Physics 212. Lecture 9. Electric Current

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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! Physics 212 Lecture 9, Slide 1

2 Conductors Charges free to move What Determines How They Move? They move until E = 0! Spheres Cylinders Infinite Planes Gauss Law E = 0 in conductor determines charge densities on surfaces Field Lines & Equipotentials Work Done By E Field b b W F dl qe dl a b a a Capacitor Networks Change in Potential Energy b U W qe dl a b a b a Series: (1/C 23 )=(1/C 2 )+(1/C 3 ) Parallel C 123 = C 1 + C Physics 212 Lecture 9, Slide 2

3 I s A L Conductivity high for good conductors. Observables: = EL I = JA Ohm s Law: J = s E I/A = s/l I = /(L/sA) R = Resistance r = 1/s I = /R R = L sa Physics 212 Lecture 9, Slide 3

4 Analogy to flow of water I is like flow rate of water is like pressure R is how hard it is for water to flow in a pipe R = L sa To make R big, make L long or A small To make R small, make L short or A big Physics 212 Lecture 9, Slide 4

5 Checkpoint 1a Checkpoint 1b Same current through both resistors Compare voltages across resistors R L A IR L A A 1 2 4A L 2 L Physics 212 Lecture 9, Slide 5 1

6 Preflight 12 The SAME amount of current I passes through three different resistors. R 2 has twice the crosssectional area and the same length as, and R 3 is three times as long as but has the same cross-sectional area as. In which case is the CURRENT DENSITY through the resistor the smallest? A. Case 1 B. Case 2 C. Case 3 J I J1 J3 2J A 2 Same Current J 1 A Physics 212 Lecture 9, Slide 6

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

8 Three resistors are connected to a battery with emf as shown. The resistances of the resistors are all the same, i.e. = R 2 = R 3 = R. Checkpoint 2a 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 R 2 in series with R 3 Current through R 2 and R 3 is the same I 23 R 2 R 3 Physics 212 Lecture 9, Slide 8

9 I 1 I 23 = R 2 = R 3 = R Checkpoint 2b Compare the current through with the current through R 2 We know: I 23 R 2 R 3 A. I 1 /I 2 = 1/2 B. I 1 /I 2 = 1/3 C. I 1 /I 2 = 1 Similarly: I 1 R 1 D. I 1 /I 2 = 2 E. I 1 /I 2 = 3 I I R R R1 2 Physics 212 Lecture 9, Slide 9

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

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

12 R 2 R 3 R 4 Calculation In the circuit shown: = 18, = 1W, R 2 = 2W, R 3 = 3W, and R 4 = 4W. What is 2, the voltage across R 2? Analysis: Ohm s Law: when current I flows through resistance R, the potential drop is given by: = IR. Resistances are combined in series and parallel combinations R series = R a + R b (1/R parallel ) = (1/R a ) + (1/R b ) Physics 212 Lecture 9, Slide 12

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

14 R 2 R 3 R 4 Calculation In the circuit shown: = 18, = 1W, R 2 = 2W, R 3 = 3W, and R 4 = 4W. What is 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) R 2, R 3, and R 4 are connected in parallel (C) R 3 and R 4 are connected in series (R 34 ) which is connected in parallel with R 2 (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 Physics 212 Lecture 9, Slide 14

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

16 R 3 R 24 Calculation In the circuit shown: = 18, = 1W, R 2 = 2W, R 3 = 3W, and R 4 = 4W. What is 2, the voltage across R 2? Combine Resistances: What is the value of R 234? R 2 and R 4 are connected in series = R 24 R 3 and R 24 are connected in parallel = R 234 (A) R 234 = 1 W (B) R 234 = 2 W (C) R 234 = 4 W (D) R 234 = 6 W R 2 and R 4 in series R 24 = R 2 + R 4 = 2W 4W 6W (1/R parallel ) = (1/R a ) + (1/R b ) 1/R 234 = (1/3) + (1/6) = (3/6) W 1 R 234 = 2 W Physics 212 Lecture 9, Slide 16

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

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

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

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

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

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