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.
-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
To solve the differential equation for Q(t), we need to find the complementary and particular solutions separately.
First, let's find the complementary solution, which represents the behavior of the system without the driving force:
2Q'' + 24Q' + 200Q = 0
The characteristic equation is:
2r^2 + 24r + 200 = 0
Solving this quadratic equation using the quadratic formula, we get:
r = (-24 ± √(24^2 - 4*2*200)) / (2*2)
r = (-24 ± √(576 - 1600)) / 4
r = (-24 ± √(-1024)) / 4
Since the discriminant is negative, the roots will be complex conjugates:
r = (-24 ± 32i) / 4
r = -6 ± 8i
The general form of the complementary solution is:
Qc(t) = e^(-6t)(c1*cos(8t) + c2*sin(8t))
Now, for the particular solution, which represents the effect of the driving force:
Qp(t) = Acos(10t) + Bsin(10t)
Q'p(t) = -10Asin(10t) + 10Bcos(10t)
Q''p(t) = -100Acos(10t) - 100Bsin(10t)
Substituting these derivatives into the differential equation, we have:
-200Acos(10t) - 200Bsin(10t) - 240Asin(10t) + 240Bcos(10t) + 200Acos(10t) + 200Bsin(10t) = 12sin(10t)
Simplifying, we get:
-240Asin(10t) + 240Bcos(10t) = 12sin(10t)
Comparing the coefficients of sin(10t), we get:
-240A = 12
A = -12/240
A = -1/20
Comparing the coefficients of cos(10t), we get:
240B = 0
B = 0
Therefore, the particular solution is:
Qp(t) = -1/20 * cos(10t)
The complete solution is the sum of the complementary and particular solutions:
Q(t) = Qc(t) + Qp(t)
Q(t) = e^(-6t)(c1*cos(8t) + c2*sin(8t)) - 1/20 * cos(10t)
To find the values of c1 and c2, we need initial conditions.
Given that the initial charge is Q = 0.001 C and the initial current is 0:
Q(0) = e^(-6*0)(c1*cos(8*0) + c2*sin(8*0)) - 1/20 * cos(10*0) = 0.001
c1 - 1/20 * 1 = 0.001
c1 - 1/20 = 0.001
c1 = 1/20 + 0.001
c1 = 0.0505
Differentiating Q(t) to find the current:
I(t) = dQ/dt
I(t) = -6e^(-6t)(c1*cos(8t) + c2*sin(8t)) - 1/2 * sin(10t)
Using the initial condition I(0) = 0:
0 = -6e^(-6*0)(c1*cos(8*0) + c2*sin(8*0)) - 1/2 * sin(10*0)
0 = -6e^0(c1 - 0) - 1/2 * 0
0 = -6c1
c1 = 0
Therefore, the corrected solutions for Q(t) and I(t) are:
Q(t) = e^(-6t)c2sin(8t) - 1/20 * cos(10t)
I(t) = -6e^(-6t)c2sin(8t) - 1/2 * sin(10t)
where c2 is an arbitrary constant.
To find the correct solution for Q(t) and I(t), let's break down the problem step by step:
First, let's write the equation for the charge at time t:
2Q" + 24Q' + 200Q = 12sin(10t)
To solve this differential equation, we'll first find the homogeneous solution by setting the right-hand side to zero:
2Qh" + 24Qh' + 200Qh = 0
The characteristic equation for this homogeneous equation is:
2r^2 + 24r + 200 = 0
Solving this quadratic equation, we find two complex conjugate roots: r = -6 +/- 8i
Therefore, the homogeneous solution is:
Qh(t) = e^(-6t)(c1*cos(8t) + c2*sin(8t))
Next, let's find the particular solution for the non-homogeneous equation. We assume a particular solution of the form:
Qp(t) = Acos(10t) + Bsin(10t)
Taking the derivatives of Qp(t), we find:
Qp'(t) = -10Asin(10t) + 10Bcos(10t)
Qp''(t) = -100Acos(10t) - 100Bsin(10t)
Substituting these derivatives into the differential equation, we get:
-200Acos(10t) - 200Bsin(10t) - 240Asin(10t) + 240Bcos(10t) + 200Acos(10t) + 200Bsin(10t) = 12sin(10t)
Simplifying the equation, we have:
40Bcos(10t) - 40Asin(10t) = 12sin(10t)
Equating the coefficients of the trigonometric functions, we find:
40B = 12
-40A = 0
Solving these equations, we find A = 0 and B = 3/10
Therefore, the particular solution is:
Qp(t) = (3/10)sin(10t)
Finally, the complete solution for Q(t) is the sum of the homogeneous and particular solutions:
Q(t) = Qh(t) + Qp(t)
= e^(-6t)(c1*cos(8t) + c2*sin(8t)) + (3/10)sin(10t)
To determine the values of c1 and c2, we'll use the initial conditions:
Q(0) = 0
I(0) = 0
From Q(0) = 0, we find c1 = 0.
Differentiating Q(t) with respect to t to find I(t):
I(t) = dQ(t)/dt
= -6e^(-6t)sin(8t) + 8e^(-6t)cos(8t) + (3/10)*cos(10t)
From I(0) = 0, we find c2 = -3/80.
Therefore, the correct solutions for Q(t) and I(t) are:
Q(t) = e^(-6t)*(3/10)*sin(10t)
I(t) = -6e^(-6t)sin(8t) + 8e^(-6t)cos(8t) + (3/10)*cos(10t)
Note that sometimes errors can occur when solving differential equations and it's important to double-check the calculations and assumptions made during the process.