1 / 17

Ohm’s Law

Ohm’s Law. One of Georg Ohm’s Many Contributions to Science and Electricity. Answers to Friday’s Lab – CIRCUIT A. Remove wire A from the circuit. What happens? The outside bulb will go out while the inside bulb will remain lit.

shyla
Download Presentation

Ohm’s Law

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. Ohm’s Law One of Georg Ohm’s Many Contributions to Science and Electricity

  2. Answers to Friday’s Lab – CIRCUIT A • Remove wire A from the circuit. What happens?The outside bulb will go out while the inside bulb will remain lit. • Why does this happen?The circuit is only broken at the outside bulb, whereas the inside bulb still has an unbroken path.

  3. Answers to Friday’s Lab – CIRCUIT A • Now put back wire A. Remove wire B from the circuit. What happens?Same as question #1 • Why does this happen?Same as question #2

  4. Answers to Friday’s Lab – CIRCUIT A • Put back wire B. Now remove wire C. What happens?Both bulbs go out. • Why does this happen?The current cannot reach either bulb; the circuit is broken/incomplete. • How is this circuit wired – in series or parallel? How do you know?It’s wired in parallel because there is more than one path for the electrons to follow.

  5. Answers to Friday’s Lab – CIRCUIT B • Close the switch. What happens?Both bulbs turn on. • Remove one of the lights from the circuit. What happens?Both bulbs go out. OROne bulb is still lit and glows brighter. • Why does this happen?The circuit is broken/incomplete. ORAll voltage/energy is going to one bulb only and is not divided between both.

  6. Answers to Friday’s Lab – CIRCUIT B • Put the light bulb back on. Now remove the battery from the circuit. What happens?Both bulbs go out. • Why does this happen?There is no source to supply electricity to bulbs.

  7. Answers to Friday’s Lab – CIRCUIT B • Disconnect wire A. What happens? Put back wire A and disconnect wire B. What happens?In both cases, both bulbs go out. • Why does this happen?The circuit is incomplete/broken. • How is this circuit wired – in series or parallel? How do you know?It is wired in series because there is only one path to follow.

  8. AC vs. DC What is the major difference between static electricity and current electricity???? Static electricity builds up on an insulator, while current electricity is the flow of electrons through a conductor.

  9. Georg Ohm • http://www.youtube.com/watch?v=oScVYw5WaFA • http://www.youtube.com/watch?v=aXrcdvWQYew

  10. Ohm’s Law • Georg Ohm described how voltage and current are affected when one of the values is changed. • Ohm’s Law states that as long as the temperature stays the same, V = IR

  11. Ohm’s LawV = IR • This means: • The resistance, R, remains constant • The current, I, is directly proportional to the voltage, V

  12. Ohm’s LawV = IR • We can rearrange this equation depending on what we don’t know. • If we are looking for the voltage, V, what would the equation be? • V = IR • If we are looking for the current, I, what would the equation be? • I = V/R • If we are looking for the resistance, R, what would the equation be? • R = V/I

  13. Ohm’s LawV = IR • A pneumonic for remembering Ohm’s Law:

  14. Ohm’s Law – Example 1 • A current of 1.5 A flows through a 30-Ωresistor that is connected across a battery. What is the battery’s voltage? • Use the GRASP method. • G – Given: current I = 1.5 A; resistance R = 30 Ω • R – Required: voltage V = ? • A and S – Analysis and Solution: • The correct equation is V = IR • Substitute the values & their units; solve the problem • V = IR = (1.5A)(30 Ω) = 45 V • P – Paraphrase:The voltage in the circuit is 45 V.

  15. Ohm’s Law – Example 2 • A fire truck has a searchlight with a resistance of 60 Ω that is placed across a 24-V battery. What is the current in this circuit? • Use the GRASP method. • G – Given: resistance R = 60 Ω; voltage V = 24 V • R – Required: current I = ? • A and S – Analysis and Solution: • The correct equation is I = V/R • Substitute the values & their units; solve the problem • I = V/R = (24V)/(60 Ω) = 0.4 A • P – Paraphrase:The current in this circuit is 0.4 A.

  16. Ohm’s Law – Example 3 • A current of 625 mA runs through a bulb that is connected to a 120-V power supply. What is the resistance of the bulb? • Use the GRASP method. • G – Given: current I = 625 mA = 0.625 A; voltage V = 120 V • R – Required: resistance R = ? • A and S – Analysis and Solution: • The correct equation is R = V/I • Substitute the values & their units; solve the problem • R = V/I = (120V)/(0.625A) = 192 Ω • P – Paraphrase:The resistance in the circuit is 192Ω.

  17. Short Circuits • When a problem develops within a wire that allows electrons to flow through a device along a different path than the one intended. • A short circuit is an accidental low-resistance connection between two points in a circuit, often causing excess current flow. • These can be dangerous – how?? • When a transmission line’s knocked down in a storm • However, it can also be helpful: • Using a wire to go across a load – current flows more easily through the wire path than through a bulb, for example – allowing work to be done on a device while not interrupting the circuit.

More Related