Using techniques that you have learned in the class (for example, Laplace transf

Question

Using techniques that you have learned in the class (for example, Laplace transform, partial fraction expansion) derive an expression for V(s) the Laplace transform of v(t). Do not assume the initial conditions are zero. The initial conditions in the previous circuit are: v(0-) = 0 and ic(0-) - k. What is the value of d/dtv(0-)? Hint: If you can not determine the value of this, take it equal to k2 and move on the next part. Part C. Now take: C = 7 L = 7 R = 1/2. (1) Using techniques that you have learned in the class (for example, Laplace transform, partial fraction expansion) derive an expression for v(t) the current in the circuit. Use the initial condition stated in the previous part and the result of the previous part.

Explanation / Answer

(A)

V(s) = IR(s)*R = IL(s)*sL - L*iL(0) = IC(s)*(1/sC) + vc(0)/s

Also,

IR(s) + IL(s) + IC(s) = 0

=>

V(s)/R + [ V(s) + L*iL(0) ] / sL + [ V(s) - vc(0)/s ] * sC = 0

V(s)*(1/R + 1/sL + sC) = (C*vc(0) - iL(0)/s)

V(s) = RL[ Cvc(0)*s - iL(0) ] / [ RLC*s^2 + L*s + R ]

(B)

ic = C*dvc/dt

=> ic(0-) = C*dvc/dt (0-)

dvc/dt (0-) = k/C

(C)

since vc(0-) = 0

=> iR(0-) = 0

=> iL(0-) = -ic(0-) = -k

V(s) = RL[ Cvc(0)*s - iL(0) ] / [ RLC*s^2 + L*s + R ]

V(s) = (3.5)*k / (24.5*s^2 + 7*s + 0.5)

V(s) = 7k/(49s^2 + 14s + 1)

V(s) = 7k/(7s+1)^2

V(s) = (k/7)/(s + 1/7)^2

=>

v(t) = (k/7)*t*e^(-t/7)*u(t)

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