Question

Exercise 3 An RLC circuit is made of a resistor, an inductor and a capacitor connected in series to a battery. The current I(t) in such a circuit satisfies the ODE LI"(t) + RI (1) + (t) = G(t) where L is the inductance (unit: henrys (H)), R is the resistance (unit: ohms (N2), C is the capacitance (unit: farads (F)), and G is the forcing term generated by an AC power (G is actually the derivative with respect to time of the AC voltage). The current is measured in amperes (A), and the time t in seconds (S). Analysing the units of each terms in the ODE, give a relation between H, s and N, and another relation between S2, F and s. In the last two questions, we take L = 1H, R= 21 and C=1/2F. Find all the solutions to the homogeneous equation. We now apply a forcing term G(t) = Go cos(t) + Ge- cos(t) (in Vs-, with v=volts). Without actually computing its exact coefficients, give the general form of all the solutions to the RLC equation.

Answer 1

I't) + RI'(t) + LI (t) = G(+) @ units of Git) us volts per second = V/s So the units of LI' (t) ,R I'(t), Į I (t) must be the same Unit of current Ilt) is Ampere = A unit of I'(t) is Ampere per second = Als Unit of I"(t) is Ampere per second = Als? unit of LI'lt) = vls H Als" - v/s HALS=v unit of LI"(t) = unit of RI' (t) HAIS = 2 Als Hls=2] or [H=25] Unit of RI' (t) = unit of - Ilt). a. Als = A F_2=5

2 L=1H, R=2-2, C= ₂ F I"(t) + 2 I'(+) + 2 I (t) = 6(t) ♡ For homogeneous equation G(t) = 0 I"(t) + 2 1'(t) +2 I (t) =0 Auxiallary equation is mp+2m +2=0 roots mi, m2 = -2 m =-1 ti m2 =-1-1 I are complex, hence the general homogeneous solution is i(t) = c et cos (t) + C et sin (t) . for any arbitrary Constants G, C2 © I"(t) + 2 I'(t) +2 I (t) = Go Ces (t) + G, ēt cos (t) Let us solving this wing Laplace teansforms. Let Y(s) be the Laplace transform of I(t) applying Laplace tlansforms on both sides of differential equation & V s'y(s) + 2 sy(s)+ 2y(s) = Y _ Gos + Gis +1 * CS+)2+1

Gis Y(s) = Gos (5+1) (sustolt (5+1) (54+2 5+2) . (54+25+2) Breaking into partial fractions Y(s) = Gols+2) 5(s'+1) - (+4) Got D 5 (84+2 5+2) GIS (S'+2 5+2) taking universe Laplace teans felm i&t) = Go Gos (t) + 260 Sin{t) - Go e Cest) – 3 Go ēt sún (t) 5 it) = Go (1-2) Cos (4) + Go Sin (t) (2-32+) to +Gil et Cos(7) e (4-7) Sin (6-7) 27. (By convolution formula) it) = Gap (1et) 68 (1) + Cog Sun It) (2–3et) 0 +Gre | Cos (7) Sin (t-1) dT.

Cos (7) Sin (t-1) dy = t sin (t) – Cos (t) + + sin(t) + 08 (4) = Not t sin (t). Cos (7) Sin (t-r) dy 11 (sin (t) + Sin (+-27)) dr oot = 1 Sin (t) t + 1 Cos (1-27) [. ł (-2) To = t sin (t) - < (cos (-+)-68 (t)) = t sin (t) i(t) = 60 (1) -et) 6s (4) + Go Sin (t) (2-3ēt) + Cutet sin (4) Total solution is (t) = ci et cos (b) +cz ēt Sun (t) + Cho (1-25) Gas (4) & Go Sin (t) (2-3et) + Git et sin (t).