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

2. 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

3. 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

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

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

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

7. 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

8. 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

9. 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)

10. 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

11. Resistivity of materials

12. 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

13. 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

14. 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

15. Power Work done/ loss of PE for (+) going with field 𝑊=−∆𝑃𝐸=𝑄𝑉
No ½ because voltage is constant
𝑃𝑜𝑤𝑒𝑟= ∆𝑃𝐸 ∆𝑡 = 𝑞𝑉 ∆𝑡 = 𝑞 𝑡𝑖𝑚𝑒 𝑉=𝐼𝑉
Alternate forms
P=IV = I2R = V2/R V +

16. 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

17. 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

18. Power examples
Problems 31, 32, 33, 38,

19. 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

20. 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

21. 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

22. 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!