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ENGR5001 / Week 3
ENGR201 2024 A2(a)-(b)Past paperOld spec ENGR2012:213 marks16 min

ENGR201 Engineering Analysis 2024 A2 (a)-(b)[PARTIAL]

Figure A2-1 shows a typical RLC circuit. After applying Kirchhoff's law, the charge q(t)q(t) can be solved from the differential equation

2d2q(t)dt2+16dq(t)dt+50q(t)=300(A1)2\frac{d^2q(t)}{dt^2} + 16\frac{dq(t)}{dt} + 50q(t) = 300 \qquad \text{(A1)}

At t=0t = 0 the charge q(0)q(0) on the capacitor is zero and the current I(0)=dq(0)dtI(0) = \dfrac{dq(0)}{dt} of the circuit is also zero.

(Parts (c) and (d) of the original question, Newton iteration for the time of maximum current, are not on the ENGR5001 syllabus and are left out.)

Figure A2-1: series RLC circuit driven by a voltage source V(t), current I(t).
Figure A2-1: series RLC circuit driven by a voltage source V(t), current I(t).
Formulas you may need
  • L[q′]=sQ−q(0)\mathcal{L}[q'] = sQ - q(0), L[q′′]=s2Q−sq(0)−q′(0)\mathcal{L}[q''] = s^2Q - sq(0) - q'(0); constant c→c/sc \to c/s (learn this; printed in the ENGR201 Laplace table)
  • eatcos⁡ωt↔s−a(s−a)2+ω2e^{at}\cos\omega t \leftrightarrow \dfrac{s - a}{(s - a)^2 + \omega^2}, eatsin⁡ωt↔ω(s−a)2+ω2e^{at}\sin\omega t \leftrightarrow \dfrac{\omega}{(s - a)^2 + \omega^2} (learn this; printed in the ENGR201 Laplace table)
  • Current I=dqdtI = \dfrac{dq}{dt}, so I(s)=sQ(s)−q(0)I(s) = sQ(s) - q(0) (learn this)
  1. (a)
    Use the Laplace transform to find the charge Q(s)Q(s) in the s-domain.
    [8]Third
  2. (b)
    Use the inverse Laplace transform to find the charge q(t)q(t) and the current I(t)I(t) in the time domain.
    [5]