1 / 21

Physics 212 Lecture 9

Physics 212 Lecture 9. Electric Current. Exam Here, Tuesday, June 26, 8 – 9:30 AM Will begin at 7:30 for those who must leave by 9. Office hours 1-7 PM, Rm 232 Loomis Bring your ID !. What Determines How They Move?. They move until E = 0 !. Spheres Cylinders Infinite Planes.

cmcclanahan
Télécharger la présentation

Physics 212 Lecture 9

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. Physics 212 Lecture 9 Electric Current Exam Here, Tuesday, June 26, 8 – 9:30 AM Will begin at 7:30 for those who must leave by 9. Office hours 1-7 PM, Rm 232 Loomis Bring your ID !

  2. What Determines How They Move? They move until E = 0 ! Spheres Cylinders Infinite Planes Gauss’ Law Field Lines & Equipotentials Capacitor Networks Work Done By E Field Equipotentials Field Lines Series: (1/C23)=(1/C2)+(1/C3) Parallel C123 = C1 + C23 Change in Potential Energy Conductors Charges free to move E = 0 in conductor determines charge densities on surfaces 05

  3. Conductivity – high for good conductors. Ohm’s Law: J = s E Observables: V = EL I = JA I/A = sV/L I = V/(L/sA) R = Resistance r = 1/s I = V/R L R = sA I s A L V Note: “Conventional current flow”, I, is opposite to direction electrons flow

  4. To make R big, make L long or A small L R = To make R small, make L short or A big sA Analogy to flow of water I is like flow rate of water V is like pressure R is how hard it is for water to flow in a pipe

  5. Checkpoint 1a Checkpoint 1b Same current through both resistors Compare voltages across resistors

  6. Preflight 12 Same Current The SAME amount of current I passes through three different resistors. R2 has twice the cross-sectional area and the same length as R1, and R3 is three times as long as R1 but has the same cross-sectional area as R1. In which case is the CURRENT DENSITY through the resistor the smallest? A. Case 1 B. Case 2 C. Case 3

  7. Resistor Summary Series Parallel R1 R2 R1 R2 Each resistor on the same wire. Each resistor on a different wire. Wiring Different for each resistor. Vtotal = V1 + V2 Same for each resistor. Vtotal = V1 = V2 Voltage Same for each resistor Itotal = I1 = I2 Different for each resistor Itotal = I1 + I2 Current Increases Req = R1 + R2 Decreases 1/Req = 1/R1 + 1/R2 Resistance

  8. Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R1= R2 = R3 = R. Checkpoint 2a Current through R2 and R3 is the same Compare the current through R2 with the current through R3: A. I2 > I3 B. I2 = I3C. I2 < I3 R2 in series with R3

  9. I1 I23 We know: Similarly: R1 = R2 = R3 = R Checkpoint 2b Compare the current through R1 with the current through R2 A. I1/I2 = 1/2 B. I1/I2 = 1/3 C. I1/I2 = 1 D. I1/I2 = 2 E. I1/I2 = 3

  10. V2 V3 V23 Consider loop R1 = R2 = R3 = R Checkpoint 2c Compare the voltage across R2 with the voltage across R3 V2 > V3 A B V2 = V3 = V C V2 = V3 < V D V2 < V3

  11. V1 V23 R1 in parallel with series combination of R2 and R3 R1 = R2 = R3 = R Checkpoint 2d Compare the voltage across R1 with the voltage across R2 A V1 = V2 = V V1 = ½ V2 = V B C V1 = 2V2 = V D V1 = ½ V2 = 1/5 V V1 = ½ V2 = ½ V E

  12. Calculation R2 R1 In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, andR4 = 4W. What is V2, the voltage across R2? V R3 R4 • Analysis: • Ohm’s Law: when current I flows through resistance R, the potential drop V is given by: V = IR. • Resistances are combined in series and parallel combinations • Rseries = Ra + Rb • (1/Rparallel) = (1/Ra) + (1/Rb)

  13. Calculation R1 and R2 are connected: (A) in series(B) in parallel(C) neither in series nor in parallel Parallel Combination Series Combination Series : Every loop with resistor 1 also has resistor 2. Parallel: Can make a loop that contains only those two resistors R2 R1 In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, andR4 = 4W. What is V2, the voltage across R2? V R3 R4 • Combine Resistances:

  14. Calculation R2 R1 In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, andR4 = 4W. What is V2, the voltage across R2? V R3 R4 • We first will combine resistances R2 + R3 + R4: • Which of the following is true? • R2, R3 and R4 are connected in series • R2, R3, and R4 are connected in parallel • R3 and R4 are connected in series (R34) which is connected in parallel with R2 • R2 and R4 are connected in series (R24) which is connected in parallel with R3 • R2 and R4 are connected in parallel (R24) which is connected in parallel with R3

  15. Calculation (A) (B) (C) R2 R1 In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, andR4 = 4W. What is V2, the voltage across R2? V R3 R4 R2 and R4 are connected in series (R24) which is connected in parallel with R3 Redraw the circuit using the equivalent resistor R24 = series combination of R2 and R4.

  16. Calculation R1 V R3 R24 What is the value of R234? (A) R234= 1 W (B) R234= 2 W (C) R234= 4 W (D) R234= 6 W R234 = 2W 1/R234 = (1/3)+ (1/6) = (3/6) W-1 R24 = R2 + R4 = 2W + 4W = 6W In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, andR4 = 4W. What is V2, the voltage across R2? • Combine Resistances: R2 and R4 are connected in series = R24 R3 and R24 are connected in parallel = R234 • R2 and R4 in series (1/Rparallel) = (1/Ra) + (1/Rb)

  17. Calculation R1 In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, andR4 = 4W. R24 = 6W R234 = 2W What is V2, the voltage across R2? V I1 = I234 R234 R1 and R234 are in series. R1234 = 1 + 2 = 3 W Our next task is to calculate the total current in the circuit Ohm’s Law tells us I1234 = V/R1234 = 18 / 3 = 6 Amps V R1234 = I1234

  18. What is Vab, the voltage across R234 ? (A) Vab= 1 V (B) Vab= 2 V (C) Vab= 9 V (D) Vab= 12 V (E) Vab= 16 V V In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, andR4 = 4W. R24 = 6W R234 = 2W I1234 = 6 A What is V2, the voltage across R2? I1234 R1234 R1 I234 = I1234 Since R1 in series w/ R234 a V R234 V234 = I234 R234 = 6 x 2 = 12 Volts = I1 = I234 b

  19. Calculation R1 R1 V V R3 R234 R24 R3 and R24 were combined in parallel to get R234 Voltages are same! V = 18V R1 = 1W R2 = 2W R3 = 3W R4 = 4W. R24 = 6W R234 = 2W I1234 = 6 Amps I234 = 6 Amps V234 = 12V What is V2? Which of the following are true? A) V234 = V24 B) I234 = I24 C) Both A+B D) None Ohm’s Law I24 = V24 / R24 = 12 / 6 = 2 Amps

  20. Calculation R2 R1 R1 I24 V V R3 R3 R24 R4 R2 and R4 where combined in series to get R24 Currents are same! Ohm’s Law V2 = I2 R2 = 2 x 2 = 4 Volts! The Problem Can Now Be Solved! V = 18V R1 = 1W R2 = 2W R3 = 3W R4 = 4W. R24 = 6W R234 = 2W I1234 = 6 Amps I234 = 6 Amps V234 = 12V V24 = 12V I24 = 2 Amps What is V2? I1234 Which of the following are true? A) V24 = V2 B) I24 = I2 C) Both A+B D) None

  21. Quick Follow-Ups R1 I1 I2 a I3 V R234 b • What is I3 ? (A) I3= 2 A (B) I3= 3 A (C) I3= 4 A I3 = V3/R3 = 12V/3W = 4A NOTE: I2 = V2/R2 = 4/2 = 2 A I1 = I2 + I3 Make Sense??? R2 R1 V = 18V R1 = 1W R2 = 2W R3 = 3W R4 = 4W. R24 = 6W R234 = 2W V234= 12V V2 = 4V I24 = 2 Amps I1234 = 6 Amps V = R3 R4 V3 = V234 = 12V • What is I1 ? We know I1 = I1234 = 6 A

More Related