Question

1) (40 pts total) Solving and order ODE using Laplace Transforms: Consider a series RLC circuit with resistor R, inductor L, and a capacitor C in series. The same current i(t) flows through R, L, and C. The voltage source v(t) is removed at t=0, but current continues to flow through the circuit for some time. We wish to find the natural response of this series RLC circuit, and find an equation for i(t). Using KVL and differentiating the equation to remove the integral from the voltage across the capacitor gives the following 2nd-order ODE: d'i(t) Rdi(t) 1 - +- +- (t) = 0 de? dt LC Use Laplace transforms to solve the ODE by the following steps below: a) (5 pts) Let R=2200, L=500mH, and C=500pF. Rewrite the 2nd-order ODE using these values b) (5 pts) Convert the 2nd order ODE to the s-domain using the Laplace transform. Use the initial conditions: i(0) - 0.95A = 950mA, :'(0) - --20A/S c) (5 pts) Write an equation for 1(s), with the denominator being a polynomial in s d) (5 pts) Factor the denominator so that it is the product of two 1 degree terms in s e) (5 pts) Use partial fraction expansion on I(s). f) (5 pts) Solve for i(t) by converting back to the time-domain. Use the Laplace key slides in BBLearn to assist you in this process. Remember to use the properties of linearity, time-shifting, and s-domain shifting as needed to assist you in this conversion 9) (5 pts) Verify that (0) = 0.95A and that i'(0) = 40 -20 A/S h) (5 pts) Find tast + Does your answer make sense?

Answer 1

2nd order ODER dzict), § dilt) + teilt)=0 (as R=220 2 L=500m H, C2 500MF. E = 220 - 440. ic - 300 XLOS X5008 16 6) 4000 ... The 2nd ordes ODE becomes- 42964 + 440 ditt 74000 ft)-01-0 (6) Applying Laplace transform on both sides of egp 0), be get ilo Tisz IB-s169-ri'(o)] +440 [SI(s) - hits 440001/8)=0 (86+4408 74000) IT)- 440 2(0)+31(0)+i (0) =) (8174408+4000)?(s) - (440%+s) 0.95)-20 -) (8 +440574000)I(s) =(0.955+ 398). © I() = 6037470394000) 20.95 5. 4398 8 +440 8 7 4000 (d(3 +4400+4000) == (s+9.287) (6+430.743)

e let, Ils) = (0,953 +398) K Sa (+9.287) (84 480.45) 649.287) * 54980.135 K = line 28 (8 + 9:287) I (5) 0.951-9.287) +398 19.2871430.713) Kz = linu -0.9235 (st 430.713) ICS) so E 430.713) 0.95(-430-713)+398 (-430.71349.284) = 0.02652. 0,02652 I(5)= 0.9235 F879.287) + (8 + 430.713) using inverse Laplace Transform, we get- Tilt = 6.9285 29:2874 +0.02652 € 430.7135 )ulty gence in FJ iro)=(0.9235 +0.02652)-0.95002 A i' (t) = 70.9205x69.287) 9.2870 +0,02652 (-430.743) 2430.7137 (0) - 0,9235X(-9,287) +0.02652(-430.713) 5- 20 A/s. [verifred]. al (9) as too, itt: Lini(t) =( makes sense as at t This answer capacitor will act ie at steady-state, the the current in the like an open ckt and RLC series ckt will become zero,