Lecture30 - Chapter 6

B sketch thevoltage current power and energy as

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b) Sketch thevoltage, current, power and energy as functions of time. c) Specify intervals when energy is being stored in/delivered by capacitor. Thevoltage pulseapplied across theterminals of a 0.5 F capacitor is μ v(t) = 0 t 0 v(t) = 4t volts 0 t 1 ≤ ≤ v(t) = 4e-(t-1) volts 1 t ≤ ≤∞
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1111 Example 1 (2/3) a) v(t) = 0 V t 0 v(t) = 4t V 0 t 1 ≤ ≤ v(t) = 4e-(t-1) V 1 t ≤ ≤∞ i = Cdv/dt = 0 A t 0 i = Cdv/dt = 4C = 2 A μ 0 t 1 ≤ ≤ i = Cdv/dt = -4e-(t-1)C = -2e-(t-1) A μ 1
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1212 Example 1 (3/3) -10.000 -8.000 -6.000 -4.000 -2.000 0.000 2.000 4.000 6.000 8.000 10.000 0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 time t v i p w b) v, i, p and w curves areplotted as functions of time c) Energy stored 0 t 1. Energy delivered 1 t ≤ ≤ ≤ ≤∞
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1313 Example 2 (1/4) An uncharged 0.2 F capacitor is driven by a triangular current pulse, μ described by: i(t) = 0 t 0 i(t) = 5000t 0 t 20 s ≤ ≤ μ i(t) = 0.2 – 5000t 20 t 40 s ≤ ≤ μ i(t) = 0 t 40 s μ a) Derivetheexpression for the capacitor voltage, power and energy for each of thefour timeintervals needed to describe thecurrent b) Plot i, v, p and w versus t Initial voltage (Initial Condition) (IC)
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1414 Example 2 (2/4) a) For t 0: v, p and w are all zero For 0 t 20 s: ≤ ≤ μ 2 t 0 9 6 t 10 x 5 . 12 0 τ d τ 5000 10 x 5 v = + = ( 29 3 12 2 9 t 10 x 5 . 62 ) t 5000 ( t 10 x 5 . 12 vi p = = = ( 29 4 12 2 2 9 6 2 t 10 x 625 . 15 t 10 x 5 . 12 ) 10 x 2 . 0 ( 2 1 Cv 2 1 w = = = - t t i 5000 ) ( =
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1515 Example 2 (3/4) For 20 t 40 s: ≤ ≤ μ ( 29 ) t 5000 2 . 0 ( 10 t 10 x 5 . 12 t 10 vi p 2 9 6 - - - = = 5 6 2 6 3 9 4 12 10 t 10 x 2 t 10 x 123 . 0 t 10 x 5 . 2 t 10 x 625 . 15 w - - + - + - = 10 t 10 x 5 . 12 t 10 5 d ) 5000 2 . 0 ( 10 x 5 v 2 t
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