Sunday
February 7, 2016

# Homework Help: Math

Posted by COFFEE on Tuesday, July 10, 2007 at 8:05pm.

A series circuit contains a resistor with R = 24 , an inductor with L = 2 H, a capacitor with C = 0.005 F, and a generator producing a voltage of E(t) = 12 sin(10t). The initial charge is Q = 0.001 C and the initial current is 0. Find the charge at time t.
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Q" + 24 Q' + 200 Q = 12 sin (10t)
r^2 + 24r + 200 = 0
r = -12 +/- 2sqrt(14)*i

Qc(t) = e^-12*(c1*cos(2sqrt(14)t)+c2*sin(2sqrt(14)t))
Qp(t) = A*cos(10t) + B*sin(10t)
Q'p(t) = -10A*sin(10t) + 10B*cos(10t)
Q"p(t) = -100A*cos(10t) - 100B*sin(10t)

(-100A*cos(10t)-100B*sin(10t))+24(-10A*sin(10t)+10B*cos(10t))+200(A*cos(10t)+B*sin(10t))
(100A+240B)*cos(10t)+(100B+240A)*sin(10t)=12*sin(10t)
100A+240B=12 -> 25A + 60B = 3
-240A+100B=0 -> -48A + 20B = 0
B = 36/845, A = 3/169
Q(t)=e^(-12t)(c1*cos(2sqrt(14)t)+c2*sin(2sqrt(14)t))+(3/169)cos(10t)+(36/845)sin(10t)
c1 = -3/169

I(t)=e^(-12t)[(-12*c1+2sqrt(14)*c2)cos(2sqrt(14)t)+(-2sqrt(14)*c1-12*c2)sin(2sqrt(14)t))]
+(10/845)(15*cos(10t)+36*cos(10t))
I(0)=-12*c1+2sqrt(14)*c2+(360/845)=0
c2 = 108/(338sqrt(14))

Q(t)=e^(-12t)((-3/169)cos(2sqrt(14)t)+(108/(338sqrt(14))sin(2sqrt(14))+(1/845)(15cos(10t)+36sin(10t))

Also,
I(t)=e^(-12t)[(144/169)cos(2sqrt(14)t)-(1128/(338sqrt(14))sin(2sqrt(14)t)]

Is this correct? Thanks.

Substitute Q(t) in the differential equation to see if it is correct. The probability of making two mistakes that exactly cancel each other out is pretty small...

Ok, its incorrect. Where am I making a mistake?

Check if Qp satisfies the inhomogeneous equation and if Qc satisfies the homogeneous part.