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

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
PSpice: * Creating circuit file "p5_35-SCHEMATIC1-sweep.sim.cir" ** WARNING: THIS AUTOMATICALLY GENERATED FILE MAY BE OVERWRITTEN BY
Image of page 2
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
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern