BME211HW7 - Homework#7 Solutions P8.3-15 At steady-state...

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Homework #7 Solutions P8.3-15 At steady-state, immediately before t = 0: ( 29 10 12 0 0.1 A 10 40 16 40||10 i = = �� + + After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be 2 2 20 1 so 0.3 A, 40 , s 40 2 Finally: ( ) (0.1 0.3) 0.3 0.3 0.2 A sc t t t t L I R R i t e e τ - - = = = = = = - + = - P8.3-16 At steady-state, immediately before t = 0 After t = 0, we have: 6 so 12 V, 200 , (200)(20 10 ) 4 ms oc t t V R R C τ - = = = = = 0.004 Finally: ( ) (12 12) 12 12 V t v t e - = - + =
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P8.6-8 For t < 0, the circuit is: After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get: ( 29 ( 29 ( 29 ( 29 / 4000 0.00005 5 15 6 15 15 9 V t c t v t e e - - = - + - - - = - + P8.6-9 The input changes abruptly at time t = 0. The voltage v ( t ) may not be continuous at t = 0, but the capacitor voltage, v C ( t ) will be continuous. We will find v C ( t ) first and then use KVL to find v ( t ). The circuit will be at steady state before t = 0 so the capacitor will act like an open circuit. ( 29 ( 29 5 0 0 7 4.375 V 5 3 v v + = - = = +
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After t = 0, we replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit. ( 29 ( 29 oc 5 5 7 14 7 4.375 V 8 8 v = - = - = - t 5 3 1.875 R = = P The time constant is t 0.8625 R C s τ = = 1 1 1.16 s τ = So ( 29 ( 29 ( 29 1.16 1.16 C 4.375 4.375 4.375 4.375 8.75 V for 0 t t v t e e t - - = - - + - = - +
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