1. Topics Time Travel Antigravity Warp Drive Teleportation Flying Saucers
Alien Abductions
2016 April 1

2. Conduction and Current
Polarization vs. Conduction
Batteries, Current, Resistance
Ohm’s Law and Examples
Resistivity vs. Resistance
Power and Examples

3. Materials can do 2 things:
Electrical Properties of Materials
Materials can do 2 things:
Store charge
Initial alignment of charge with applied voltage
Charge proportional to voltage
Temporary short-range alignment
Conduct charge
Continuous flow of charge with applied voltage
Current proportional to voltage
Continuous long-range movement

4. Charge Storage vs Conduction
Q = CV
Charge in Coulombs
Energy stored in Joules
Conduction
V= IR (I=GV G=1/R)
Charge flow in Coulombs/second (amps)
Power created or expended in Watts

5. Batteries Battery Electrochemical Source of voltage
Positive and negative
1.5 volt, 3 volt, 9 volt, 12 volt
Circuit symbol

6. Current Current Coulombs/second = amperes I = ΔQ / Δt Example 18-1
Requires complete circuit
Circuit diagram
Positive vs. negative flow

7. Resistance Resist flow of current (regulate)
Atomic scale collisions dissipate energy
Energy appears in other forms (heat, light)
Applications
Model appliance behavior
Heater (collisions cause heat)
Regulate current/voltage on circuit board
Resistors and color code

8. Resistance and Ohm’s Law
Storage vs. Conduction
Q = CV (storage)
I = GV (conduction)
Current proportional to voltage
Proportionality is conductance
Use inverse relation
V = IR
Resistance
Units volts/amps = ohms
Ohm’s law
If V proportional to I, ohmic
Otherwise non-ohmic
Example 18-3

9. Ohm’s Law Examples 𝐼= 9 𝑉 1.6 Ω =5.625 𝐴 (5.625 𝐶 𝑠) 60𝑠 =337.5 𝐶
𝐼= 𝑉 𝑅 = 240 𝑉 9.6 Ω =25 𝐴
25 𝐴=25 𝐶 𝑆 (25 𝐶 𝑠)(60 𝑠 min⁡)(50 min⁡)=75,000 𝐶
𝐼= 9 𝑉 1.6 Ω =5.625 𝐴 (5.625 𝐶 𝑠) 60𝑠 =337.5 𝐶
⋕𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠= 𝐶 1.6∙ 10 −19 𝐶 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 =2.11∙ 10 21

10. Ohm’s Law Examples OK, but do not touch the other wire!
𝑅= 2.5∙ 10 −5 Ω 𝑚 𝑚 = 10 −6 Ω
𝑉=𝐼𝑅=2800 𝐴 ∙ 10 −6 𝑉 𝐴 =2.8 𝑚𝑉
OK, but do not touch the other wire!
(heat of vaporization of squirrel)

11. Resistivity vs. Resistance
Property of material vs. property of device
Similar to dielectric constant vs. capacitance
Becomes resistance vs. resistivity
We use reciprocals, resistance and resistivity
ρ published for materials, like K.
High ρ poor conductor, σ good conductor (similar to K for storage)
Storage
Conduction
𝐶= 𝐾 𝜀 𝑜 𝐴 𝑑
𝑅= 𝜌𝑑 𝐴 𝐺= 𝜎𝐴 𝑑

12. Resistivity of materials

13. Resistivity Example Area of wire from resistivity and length
𝑅= 𝜌𝑑 𝐴 𝐴= 𝑝𝑑 𝑅 = (1.68 ∙ 10 −8 Ω 𝑚)(20 𝑚) 0.1Ω =3.4∙ 10 −6 𝑚 2
Radius of wire
𝐴=𝜋 𝑟 𝑟= 𝐴 𝜋 =1.04 𝑚𝑚
Voltage Drop along wire
𝑉=𝐼𝑅=4𝐴 ∙0.1 Ω=0.4 𝑉

14. Resistivity examples
𝑅 𝑎𝑙 = 𝜌 𝑎𝑙 𝑑 𝐴 = 2.65∙ 10 −8 Ω 𝑚 (10 𝑚) 3.14∙ 10 −6 𝑚 2 =0.084 Ω
𝑅 𝑐𝑢 = 𝜌 𝑐𝑢 𝑑 𝐴 = 1.68∙ 10 −8 Ω 𝑚 (20 𝑚) 4.91∙ 10 −6 𝑚 2 =0.068 Ω
𝑅= 𝜌𝑑 𝐴 = 1.68∙ 10 −8 Ω 𝑚 (26 𝑚) 2.08∙ 10 −6 𝑚 2 =0. 21Ω
𝑉=𝐼𝑅=12 𝐴 ∙0.21 Ω=2.52 𝑉

15. Resistivity example Along x Along y Along z
𝑅= 𝜌𝑑 𝐴 = 3.0∙ 10 −8 Ω 𝑚 (.01 𝑚) .02 𝑚 (.04 𝑚) =3.8 ∙ 10 −4 Ω
Along y
𝑅= 𝜌𝑑 𝐴 = 3.0∙ 10 −8 Ω 𝑚 (.02 𝑚) .01 𝑚 (.04 𝑚) =1.5 ∙ 10 −3 Ω
Along z
𝑅= 𝜌𝑑 𝐴 = 3.0∙ 10 −8 Ω 𝑚 (.04 𝑚) .01 𝑚 (.02 𝑚) =6.0 ∙ 10 −4 Ω

16. Power - Created in Battery
Work done on + charge moving across V in battery
𝑊=𝑄 𝑉 𝑎𝑣𝑔 =𝑄𝑉
No “½” because voltage is constant
Since flow is continuous, interested in rate
𝑊𝑜𝑟𝑘 ∆𝑡𝑖𝑚𝑒 = 𝑃𝑜𝑤𝑒𝑟 = ∆𝑞 ∆𝑡 𝑉 = 𝐼𝑉
This is the work done on the charge by electrochemical action V +

17. Power - Consumed in Resistor
Work done by + charge moving across V in resistor
𝑊=𝑄 𝑉 𝑎𝑣𝑔 =𝑄𝑉
Where is the work going?
Collisions
Heat
Since flow is continuous, interested in rate
𝑊𝑜𝑟𝑘 ∆𝑡𝑖𝑚𝑒 = 𝑃𝑜𝑤𝑒𝑟 = ∆𝑞 ∆𝑡 𝑉 = 𝐼𝑉
Alternate forms
𝑃=𝐼𝑉= 𝐼 2 𝑅= 𝑉 2 𝑅

18. Power example Calculate current 𝑃=𝐼𝑉 𝐼= 𝑃 𝑉 = 40 𝑉 𝐴 12 𝑉 =3.33 𝐴
𝑃=𝐼𝑉 𝐼= 𝑃 𝑉 = 40 𝑉 𝐴 12 𝑉 =3.33 𝐴
Calculate Resistance
R= 𝑉 𝐼 = 12 𝑉 3.33 𝐴 =3.6 Ω
In one step
𝑃= 𝑉 2 𝑅 𝑅= 𝑉 2 𝑃 = 12 𝑉 𝑉 𝐴 =3.6Ω

19. Power example Power 𝑃=𝐼𝑉=15 𝐴 ∙120 𝑉=1800 𝑊=1.8 𝑘𝑊
Electric Company charges for energy not power
𝐸𝑛𝑒𝑟𝑔𝑦=𝑝𝑜𝑤𝑒𝑟 ∙𝑡𝑖𝑚𝑒
But instead of using Joules, they use kW-hours
$= 90 ℎ 1.8 𝑘𝑊 (.092 $ 𝑘𝑊∙ℎ)=$15

20. Power examples
Problems 33, 36, 37, 42,

21. Problem 33 950 W 1450 W 𝑃=𝐼𝑉 𝐼= 𝑃 𝑉 = 1450 𝑊 120 𝑉 =12.08 𝐴
𝑃=𝐼𝑉 𝐼= 𝑃 𝑉 = 950 𝑊 120 𝑉 =7.92 𝐴
𝑉=𝐼𝑅 𝑅= 𝑉 𝐼 = 120 𝑉 7.92 𝐴 =15.15 Ω
1450 W
𝑃=𝐼𝑉 𝐼= 𝑃 𝑉 = 𝑊 120 𝑉 =12.08 𝐴
𝑉=𝐼𝑅 𝑅= 𝑉 𝐼 = 120 𝑉 𝐴 =9.93 Ω
Higher Power – Lower resistance

22. Problem 36 In Europe In North America
𝑃=𝐼𝑉 𝐼= 𝑃 𝑉 = 75 𝑊 240 𝑉 = 𝐴
𝑉=𝐼𝑅 𝑅= 𝑉 𝐼 = 240 𝑉 𝐴 =768 Ω
In North America
𝑉=𝐼𝑅 𝐼= 𝑉 𝑅 = 120 𝑉 768 Ω = 𝐴
𝑃=𝐼𝑉= 𝐴 120 𝑉 =18.75 𝑊
Go the other way, bulb burns out!

23. Problem 37 Power and Energy
𝐽𝑜𝑢𝑙𝑒𝑠=𝑊𝑎𝑡𝑡𝑠∙𝑠𝑒𝑐𝑜𝑛𝑑𝑠 𝑒𝑛𝑒𝑟𝑔𝑦=𝑘𝑊∙ℎ𝑜𝑢𝑟𝑠 (𝑈𝑆 𝑢𝑛𝑖𝑡𝑠)
Time (hours)
𝑡𝑖𝑚𝑒= 5 𝑚𝑖𝑛 𝑑𝑎𝑦 × 4 𝑑𝑎𝑦 𝑤𝑒𝑒𝑘 × 4 𝑤𝑒𝑒𝑘 𝑚𝑜𝑛𝑡ℎ =80 𝑚𝑖𝑛=1.3 ℎ𝑜𝑢𝑟𝑠
Energy in kW-h
𝑒𝑛𝑒𝑟𝑔𝑦=550 𝑊∙1.3 ℎ𝑜𝑢𝑟𝑠=715 𝑊ℎ=0.715 𝑘𝑊ℎ
Cost
𝑐𝑜𝑠𝑡=9 𝑐𝑒𝑛𝑡𝑠 𝑘𝑊ℎ ∙0.715 𝑘𝑊ℎ=6.4 𝑐𝑒𝑛𝑡𝑠

24. Problem 42 Resistance of each wire
𝑅= 𝜌𝑙 𝐴 = 1.68 ∙ 10 −8 Ω 𝑚 ∙2.7 𝑚 𝜋 (0.645∙ 10 −3 𝑚) 2 =3.47∙ 10 −2 Ω
Power loss in each wire
𝑃=𝐼𝑉= 𝐼 2 𝑅= 18 𝐴 2 ∙3.47∙ 10 −2 Ω=11.24 Watts
Power loss in both wires 22.5 Watts

25. Power transmission
All customer cares about to run his home or factory: 𝑃=𝑉𝐼=620 𝑘𝑊
Can do low voltage at high current
High voltage at low current
Transformers can switch back and forth

26. Power transmission – 12,000 V 𝑉=𝐼𝑅=51.67 𝐴 ∙3Ω=155 𝑉
At 12,000 V current must be
𝐼= 620 𝑘𝑊 12 𝑘𝑉 =51.67 𝐴
Voltage drop along wire will be
𝑉=𝐼𝑅=51.67 𝐴 ∙3Ω=155 𝑉
Power wasted in wire will be
P=𝐼𝑉=51.67 𝐴 ∙155 𝑉=8 𝑘𝑊

27. Power transmission – 50,000 V 𝑉=𝐼𝑅=12.4 𝐴 ∙3Ω=37.2 𝑉
At 50,000 V current must be
𝐼= 620 𝑘𝑊 50 𝑘𝑉 =12.4 𝐴
Voltage drop along wire will be
𝑉=𝐼𝑅=12.4 𝐴 ∙3Ω=37.2 𝑉
Power wasted in wire will be
P=𝐼𝑉=12.4 𝐴 ∙37.2 𝑉=0.46 𝑘𝑊

28. Power transmission – 2,000 V 𝑉=𝐼𝑅=310 𝐴 ∙3Ω=930 𝑉
At 2,000 V current must be
𝐼= 620 𝑘𝑊 2 𝑘𝑉 =310 𝐴
Voltage drop along wire will be
𝑉=𝐼𝑅=310 𝐴 ∙3Ω=930 𝑉
Power wasted in wire will be
P=𝐼𝑉=310 𝐴 ∙930 𝑉=288 𝑘𝑊
<< Half the voltage and power are wasted!