1 / 36

Current and Resistance

Current and Resistance. CHAPTER 19. The nature of electricity. How does the energy generated by wind farms get to people’s houses to power their appliances?. Current. Current is the rate of change of electric charge

samson
Download Presentation

Current and Resistance

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Current and Resistance CHAPTER 19

  2. The nature of electricity • How does the energy generated by wind farms get to people’s houses to power their appliances?

  3. Current • Current is the rate of change of electric charge • A current exits whenever there is a net movement of electric charge through a medium • The unit for current is the ampere • 1 ampere= 1 Coulomb second

  4. Sample Question p. 695 #2 • In a particular television tube, the beam current is 60 µA. How long does it take for 3.75 x 1014 electrons to strike the screen? • First calculate the electric charge of 3.75 x 1014 electrons. • 1 electron has a charge of 1.60 x 10-19 C • (3.75 x 1014 ) (1.60 x 10-19 C)= 6 x 10-5 C

  5. Sources and Types of Current • Batteries and generators work by converting other forms of energy into electrical potential energy • Batteries convert chemical energy into electrical potential energy • Generators convert mechanical energy (KE and PE) into electrical potential energy

  6. Potential Difference • Potential Difference, ΔV, is the driving force behind current • Increasing potential difference results in a greater current • i.e. using a 9.0 V battery generates a greater current than a 6.0 V battery • V is measured in volts • 1 volt= 1 Joule/Coulomb

  7. Resistance • Some conductors allow charges to move through them more easily than others • The opposition to the motion of charge through a conductor is the conductor’s resistance • The unit for resistance is the ohm (Ω) • Ohm’s Law:

  8. Ohm’s Law • Resistance is inversely proportional to current • As the resistance increases, the current decreases • For most materials, resistance is independent of V. • Therefore, changing V affects the current, not the resistance

  9. Factors that affect resistance (p. 703)

  10. Sample Problem p. 703 #6 • The current in a certain resistor is 0.50 A when it is connected to a potential difference of 110 v. What is the current in this same resistor if • a. The operating potential difference is 90.0 V? • b. The operating potential difference is 130 V?

  11. p. 703 #6 • I= 0.50 A, V = 110 V • We’re looking for the new current if the potential difference is changed • According to Ohm’s Law: • We’re missing R. Let’s find it 

  12. p. 703 #6 • Let’s find the new current for each potential difference • A. • B.

  13. Superconductors • Superconductors have zero resistance below a certain temperature called the critical temperature. • Once a current is established in a superconductor it will continue even if the potential difference is removed

  14. Electric Power • Electric power is the rate at which electrical energy is converted to other types of energy • Power is measured in Watts

  15. Circuits and Circuit Elements Chapter 20

  16. Schematic Diagrams (p. 731) • A diagram that depicts the construction of an electrical apparatus is a schematic diagram

  17. Electric Circuits • An electric circuit is a path through which charges can be conducted

  18. Necessary Parts of an electrical circuit • The wire: Current flows through the wire • The resistor: Can be a light bulb • The emf source: Provides the potential difference…it’s usually a battery

  19. Series Circuits • When resistors are connected in series, all the charges have to follow a single path • When one light bulb goes out, they all go out 

  20. Series Circuits • When resistors are connected in series, the current in each resistor is the same!!! • The total current in the circuit depends on how many resistors are present • The equivalent resistance is the sum of the circuit’s resistances • THE EQUIVALENT RESISTANCE SHOULD ALWAYS BE GREATER THAN THE LARGEST RESISTANCE IN THE CIRCUIT

  21. Series Current • To find the total current in the circuit, first find the equivalent resistance and then use Ohm’s Law • Although the current in each resistor has to be the same, the potential difference doesn’t have to be the same.

  22. Sample Problem p. 739 #2 • A 4.0 Ω resistor, an 8.0 Ω resistor and a 12.0 Ω resistor are connected in series with a 24.0 V battery • A. Calculate the equivalent resistance • B. Calculate the current in the circuit • What is the current in each resistor? • For resistors in series, the current in each resistor is the same…so the answer is 1.0 A

  23. Parallel Circuits • A parallel circuit is a wiring arrangement that provides alternative pathways for the movement of charges

  24. Parallel Circuits • The total current in a parallel circuit is equal to the sum of the current in each resistor • The equivalent resistance in a parallel circuit is calculated using the following equation • The potential difference across each resistor is the same

  25. Sample Problem p. 744 # 2 • An 18.0 Ω, 9.00 Ω, and 6.00 Ω resistor are connected in parallel to an emf source. A current of 4.0 A is in the 9.00 Ω resistor. • a. Calculate the equivalent resistance of the circuit. • B. What is the potential difference across the source? • C. Calculate the current in the other resistors

  26. Complex Circuits • Most circuits have both series and parallel components

  27. Complex Circuits (p. 747) • To determine the equivalent resistance for a complex circuit, you have to simplify the circuit into groups of series and parallel resistors • Sample Problem 20C (p. 747) Since the 6.0 Ω and 2.0 Ω resistor are connected in series, their equivalent resistance is 8.0 Ω

  28. Sample Problem 20C (p. 747) • The new 8.0 Ω resistor and 4.0 Ω resistor are connected in parallel. Their equivalent resistance can be found using the following equation: Req= 2.7 Ω

  29. Sample Problem 20 C (p. 747) • Finally, the last three resistors are connected in series so their equivalent resistance= 9.0 Ω + 2.7 Ω + 1.0 Ω= 12.7 Ω • The circuit can now be redrawn with the equivalent resistance connected to the original emf source

  30. Determining I and V for a resistor in a complex circuit • To find the current and/or potential difference across a particular resistor in a complex circuit you must first find the equivalent resistance for the circuit • Then you must rebuild the circuit in steps and calculate the current and potential difference for each group

  31. Sample Problem 20D (p. 749) • Sample problem 20D is a continuation of sample problem 20C. • We already determined the equivalent resistance for the circuit…12.7 Ω • Next we need to rebuild the circuit and find the potential difference and current for each group.

  32. The total current in the circuit

  33. Rebuilding the circuit • Work backward to find the current and potential difference for the next group. • These three resistors are connected in series. That means the current across all three resistors is the same (I=0.71 A). • We only care about the middle resistor because it’s the only one that leads to the 2.0 Ω resistor

  34. Rebuilding the circuit • Work backward to find the current and potential difference for the next group. • The 2.7 Ω resistor is composed of the 8.0 Ω and 4.0 Ω resistors in parallel • This means they have the same potential difference. (V=1.9 V) • We only care about the 8.0 Ω resistor because it’s the only one that leads to the 2.0 Ω resistor

  35. Rebuilding the circuit • Work backward to find the current and potential difference for the next group. • The 8.0 Ω resistor is composed of the 6.0 Ω and 2.0 Ω resistors connected in series. • This means they share the same current (I=0.24 A) • Solve for the potential difference and you’re done 

  36. http://alkalinebatteries.us/images/batteries3.jpg • http://www.renewablepowernews.com/wp-content/uploads/2008-04-wind-farm-hawaii22.jpg

More Related