressure”),  The resistance is the opposition encountered.

22. Ohm’s Law  Developed in 1827 by Georg Simon Ohm  For a fixed resistance, the greater the voltage (or pressure) across a resistor, the more the current.  The more the resistance for the same voltage, the less the current.  Current is proportional to the applied voltage and inversely proportional to the resistance.

23. Ohm’s Law (Golden Triangle)

24. Plotting Ohm’s Law

25. Insert Fig 4.8

26. Example : When a 3-V battery is connected to a light, a current of 6 mA is observed. What is the resistance of the light filament? Source of EMF R I +- V = 3 V 6 mA R = 500 R = 500 

27. Copyright © Texas Education Agency, 2011. All rights reserved.27 In this first example, we will calculate the amount of current (I) in a circuit, given values of voltage (E) and resistance (R): What is the amount of current (I) in this circuit?

28. Copyright © Texas Education Agency, 2011. All rights reserved.28 In the last example, we will calculate the amount of voltage supplied by a battery, given values of current (I) and resistance (R): What is the amount of voltage provided by the battery?

29. Energy  Energy ( W ) lost or gained by any system is determined by: W = P×t  Since power is measured in watts (or joules per second) and time in seconds, the unit of energy is the wattsecond (Ws) or joule (J)

30. Power  Power is an indication of how much work (the conversion of energy from one form to another) can be done in a specific amount of time; that is, a rate of doing work. Power is the rate of doing work –Power = Work / time Power is measured in watts (W) Work and energy measured in joules (J) One watt = One joule per second

31. Power in Electrical Systems From V = W /Q → W = V × Q and I = Q / t → Q = I× t W = V× I × t, P = W / t P = V× I × t / t → P = V× I From Ohm’s Law ( I = V / R ), we can also find that P = I 2 R and P = V 2 / R

32. 32 Power in Electrical Systems We should be able to use any of the power equations to solve for V, I, or R if P is given For example:

33. Diagnosis and Troubleshooting of Automotive Electrical, Electronic, and Computer Systems, Fifth Edition By James D. Halderman © 2010 Pearson Higher Education, Inc. Pearson Prentice Hall - Upper Saddle River, NJ 07458 WATT’S LAW ( Magic Circle )

34. Efficiency  Efficiency (  ) of a system is determined by the following equation:  = P o / P i Where:  = efficiency, (η is Greek letter eta) P o = power output P i = power input - Poor efficiency in energy transfers results in wasted energy, generates more heat. - Heat must be removed (requires fans and heat sinks)

35. Efficiency To find the total efficiency of a system –Obtain product of individual efficiencies of all subsystems:  Total =  1 ×  2 ×  3 × ∙∙∙

36. Efficiency  The basic components of a generating (voltage) system are depicted below, each component has an associated efficiency, resulting in a loss of power through each stage. Insert Fig 4.19