R = 75 Ohms
L = 70 mH
C = 15 millifarads
t_end = 0.1 second
Voltage rating = 1.5 kV
1A. Assume that no initial energy is stored in the circuit when the switch, which is shorting the DC current source, is opened. The problem is to find the time-domain expression for v(t) when t > 0 and determine if the voltage rating threshold of the capacitor has been exceeded when the DC current source is first switched on.
v(t) = I * u(t) (Equation 1)
Notes:
- u(t) is the unit step function defined as u(t) = 0, t < 0 and u(t) = 1, t > 0.
- Initial condition: The initial voltage across the circuit is v(0) = 0 because the initial energy stored in the circuit is zero.
- The DC current value is I = 1 A.
- Use the specific values for R, L, C, and the capacitor's voltage rating threshold assigned to you. You can introduce those values right away in part (a) or later.
(a) Using Laplace transforms, transform the integrodifferential equation (Equation 1) into an algebraic equation in the s-domain, where V(s) is the unknown variable. (10 points)
(b) Solve the algebraic equation for the unknown V(s). The final result will be a rational function. (10 points)
(c) Find the partial fraction expansion of V(s). Find the roots of the polynomial in the denominator. (5 points)
II. Solve for the unknown coefficients for each partial fraction factor. (22 points)
(d) Take the inverse Laplace transform of the partial fraction expansion of V(s) to find the time-domain expression for the voltage response: v(t) = 1/V(s). (10 points)
