hw2-sol

hw2-sol - 2 i(t = is(t ic(t ic(t = C d(Vc(t dt is(t = Vs R1...

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2. () sc it i t =+ 1 s s V R = ( () ) cc d it C Vt dt = 2 0 Vt itR + = , 2 ( ()) 0 d Vt C VtR dt += The eigenvalue of the solution is 2 1 CR λ =− , so the solution is 2 t t CR Ke Ke = , where K is a constant we will get from initial condition. Initial condition: The power switch is open before 0 t = , so it’s just a simple series circuit containing s V , 1 R , C and 2 R . The current of this series circuit is 0 and the capacitor store charges coming from s V . Positive charges go to the left side of the capacitor while negative charges go to its right. (0 ) (0 ) 10 s VV +− == = 2 18 (0) 10 t t CR c c V t Ke Ke 10 K = − , 18 10 t c e () 5 s = 18 5 ) 3 ( () ) 3 t ccc dd e dt dt = 18 5 3 t e =+= +
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13. (a) V : the volume with unit l ; () Ct : the total amount of drug concentration, with unit mg/l ; 0 k : the amount of drug enters the compartment per hour, with unit mg/hr ; e k : the percentage of total concentrated drug that runs out of the compartment per hour, with unit -1 hr . At some time point, drug enters the compartment is 0 k ; While total concentration amount is CtV , concentration created at this time point is (( )) d dt , with unit mg/hr ; Excretion is the percentage per hour e k multiply by the total concentrated amount . Then, Concentration = Infusion - Excretion 0 ) ) ( ) e V d k kCtV dt =− (b) Substitute with parameters 20 ) )8( ) 2 0 0 d dt += Homogeneous solution: 20 ) ) 0 d dt + = The eigenvalue of the solution is 8 0.4 20 λ = −= , so the homogeneous solution should be in the form of 1 0.4 h t Ct K e = , where 1 K is a constant. Particular solution: As the right side of the equation is 200, the particular
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solution should in a form of a constant 2 K .
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hw2-sol - 2 i(t = is(t ic(t ic(t = C d(Vc(t dt is(t = Vs R1...

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