-200Acos(10t)-200Bsin(10t)-240Asin(10t)+240Bcos(10t)+200Acos(10t)+200Bsin(10t)=12sin(10t)
do that step again, you did your math wrong. You should get A=-1/20 and B=0
A series circuit contains a resistor with R = 24 ohms, 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.
(a) Find the charge at time t.
2Q"+24Q'+200Q=12sin(10t)
2r^2+24r+200=0, r=-6 +/i 8*i
Qc(t)=e^-6t*[c1*cos(8t)+c2*sin(8t)
Qp(t)=Acos(10t)+Bsin(10t)
Q'p(t)=-10Asin(10t)+10Bcos(10t)
Q"p(t)=-100Acos(10t)-100Bsin(10t)
-200Acos(10t)-200Bsin(10t)-240Asin(10t)+240Bcos(10t)+200Acos(10t)+200Bsin(10t)=12sin(10t)
240B=12, -240A=0, B=1/20, A=0 ???
This seems unlikely, yes?
Qp(t)=(1/20)sin(10t)
Q(t)=e^-6t[c1*cos(8t)+c2*sin(8t)]+(1/20)sin(10t)
Q(0)=0, c1=0
I=dQ/dt=e^-6t[(-6*c1+8*c2)cos(8t)+(-8*c1-6*c2)sin(8t)]+(1/2)cos(10t)
I(0)=0, 8*c2+1/2=0, c2=-1/16
Q(t)=e^-6t((-1/16)sin(8t))+(1/20)sin(10t)
which is incorrect.
Please help me find my error so that I can find the correct answer for Q(t) and I(t). Thanks.
1 answer
0 answers