20081ee10_1_EE10HW6SOL

20081ee10_1_EE10HW6SOL - 5-e-t 11.25e-2.25t =-105e-t...

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PSpice: * Creating circuit file "p5_35-SCHEMATIC1-sweep.sim.cir" ** WARNING: THIS AUTOMATICALLY GENERATED FILE MAY BE OVERWRITTEN BY
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SUBSEQUENT SIMULATIONS *Libraries: * Local Libraries : * From [PSPICE NETLIST] section of C:\Program Files\OrcadLite\PSpice\PSpice.ini file: .lib "nom.lib" *Analysis directives: .DC LIN PARAM RLOAD 1k 8k 100 .PROBE V(*) I(*) W(*) D(*) NOISE(*) .INC ".\p5_35-SCHEMATIC1.net" **** INCLUDING p5_35-SCHEMATIC1.net **** * source P5_35 R_R1 N00894 VP 47k R_R2 0 VN 1.5k E_U1 VOUT 0 VALUE {LIMIT(V(VP,VN)*1E6,-9,9)} R_RL VN VOUT {RLOAD} V_V1 N00894 0 3 R_RIN VP VN 500k .PARAM RLOAD=2k **** RESUMING p5_35-SCHEMATIC1-sweep.sim.cir **** .END JOB CONCLUDED TOTAL JOB TIME .08 di g /dt = -10e -t 5 (-10e -t ) + 40 (-e -t + 11.25e -2.25t ) + 90 (e -t -5e -2.25t ) = 0 -90 e -t + 90e -t + 450 e -2.25t - 450 e -2.25t = 0 c) v 1 (t) = v L1 + v M (by KVL) = L 1 di g /dt + M di 2 /dt = 10(-10e -t
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Unformatted text preview: ) + 5 (-e-t + 11.25e-2.25t ) =-105e-t + 56.25e-2.25t d) Evaluate when t=0 v 1 (0) =-105e-t + 56.25e-2.25t =-105 + 56.25 =-48.75 V Notes: To solve the system, we needed to find the initial inductor current (the part on the left side of the switch), and the time constant (the part to the right). Once we have these, since it’s a first order system (there is only one component with a d/dt characteristic), we can apply the relevant exponential solution. To find the equivalent resistance connected to the inductor, after the switch is opened, we calculated the Thevinin resistance by applying a test current source. A test current source is easier than applying a voltage as we see the current divides into the 100 and 200 Ω resistors, then using i ∆ we work backwards to get v T , and R TH ....
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20081ee10_1_EE10HW6SOL - 5-e-t 11.25e-2.25t =-105e-t...

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