ECE 231 -7

ECE 231 -7 - ECE-231 Circuits and Systems I Fall 2011...

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ECE-231 Circuits and Systems I Fall 2011 Session 7 Dr. Stewart Personick Office: ECEC Room 331 sdp2@verizon.net
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Review of the last part of the Session 6 notes
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Network component: Capacitor + - Rs (ohms) a b At t=0, Vab(t)= 0V (there is no charge on either plate of the capacitor) When the switch is closed (at t=0), current starts to flow Vab(t) + RsC x d[Vab(t)]/dt - V = 0 ; for t>0 seconds i(t) Vab(t ) = V [1 - e -(t/ R s C ) ] For t> 0 seconds RsC = τ , where τ (seconds) is called the “time constant of the circuit” t V Vs= V C (farads)
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Simulation (see posted file: RC circuit.asc) Vs = 18V, Rs=20 Ohms, and C= 0.015F Rs x C = 0.3 seconds n003
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RsC = 20 ohms x 0.15 farads = 0.3 seconds 1.0s = 0.1s + 3RsC 18V x [1- exp(-3)] = 18V x 0.9502 =17.10V The switch closes at t= 0.1s
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The current i(t), at t= 0.1+ seconds = 18V/20 Ohms = 900mA
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Network component: Capacitor + - Vs= V - e - e - e - e e e e e - + The energy stored in the capacitor, at any time t , = ½ x C x [Vab(t)]2 (Joules) b i(t) = 0 Vab= V t>> RsC Rs
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New materials
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Network component: Capacitor + - Vs= V Rs Comparing to an open circuit: The capacitance of an ideal open circuit is 0 farads The capacitance of a real open circuit is typically << 1 picofarad In most cases (but not all cases) we can consider an open circuit to have a capacitance of 0 farads… and still predict the correct behavior of the actual circuit a b
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Network component: Capacitor + - Vs= V i(t) = 0 - + Comparing to an open circuit: The capacitance of an open circuit is ~ 0 farads When the switch is closed, current flows for ~ 0 seconds The total charge stored in the open circuit is ~ 0 Coulombs The stored energy in the open circuit is ~ 0 Joules V a b Rs
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Network component: Capacitor + - Vs= 250V a b Let’s assume that, for t<0 seconds (when the switch is open) the capacitor is charged to 100 Volts . I.e. there is charge on each plate of the capacitor before the switch is closed Vab = 100 Volts; for t<0 seconds t<0 seconds C (farads) Rs
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Network component: Capacitor + - a b (KVL): 250V = Rs x i(t) + Vab(t) Vab(t) = Q(t)/C Therefore: 250V= Rs x i(t) + Q(t)/C; for t>0 seconds i(t) - + Vab Vs= 250V t>0 seconds C (farads) Rs
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Network component: Capacitor + - Vs= 250V i(t) - e - e - e - e e e e e electrons i(t) = d[Q(t)]/dt = -1.6 x 10-19 x the number of electrons flowing (counterclockwise) per second through any node in the mesh Q(t)= C x Vab(t) Therefore: i(t) = C d[Vab(t)]/dt a b electrons C (farads) Rs
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Network component: Capacitor + - a b KVL: Rs x i(t) + Q(t)/C -250V = 0 also: i(t) = d[Q(t)]/dt = C x d[Vab(t)]/dt ; and Q(t) /C = Vab(t) Therefore: Rs d[Q(t)]/dt + Q(t)/C - 250V = 0; for t>0 seconds Rs C x d[Vab(t)]/dt + Vab(t) - 250V = 0; for t>0 seconds and Vab (0+) = 100V i(t) Vs= 250V t>0 seconds - + Vab C (farads) Rs
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Network component: Capacitor + - a b At t=0+, Vab(t) = 100V When the switch is closed (at t=0s), current starts to flow Vab(t) + Rs C x d[Vab(t)]/dt - 250V = 0; for t>0 seconds i(t) Vab(t ) = 100V + (250V -100V) [1 - e -(t/ R s C ) ] RsC = τ , where τ (seconds) is called the “time constant of the circuit” t 250 V Vs= 250V C 100 V Rs
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Network component: Capacitor + - a b i(t) Vab(t ) = 100V e -(t/ R s C ) + (250V) [1 - e -(t/ R s C ) ] RsC = τ , where τ (seconds) is called the “time constant of the circuit” t 250 V Vs= 250V C 100 V Rs At t=0+, Vab(t) = 100V When the switch is closed (at t=0s), current starts to flow Vab(t) + Rs C x d[Vab(t)]/dt - 250V = 0; for t>0 seconds
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Network component: Capacitor + - Vs=
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This note was uploaded on 03/01/2012 for the course ECE 231 taught by Professor Pietrucha during the Spring '08 term at NJIT.

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ECE 231 -7 - ECE-231 Circuits and Systems I Fall 2011...

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