Natural and step response RLC find iL(t)

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SUMMARY

The discussion focuses on solving the circuit problem involving an RLC circuit to find the inductor current iL(t) for t > 0 after a switch closure at t=0. The circuit parameters include a 4A current source, 2 kΩ resistor, 62.5 H inductor, and 2.5 µF capacitor, all in parallel. The solution derives the expression iL(t) = 4 - 12e^(-40t) + 3e^(-160t), with initial conditions iL(0-) = -5mA and iL(∞) = 4mA, using characteristic equations and Laplace transforms.

PREREQUISITES
  • Understanding of RLC circuit analysis
  • Familiarity with Laplace transforms
  • Knowledge of differential equations
  • Ability to interpret circuit diagrams
NEXT STEPS
  • Study RLC circuit transient response analysis
  • Learn about Laplace transform techniques for circuit analysis
  • Explore the use of characteristic equations in solving differential equations
  • Review initial and final value theorems in circuit theory
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing transient responses in RLC circuits will benefit from this discussion.

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Homework Statement




The switch in the circuit in the figure has been open a long time before closing at t=0.
Find iL(t) t>0
Express your answer in terms of t, where t is in milliseconds.

https://www.flickr.com/photos/84781786@N03/8899251427/

The Attempt at a Solution



We simplify to circuit to get

4A current , 2kohms,62.5H,2.5uF (All in parallel with each other).

iL(0-)= -15/3000 =-5mA ,
Vc(0-)=0,
iL(∞)= 4mA,

wo^2= 1/LC=10^6/62.5*.5 =6400
wo=80 rad/s

a= 1/2RC =10^6/4000*2.5 =100

s1,s2=-100±√100^2-80^2=-100±60
s1=-40 , s2=-160

iL=If+A'1e^-40t + A'2e^-160t

iL(∞)=If=4mA
iL(0)=A'1+A'2+If=-5mA

so A'1+A'2=-9mA

diL/dt(0) =0=-40A1 - 160A'2

A'1=12mA A'2=3mA

iL=4-12e^-40t+3e^-160t t≥0
 

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Your diagram doesn't appear to match your problem statement.
 

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