1. Physics 212 Lecture 9, Slide 1 Physics 212 Lecture 9 Electric Current Exam Here, Tuesday, June 26, 8 – 9:30 AM 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. Physics 212 Lecture 9, Slide 2 05 Conductors Charges free to move E = 0 in conductor determines charge densities on surfaces What Determines How They Move? They move until E = 0 ! SpheresCylinders Infinite Planes Infinite Planes Gauss’Law Field Lines & Equipotentials FieldLines Equipotentials Work Done By E Field Change in Potential Energy Capacitor Networks Series: (1/C 23 )=(1/C 2 )+(1/C 3 ) Parallel C 123 = C 1 + C 23

3. Physics 212 Lecture 9, Slide 3 I A V L  V = EL I = JA Observables: I/A =  V/L I = V/(L/  R = L AAAA I = V/R R = Resistance  = 1/  Ohm’s Law: J =  E Conductivity – high for good conductors. Note: “Conventional current flow”, I, is opposite to direction electrons flow

4. Physics 212 Lecture 9, Slide 4 R = L AAAA Analogy to flow of water 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 To make R big, make L long or A small To make R small, make L short or A big

5. Physics 212 Lecture 9, Slide 5 Checkpoint 1a Checkpoint 1b Same current through both resistors Compare voltages across resistors

6. Physics 212 Lecture 9, Slide 6 Preflight 12 Same Current The SAME amount of current I passes through three different resistors. R 2 has twice the cross- sectional area and the same length as R 1, and R 3 is three times as long as R 1 but has the same cross-sectional area as R 1. In which case is the CURRENT DENSITY through the resistor the smallest? A. Case 1B. Case 2C. Case 3

7. Physics 212 Lecture 9, Slide 7 Voltage Current Resistance Series Parallel Resistor Summary Different for each resistor. V total = V 1 + V 2 Increases R eq = R 1 + R 2 Same for each resistor I total = I 1 = I 2 Same for each resistor. V total = V 1 = V 2 Decreases 1/R eq = 1/R 1 + 1/R 2 Wiring Each resistor on the same wire. Each resistor on a different wire. Different for each resistor I total = I 1 + I 2 R1R1 R2R2 R1R1 R2R2

8. Physics 212 Lecture 9, Slide 8 R 2 in series with R 3 Checkpoint 2a Current through R 2 and R 3 is the same Compare the current through R 2 with the current through R 3 : A. I 2 > I 3 B. I 2 = I 3 C. I 2 < I 3 Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R 1 = R 2 = R 3 = R.

9. Physics 212 Lecture 9, Slide 9 Checkpoint 2b R 1 = R 2 = R 3 = R Compare the current through R 1 with the current through R 2 A. I 1 /I 2 = 1/2 B. I 1 /I 2 = 1/3 C. I 1 /I 2 = 1 D. I 1 /I 2 = 2 E. I 1 /I 2 = 3 We know: I 23 Similarly: I1I1

10. Physics 212 Lecture 9, Slide 10 R 1 = R 2 = R 3 = R Checkpoint 2c Compare the voltage across R 2 with the voltage across R 3 V 2 > V 3 V 2 = V 3 = V V 2 = V 3 < V V 2 < V 3 A B C D Consider loop V2V2 V3V3 V 23

11. Physics 212 Lecture 9, Slide 11 R 1 = R 2 = R 3 = R Checkpoint 2d Compare the voltage across R 1 with the voltage across R 2 V 1 = V 2 = V V 1 = ½ V 2 = V V 1 = 2V 2 = V V 1 = ½ V 2 = 1/5 V V 1 = ½ V 2 = ½ V A B C D E R 1 in parallel with series combination of R 2 and R 3 V1V1 V 23

12. Physics 212 Lecture 9, Slide 12Calculation In the circuit shown: V = 18V, R 1 = 1  R 2 = 2  R 3 = 3  and  R 4 = 4  What is V 2, the voltage across R 2 ? 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 R series = R a + R b (1/R parallel ) = (1/R a ) + (1/R b ) V R1R1 R2R2 R4R4 R3R3

13. Physics 212 Lecture 9, Slide 13Calculation In the circuit shown: V = 18V, R 1 = 1  R 2 = 2  R 3 = 3  and  R 4 = 4  What is V 2, the voltage across R 2 ? Combine Resistances: (A) (B) (C) (A) in series (B) in parallel (C) neither in series nor in parallel R 1 and R 2 are connected: V R1R1 R2R2 R4R4 R3R3 Parallel: Can make a loop that contains only those two resistors Parallel Combination Series Combination Series : Every loop with resistor 1 also has resistor 2.

14. Physics 212 Lecture 9, Slide 14Calculation In the circuit shown: V = 18V, R 1 = 1  R 2 = 2  R 3 = 3  and  R 4 = 4  What is V 2, the voltage across R 2 ? We first will combine resistances R 2 + R 3 + R 4 : Which of the following is true? (A) R 2, R 3 and R 4 are connected in series (B) (B) R 2, R 3, and R 4 are connected in parallel (C) (C) R 3 and R 4 are connected in series (R 34 ) which is connected in parallel with R 2 (D) (D) R 2 and R 4 are connected in series (R 24 ) which is connected in parallel with R 3 (E) R 2 and R 4 are connected in parallel (R 24 ) which is connected in parallel with R 3 V R1R1 R2R2 R4R4 R3R3

15. Physics 212 Lecture 9, Slide 15Calculation In the circuit shown: V = 18V, R 1 = 1  R 2 = 2  R 3 = 3  and  R 4 = 4  What is V 2, the voltage across R 2 ? V R1R1 R2R2 R4R4 R3R3 Redraw the circuit using the equivalent resistor R 24 = series combination of R 2 and R 4. (A) (B) (C) (A) (B) (C) R 2 and R 4 are connected in series (R 24 ) which is connected in parallel with R 3

16. Physics 212 Lecture 9, Slide 16Calculation In the circuit shown: V = 18V, R 1 = 1  R 2 = 2  R 3 = 3  and  R 4 = 4  What is V 2, the voltage across R 2 ? Combine Resistances: R 2 and R 4 are connected in series = R 24 R 3 and R 24 are connected in parallel = R 234 (A) (B) (C) (D) (A) R 234 = 1  (B) R 234 = 2  (C) R 234 = 4  (D) R 234 = 6  What is the value of R 234 ? (1/R parallel ) = (1/R a ) + (1/R b ) 1/R 234 = (1/3) + (1/6) = (3/6)   V R1R1 R 24 R3R3 R 234 = 2  R 2 and R 4 in series R 24 = R 2 + R 4 = 2 

17. Physics 212 Lecture 9, Slide 17Calculation In the circuit shown: V = 18V, R 1 = 1  R 2 = 2  R 3 = 3  and  R 4 = 4  R 24 = 6  R 234 = 2  What is V 2, the voltage across R 2 ? V R1R1 R 234 V R 1234 R 1 and R 234 are in series. R 1234 = 1 + 2 = 3  Ohm’s Law tells us I 1234 = V/R 1234 = 18 / 3 = 6 Amps = I 1234 I 1 = I 234 Our next task is to calculate the total current in the circuit

18. Physics 212 Lecture 9, Slide 18 In the circuit shown: V = 18V, R 1 = 1  R 2 = 2  R 3 = 3  and  R 4 = 4  R 24 = 6  R 234 = 2  I 1234 = 6 A What is V 2, the voltage across R 2 ? What is V ab, the voltage across R 234 ? (A) (B) (C) (D) (E) (A) V ab = 1 V (B) V ab = 2 V (C) V ab = 9 V (D) V ab = 12 V (E) V ab = 16 V V R1R1 R 234 V R 1234 I 234 = I 1234 Since R 1 in series w/ R 234 V 234 = I 234 R 234 = 6 x 2 = 12 Volts I 1234 = I 1 = I 234 a b

19. Physics 212 Lecture 9, Slide 19Calculation V = 18V R 1 = 1  R 2 = 2  R 3 = 3  R 4 = 4  R 24 = 6  R 234 = 2  I 1234 = 6 Amps I 234 = 6 Amps V 234 = 12V What is V 2 ? V R1R1 R 234 V R1R1 R 24 R3R3 Which of the following are true? A) V 234 = V 24 B) I 234 = I 24 C) Both A+BD) None Ohm’s Law I 24 = V 24 / R 24 = 12 / 6 = 2 Amps R 3 and R 24 were combined in parallel to get R 234 Voltages are same!

20. Physics 212 Lecture 9, Slide 20Calculation V = 18V R 1 = 1  R 2 = 2  R 3 = 3  R 4 = 4  R 24 = 6  R 234 = 2  I 1234 = 6 Amps I 234 = 6 Amps V 234 = 12V V 24 = 12V I 24 = 2 Amps What is V 2 ? V R1R1 R 24 R3R3 Which of the following are true? A) V 24 = V 2 B) I 24 = I 2 C) Both A+BD) None V R1R1 R2R2 R4R4 R3R3 I 1234 R 2 and R 4 where combined in series to get R 24 Currents are same! I 24 Ohm’s Law V 2 = I 2 R 2 = 2 x 2 = 4 Volts! The Problem Can Now Be Solved!

21. Physics 212 Lecture 9, Slide 21 Quick Follow-Ups V = 18V R 1 = 1  R 2 = 2  R 3 = 3  R 4 = 4  R 24 = 6  R 234 = 2  V 234 = 12V V 2 = 4V I 24 = 2 Amps I 1234 = 6 Amps What is I 3 ? (A) (B) (C) (A) I 3 = 2 A (B) I 3 = 3 A (C) I 3 = 4 A V R1R1 R 234 a b V R1R1 R2R2 R4R4 R3R3 = V 3 = V 234 = 12V I 3 = V 3 /R 3 = 12V/3  = 4A What is I 1 ? We know I 1 = I 1234 = 6 A NOTE: I 2 = V 2 /R 2 = 4/2 = 2 A I 1 = I 2 + I 3 Make Sense??? I1I1 I2I2 I3I3