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Lecture_18_Part1

# Lecture_18_Part1 - Macroscopic Analysis of Circuits...

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Microscopic treatment: insight into the fundamental physical mechanism of circuit behavior. Not easy to measure directly E , u , Q, v . It is easier to measure conventional current, potential difference macroscopic parameters Need a link between microscopic and macroscopic quantities. Macroscopic Analysis of Circuits

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Many elements in a circuit act as resistors: prevent current from rising above a certain value. Goal: find a simple parameter which can predict Δ V and I in such elements. Need to combine the properties of material and geometry. Resistance
Conventional current: nAuE q v nA q I = = Different properties of the material Geometry Group the material properties together: I = q nu ( ) AE J = I A = q nu ( ) E = σ E (A/m 2 ) Current density: Conductivity = m V A nu q σ Conductivity Combining the properties of a material

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The conductivity of tungsten at RT is σ =1.8 . 10 7 (A/m 2 )/(V/m) and it decreases 18 times at a temperature of a glowing filament (3000 K). The tungsten filament has a radius of 0.015 mm. What is E required to dive 0.3A through it? E A I J σ = = A I E σ = E = 0.3 A 1.8 × 10 7 (A/m 2 )/(V/m) / 18 ( ) 7.1 × 10 10 m 2 ( ) = 424 V/m Exercise
E Au n q v A n q I Cl 1 1 1 1 1 1 = = E Au n q v A n q I Na 2 2 2 2 2 2 = = E Au n q E Au n q I I I Cl Na 1 1 1 2 2 2 + = + = J = I A = q 2 n 2 u 2 + q 1 n 1 u 1 ( ) E = σ E 1 1 1 2 2 2 u n q u n q + = σ Conductivity with two Kinds of Charge Carriers

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= Δ f i l d E V EL V = Δ L V E Δ = E A I J σ = = AE I σ = R V V R V L A I Δ = Δ = Δ = 1 σ Conventional current: R V I Δ =
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