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Announcements  Homework for tomorrow… Ch. 30: CQ 11, Probs. 20, 22, & 32 CQ3: The answer lies in your (recently used) garden hose (look inside while turning.

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Presentation on theme: "Announcements  Homework for tomorrow… Ch. 30: CQ 11, Probs. 20, 22, & 32 CQ3: The answer lies in your (recently used) garden hose (look inside while turning."— Presentation transcript:

1 Announcements  Homework for tomorrow… Ch. 30: CQ 11, Probs. 20, 22, & 32 CQ3: The answer lies in your (recently used) garden hose (look inside while turning it on). CQ4: No 30.4: 9.3 x 10 -4 m 30.6: a) 4.6 x 10 21 b) 4.3 x 10 -12 m  Office hours… MW 12:30-1:30 pm TR 9-10 am F 1-2 pm  Tutorial Learning Center (TLC) hours: MW 8-10 am, 12-6 pmR 8-12 pm, 2-6 pm T 8-1 pm, 2-6 pmF 8-10 am, 2-5 pm

2 Chapter 30 Current & Resistance (Conductivity and Resistivity & Resistance and Ohm’s Law)

3  Current..  Current Density..  Kirchoff’s Junction Rule..  Current Density related to the E-field.. Review…

4 The current density in this wire is 1. 4  10 6 A/m 2. 2. 2  10 6 A/m 2. 3. 4  10 3 A/m 2. 4. 2  10 3 A/m 2. 5. Some other value. Quiz Question 1

5 How the current density, J, is related to the E-field driving the current: where SI Units? 30.4: Conductivity and Resistivity

6 How the current density, J, is related to the E-field driving the current: Where SI Units? 30.4: Conductivity and Resistivity

7 Both segments of the wire are made of the same metal. Current I 1 flows into segment 1 from the left. How does current I 1 in segment 1 compare to current I 2 in segment 2? 1. I 1 > I 2. 2. I 1 = I 2. 3. I 1 < I 2. 4. There’s not enough information to compare them. Quiz Question 2

8 Both segments of the wire are made of the same metal. Current I 1 flows into segment 1 from the left. How does current density J 1 in segment 1 compare to current density J 2 in segment 2? 1. J 1  J 2. 2. J 1  J 2. 3. J 1  J 2. 4. There’s not enough information to compare them. Quiz Question 3

9 Both segments of the wire are made of the same metal. Current I 1 flows into segment 1 from the left. How does the electric field E 1 in segment 1 compare to the electric field E 2 in segment 2? 1. E 1 > E 2. 2. E 1 = E 2 but not zero. 3. E 1 < E 2. 4. Both are zero because metal is a conductor. 5. There’s not enough information to compare them. Quiz Question 4

10 Define the resistivity… SI Units? 30.4: Conductivity and Resistivity

11 Define the resistivity… SI Units? 30.4: Conductivity and Resistivity

12 Define the resistivity… SI Units? 30.4: Conductivity and Resistivity

13 A 2.0 mm diameter aluminum wire carries a current of 800 mA. What is the E-field strength inside the wire? i.e. 30.6: The E-field in a wire

14 How is the current, I, related to the potential difference, ΔV in a wire? 30.5: Resistance and Ohm’s Law

15 How is the current, I, related to the potential difference, ΔV in a wire? SI Units of R? 30.5: Resistance and Ohm’s Law - Ohm’s Law

16 How is the current, I, related to the potential difference, ΔV in a wire? SI Units of R? 30.5: Resistance and Ohm’s Law - Ohm’s Law

17 Notice:  Resistivity, ρ, describes just the material.  Resistance, R, characterizes a specific piece of the conductor with a specific geometry. 30.5: Resistance and Ohm’s Law Resistance of a wire

18 Wire 2 has twice the length and twice the diameter of wire 1. What is the ratio R 2 /R 1 of their resistances? 1. 1/4. 2. 1/2. 3. 1. 4. 2. 5. 4. Quiz Question 5

19  A battery is a source of potential difference  V bat.  The battery creates a potential difference  V wire =  V bat between the ends of the wire.  The potential difference in the wire  V wire generates an E-field in the wire.  The E-field establishes a current I = JA = σAE in the wire.  The current in the wire is determined jointly by the battery and the wire’s resistance, R to be:  I   V wire /R Resistance and Ohm’s Law


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