Consider an electrical circuit comprising an inductor and a capacitor, with no e
ID: 1888776 • Letter: C
Question
Consider an electrical circuit comprising an inductor and a capacitor, with no external applied voltage, and no electrical resistance. The current, i, through the inductor is related to the charge, q, on the capacitor by the differential equation: where the inductance, L, and capacitance C, are positive constants.
L(di/dt)+q/c=0
(a)Given that the charge and the current are related through i=dq/dt derive a second order differential equation for q(t).
i know that the second order differential is
L(d2q/dt2)+q/c=0
please help me solve part b
b)assuming that the initial charge q(0)=q0 and the initial current i(0)=i0, show that
q(t)=q0cos(wt)+(i0/w)sin(wt)
where w^2=1/LC
Explanation / Answer
B) We have to solve the eq. Ld^2q/dt^2 + q/C = 0 LDq +q/C =0 (LD+1/C)q =0 here D = d2/dt2 it is a 2nd order differential eq. So we have to take A.E. = Lm^2 + 1/C =0 or m^2 = -1/LC or m = +i/sqrt(LC) or -i/sqrt(LC) so the solution is q = C1cos (t/sqrt(LC)) + C2 sin (t/sqrt(LC)) given w^2 = 1/LC or w = 1/sqrt(LC) substitute q =C1cos (wt) + C2 sin (wt) apply initial conditions given at t=0 , q(0)=q0 q0= C1cos (w0) + C2sin (w0) = C1 + C2*0 or C1 = q0 differentiate w.r.t t dq/dt = -wC1sinwt + wC2coswt we know dq/dt = i i = -wC1sinwt +wC2coswt apply initial conditinos t=0 , i(0)=i0 i0 = wC2 or C2 = i0/w so the final solutions is q=q0coswt+(i0/w)sinwt