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Transcript
Conduction and Current • Polarization vs. Conduction • Batteries, Current, Resistance • Ohm’s Law and Examples • Resistivity and Examples • Power and Examples 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 • Storage • Conduction • Q = CV • V= IR (I=GV G=1/R) • Charge in Coulombs • Charge flow in Coulombs/second (amps) • Energy stored in Joules • 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= 𝑉 𝑅 = 240 𝑉 9.6 Ω = 25 𝐴 25 𝐴 = 25 𝐶 𝑆 𝐼= 9𝑉 1.6 Ω = 5.625 𝐴 ⋕ 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 = (25 𝐶 𝑠)(60 𝑠 min)(50 min) = 75,000 𝐶 (5.625 𝐶 𝑠) 60𝑠 = 337.5 𝐶 337.5 𝐶 1.6∙10−19 𝐶 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 = 2.11 ∙ 1021 Ohm’s Law Examples 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 Storage 𝐾𝜀𝑜 𝐴 𝐶= 𝑑 Conduction 𝜌𝑑 𝑅= 𝐴 𝜎𝐴 𝐺= 𝑑 • We use reciprocals, resistance and resistivity • ρ published for materials, like K. • High ρ poor conductor, σ good conductor (similar to K for storage) 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 • Radius of wire 𝐴 = 𝜋𝑟 2 𝑟= 𝐴 𝜋 = 1.04 𝑚𝑚 • Voltage Drop along wire 𝑉 = 𝐼𝑅 = 4𝐴 ∙ 0.1 Ω = 0.4 𝑉 = 3.4 ∙ 10−6 𝑚2 Resistivity examples 𝑅𝑎𝑙 = 𝜌𝑎𝑙 𝑑 𝐴 𝑅𝑐𝑢 = 𝜌𝑐𝑢 𝑑 𝐴 𝑅= 𝜌𝑑 𝐴 = = = 2.65∙10−8 Ω 𝑚 (10 𝑚) 3.14∙10−6 𝑚2 1.68∙10−8 Ω 𝑚 (20 𝑚) 4.91∙10−6 𝑚2 1.68∙10−8 Ω 𝑚 (26 𝑚) 2.08∙10−6 𝑚2 = 0.084 Ω = 0.068 Ω = 0. 21Ω 𝑉 = 𝐼𝑅 = 12 𝐴 ∙ 0.21 Ω = 2.52 𝑉 Resistivity example • Along x 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 = • Alternate forms P=IV = I2R = V2/R 𝑞 𝑡𝑖𝑚𝑒 + V 𝑉 = 𝐼𝑉 0 Power example • Calculate current 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= 𝑉2 𝑅 𝑅= 𝑉2 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 • 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 • 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 • 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!