1. Physics 212 Lecture 9, Slide 1 Physics 212 Lecture 9 Today's Concept: Electric Current Ohm’s Law & resistors Resistors in circuits Power in circuits

2. Main Point 1 First, we defined the electric current in a conductor as the amount of charge that passes through a cross-section of the conductor per unit time (i.e., I = dq/dt). We developed a microscopic view of current in which the charge carriers are dissociated electrons in the conductor that move randomly at high speeds (~10^ 5 m/s), but when a potential difference is introduced across the conductor, the resultant electric field gives these electrons a non-zero average velocity which is the source of the electric current. Physics 212 Lecture 9, Slide 2

3. Main Point 2 Second, we introduced Ohm’s law that states that, for a wide range of materials, over a wide range of field strengths, the current density is proportional to the electric field that gives rise to it. The constant of proportionality is called the conductivity of the material. We used this constant to characterize the resistance of a resistor, obtaining the more common form of Ohm’s law in which the voltage across the resistor is equal to the product of the resistance and the current that flows through the resistor. Physics 212 Lecture 9, Slide 3

4. Main Point 3 Third, we obtained expressions for the equivalent resistance of two resistors connected either in series or in parallel. Namely, the equivalent resistance of two resistors in series is simply the sum of the individual resistances, while the inverse of the equivalent resistance of two resistors connected in parallel is equal to the sum of the individual inverse resistances. Physics 212 Lecture 9, Slide 4

5. Main Point 4 Fourth, we determined that the power, the time rate of change of the energy of a circuit component is always equal to the product of the voltage drop across the component and the current that flows through the component. In a simple circuit composed of a single resistor connected to the terminals of the battery, we found that all of the energy that is supplied by the battery is dissipated as heat in the resistor. Physics 212 Lecture 9, Slide 5

6. Physics 212 Lecture 9, Slide 6 J =  E I A V R = L AAAA L  I = V/R same as where Conductivity – high for good conductors. LLLL A resistivity – high for bad conductors. =

7. Physics 212 Lecture 9, Slide 7

8. Physics 212 Lecture 9, Slide 8 Checkpoint 1a Checkpoint 1b Same current through both resistors Compare voltages across resistors

9. Physics 212 Lecture 9, Slide 9

10. Physics 212 Lecture 9, Slide 10 Checkpoint 3 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

11. Physics 212 Lecture 9, Slide 11

12. Physics 212 Lecture 9, Slide 12 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 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.

13. Physics 212 Lecture 9, Slide 13

14. Physics 212 Lecture 9, Slide 14 Checkpoint 2b R 1 = R 2 = R 3 = R 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

15. Physics 212 Lecture 9, Slide 15

16. Physics 212 Lecture 9, Slide 16 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

17. Physics 212 Lecture 9, Slide 17

18. Physics 212 Lecture 9, Slide 18 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

19. Physics 212 Lecture 9, Slide 19

20. Physics 212 Lecture 9, Slide 20 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

21. Physics 212 Lecture 9, Slide 21Calculation 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 ? Conceptual 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 ) Strategic Analysis – –Combine resistances to form equivalent resistances – –Evaluate voltages or currents from Ohm’s Law – –Expand circuit back using knowledge of voltages and currents V R1R1 R2R2 R4R4 R3R3

22. Physics 212 Lecture 9, Slide 22