Lesson+7 - Whats really happening in class during those...

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Lesson 7 1 What’s really happening in class during those laptop moments.
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RLC Circuits Ch. 8.1-2 Lesson 7 Lesson 7 2
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V R L Lesson 7 Challenge 6 For V =R=L=1, and 0 =1, what is the phase difference between v (t) and i (t)? What lags or leads what? V L (t) RL Circuit v(t) =cos(t)  0 =1  r/s i(t) 3
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Determine the voltage across the resistor (i.e., v R (t)) in phasor form 1: Phasor V =10 @ 0 =1.0 r/s. Find the series current i L (t) 2: Determine the circuit ‘s impedances. R =10 ; Z L =jX L = j 0 L = j1=190 Z=R + jX L 1.41445 ) Lesson 7 V R L V L (t) RL Circuit v(t) =cos(t)  0 =1  r/s i(t) 4
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3: Express i L (t) as a phasor: I L = V / Z ckt = V /( R + jX L ) (Ohm’s Law) =10 /(   10 +   190 ) =10 /(   1.41445 ) = 0.707-45 (series current) Lesson 7 V R L V L (t) RL Circuit v(t) =cos(t)  0 =1  r/s i(t) 5
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4: From the complex Ohm’s Law: V R =R* I L =(10 )(0.707-45 ) = 0.707 -45 (phasor) Therefore v R (t) = 0.707 cos(t-45 ) Lesson 7 V R L V L (t) RL Circuit v(t) =cos(t)  0 =1  r/s i(t) 6
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5: From complex Ohm’s Law: V L =Z L * I L =(190 )(0.707-45 ) = 0.70745 (phasor form) Therefore v L (t)= 0.707 cos(t+45 ) Lesson 7 V R L V L (t) RL Circuit v(t) =cos(t)  0 =1  r/s i(t) 7
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Lesson 7 V L  leads  V R  by 90 V L  leads  V  by 45 V  leads  V R  by 45 V R L V L (t) RL Circuit v(t) =cos(t)  0 =1  r/s i(t) V R  and I in-phase. I R 8
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Lesson 7 V  leads I by 45 9 What is the phase difference between v (t) and i (t)? I = I R = I L = 0.707 -45 V = 1 0 ELI
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Lesson 7 Lesson 7 RLC 2 nd order systems. 10
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Lesson 7 Parallel RLC circuit (current source) 1 nodal equation in 1 unknown. Series RLC circuit (voltage source) 1 loop equation in 1 unknown. 11
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Lesson 7 Lesson 6 provides a model for analyzing the steady state behavior of a circuit (assume an input sinusoid at a specified frequency 0 ). You can use this theory to analyze the circuit shown below. Data: 0 = 3000r/s, R 1 = 1.5k , R 2 = 1k , L = 1/3H, C = 1/6 F Conduct a study using PSPICE.. V=-40cos(3000t) R=1.5k R=1k L=1/3 C=1/6u 12
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Lesson 7 V=-40cos(3000t) R=1.5k R=1k L=1/3 C=1/6u PSPICE: PSPICE sinusoidal generator produces only a sinewave: s(t)= VAMPL sin(2*p* FREQ + PHASE ), To create v(t) = -40 cos(3000t), use the following parameters: PHASE = -90, VAMPL = 40, and FREQ = 447.5. This will result in v(t)= 40 sin(3000t-90 ) [PSPICE] = -40 cos(3000t) 13
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Simulation Lesson 7 i(t) v(t) Outcome: Data suggests a  circuit-level   ELI   relationship.   What does an  ELI claim mean  in this instance? ~30-40 14
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Determine the series current  i(t)  (only).
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