Math  /  Calculus

QuestionProblem \#2 (30 points) In this circuit, the switch has been open for a long time and at t=0\mathrm{t}=0 it closes. It is given that at t=0\mathrm{t}=0 - there is no energy stored in the inductor or the capacitor. a) Show that when the switch is closed, the current I( s)I(\mathrm{~s}) in the s-domain is given by: b) Perform an inverse Laplace transform and determine the current i(t)i(\mathrm{t}) in the time-domain for t0\mathrm{t} \geq 0. I(s)=5s2+s200+105I(s)=\frac{5}{s^{2}+s 200+10^{5}} c) Plot the current i(t)i(\mathrm{t}) from t=0\mathrm{t}=0 until a time at which the current reaches a steady state value. d) On the plot of part (c), indicate the transient and steady state periods of time. e) Using the known behavior of the inductor and capacitor explain the steady state value of the current. a)

Studdy Solution

STEP 1

1. The switch in the circuit has been open for a long time, implying the circuit was at steady state before t=0 t = 0 .
2. At t=0 t = 0 , the switch closes, and there is no initial energy stored in the inductor or capacitor.
3. We need to derive the expression for the current I(s) I(s) in the s-domain when the switch is closed.

STEP 2

1. Analyze the circuit in the s-domain.
2. Apply circuit laws to derive the expression for I(s) I(s) .

STEP 3

Convert the circuit elements to their s-domain equivalents: - Resistors remain unchanged. - Inductors are represented as sL sL (where L L is the inductance). - Capacitors are represented as 1sC \frac{1}{sC} (where C C is the capacitance).

STEP 4

Apply Kirchhoff's laws (KVL or KCL) in the s-domain to form the equations governing the circuit.

STEP 5

Solve the equations for the current I(s) I(s) in terms of the given circuit parameters.

STEP 6

Simplify the expression to match the given form I(s)=5s2+s200+105 I(s) = \frac{5}{s^2 + s \cdot 200 + 10^5} .
The expression for the current I(s) I(s) in the s-domain is:
I(s)=5s2+s200+105 I(s) = \frac{5}{s^2 + s \cdot 200 + 10^5}

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