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

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

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

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

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

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

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

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= 337.5 𝐶 1.6∙ 10 −19 𝐶 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 =2.11∙ 10 21

Ohm’s Law Examples OK, but do not touch the other wire! 𝑅= 2.5∙ 10 −5 Ω 𝑚 .04 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)

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

Resistivity of materials

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 2 𝑟= 𝐴 𝜋 =1.04 𝑚𝑚 Voltage Drop along wire 𝑉=𝐼𝑅=4𝐴 ∙0.1 Ω=0.4 𝑉

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

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

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 +

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 2 40 𝑉 𝐴 =3.6Ω

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

Power examples Problems 31, 32, 33, 38,

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

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

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

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!