Differential equation RLC
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I'm trying to find the current at $t=frac{pi}{4}$
The series RLC circuit has a voltage source given by $E(t)=sin(100t)$ with a capacitor of 2 F, a resistor of 0.02 ohms, and an inductor of 0.001H. If the initial current and the initial charge on the capacitor are both zero,
determine the current in the circuit for $t=frac{pi}{4}$
Based on the information given I have the equation as follows:
$$frac{d^2I}{dt^2}+20frac{dI}{dt}+500I=100000cos(100t)$$
so the homogeneous equation associated is: $r^2+20r+500=(r+10)^2+20^2=0$
whose roots are $-10 pm 20i$. Hence, the solution to the homogeneous equation is:
$$I_h=C_1e^{-10t}cos(20t)+C_2e^{-10t}sin(20t)$$
To find a particular solution, I use the method of undetermined coefficients.
$$I_p=-10.080 cos(100t)+2.122 sin(100t).$$
And after finding $C_1$ and $C_2$ I have:
$$I(t)=e^{-10t}(10.080 cos(100t) -5.570 sin(20t))-10.080cos(20t)+2.122sin(20t)$$
But $I(frac{pi}{4})=0.080111$, and the answer the professor give was 10.076. Am I doing something wrong?
differential-equations
edited Nov 23 at 14:30
Paul Sinclair
19.1k21441
asked Nov 23 at 1:06
Lexie Walker
1597
add a comment |
up vote
0
down vote
favorite
I'm trying to find the current at $t=frac{pi}{4}$
The series RLC circuit has a voltage source given by $E(t)=sin(100t)$ with a capacitor of 2 F, a resistor of 0.02 ohms, and an inductor of 0.001H. If the initial current and the initial charge on the capacitor are both zero,
determine the current in the circuit for $t=frac{pi}{4}$
Based on the information given I have the equation as follows:
$$frac{d^2I}{dt^2}+20frac{dI}{dt}+500I=100000cos(100t)$$
so the homogeneous equation associated is: $r^2+20r+500=(r+10)^2+20^2=0$
whose roots are $-10 pm 20i$. Hence, the solution to the homogeneous equation is:
$$I_h=C_1e^{-10t}cos(20t)+C_2e^{-10t}sin(20t)$$
To find a particular solution, I use the method of undetermined coefficients.
$$I_p=-10.080 cos(100t)+2.122 sin(100t).$$
And after finding $C_1$ and $C_2$ I have:
$$I(t)=e^{-10t}(10.080 cos(100t) -5.570 sin(20t))-10.080cos(20t)+2.122sin(20t)$$
But $I(frac{pi}{4})=0.080111$, and the answer the professor give was 10.076. Am I doing something wrong?
differential-equations
edited Nov 23 at 14:30
Paul Sinclair
19.1k21441
asked Nov 23 at 1:06
Lexie Walker
1597
$0.001I_h''+0.02I_h'+500I_h equiv I_h''+20I_h'+500000I_h=0$
– Cesareo
Nov 23 at 1:22
@Cesareo, it wa a mistake. Now it's fixed.
– Lexie Walker
Nov 23 at 1:28
@Moo I didn't know if writing everything was crucial. I wrote cosine becuase with that form of equation is the derivate of E(t).
– Lexie Walker
Nov 23 at 2:37
What did you use for the initial conditions of $I$? We know the initial charge and the initial current, so isn't it better so find $q$ first?
– Dylan
Nov 24 at 13:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to find the current at $t=frac{pi}{4}$
The series RLC circuit has a voltage source given by $E(t)=sin(100t)$ with a capacitor of 2 F, a resistor of 0.02 ohms, and an inductor of 0.001H. If the initial current and the initial charge on the capacitor are both zero,
determine the current in the circuit for $t=frac{pi}{4}$
Based on the information given I have the equation as follows:
$$frac{d^2I}{dt^2}+20frac{dI}{dt}+500I=100000cos(100t)$$
so the homogeneous equation associated is: $r^2+20r+500=(r+10)^2+20^2=0$
whose roots are $-10 pm 20i$. Hence, the solution to the homogeneous equation is:
$$I_h=C_1e^{-10t}cos(20t)+C_2e^{-10t}sin(20t)$$
To find a particular solution, I use the method of undetermined coefficients.
$$I_p=-10.080 cos(100t)+2.122 sin(100t).$$
And after finding $C_1$ and $C_2$ I have:
$$I(t)=e^{-10t}(10.080 cos(100t) -5.570 sin(20t))-10.080cos(20t)+2.122sin(20t)$$
But $I(frac{pi}{4})=0.080111$, and the answer the professor give was 10.076. Am I doing something wrong?
differential-equations
edited Nov 23 at 14:30
Paul Sinclair
19.1k21441
asked Nov 23 at 1:06
Lexie Walker
1597
I'm trying to find the current at $t=frac{pi}{4}$
The series RLC circuit has a voltage source given by $E(t)=sin(100t)$ with a capacitor of 2 F, a resistor of 0.02 ohms, and an inductor of 0.001H. If the initial current and the initial charge on the capacitor are both zero,
determine the current in the circuit for $t=frac{pi}{4}$
Based on the information given I have the equation as follows:
$$frac{d^2I}{dt^2}+20frac{dI}{dt}+500I=100000cos(100t)$$
so the homogeneous equation associated is: $r^2+20r+500=(r+10)^2+20^2=0$
whose roots are $-10 pm 20i$. Hence, the solution to the homogeneous equation is:
$$I_h=C_1e^{-10t}cos(20t)+C_2e^{-10t}sin(20t)$$
To find a particular solution, I use the method of undetermined coefficients.
$$I_p=-10.080 cos(100t)+2.122 sin(100t).$$
And after finding $C_1$ and $C_2$ I have:
$$I(t)=e^{-10t}(10.080 cos(100t) -5.570 sin(20t))-10.080cos(20t)+2.122sin(20t)$$
But $I(frac{pi}{4})=0.080111$, and the answer the professor give was 10.076. Am I doing something wrong?
differential-equations
differential-equations
edited Nov 23 at 14:30
Paul Sinclair
19.1k21441
asked Nov 23 at 1:06
Lexie Walker
1597
edited Nov 23 at 14:30
Paul Sinclair
19.1k21441
asked Nov 23 at 1:06
Lexie Walker
1597
edited Nov 23 at 14:30
Paul Sinclair
19.1k21441
edited Nov 23 at 14:30
Paul Sinclair
19.1k21441
edited Nov 23 at 14:30
Paul Sinclair
19.1k21441
19.1k21441
asked Nov 23 at 1:06
Lexie Walker
1597
asked Nov 23 at 1:06
Lexie Walker
1597
asked Nov 23 at 1:06
Lexie Walker
1597
1597
$0.001I_h''+0.02I_h'+500I_h equiv I_h''+20I_h'+500000I_h=0$
– Cesareo
Nov 23 at 1:22
@Cesareo, it wa a mistake. Now it's fixed.
– Lexie Walker
Nov 23 at 1:28
@Moo I didn't know if writing everything was crucial. I wrote cosine becuase with that form of equation is the derivate of E(t).
– Lexie Walker
Nov 23 at 2:37
What did you use for the initial conditions of $I$? We know the initial charge and the initial current, so isn't it better so find $q$ first?
– Dylan
Nov 24 at 13:46
add a comment |
$0.001I_h''+0.02I_h'+500I_h equiv I_h''+20I_h'+500000I_h=0$
– Cesareo
Nov 23 at 1:22
@Cesareo, it wa a mistake. Now it's fixed.
– Lexie Walker
Nov 23 at 1:28
@Moo I didn't know if writing everything was crucial. I wrote cosine becuase with that form of equation is the derivate of E(t).
– Lexie Walker
Nov 23 at 2:37
What did you use for the initial conditions of $I$? We know the initial charge and the initial current, so isn't it better so find $q$ first?
– Dylan
Nov 24 at 13:46
$0.001I_h''+0.02I_h'+500I_h equiv I_h''+20I_h'+500000I_h=0$
– Cesareo
Nov 23 at 1:22
$0.001I_h''+0.02I_h'+500I_h equiv I_h''+20I_h'+500000I_h=0$
– Cesareo
Nov 23 at 1:22
@Cesareo, it wa a mistake. Now it's fixed.
– Lexie Walker
Nov 23 at 1:28
@Cesareo, it wa a mistake. Now it's fixed.
– Lexie Walker
Nov 23 at 1:28
@Moo I didn't know if writing everything was crucial. I wrote cosine becuase with that form of equation is the derivate of E(t).
– Lexie Walker
Nov 23 at 2:37
@Moo I didn't know if writing everything was crucial. I wrote cosine becuase with that form of equation is the derivate of E(t).
– Lexie Walker
Nov 23 at 2:37
What did you use for the initial conditions of $I$? We know the initial charge and the initial current, so isn't it better so find $q$ first?
– Dylan
Nov 24 at 13:46
What did you use for the initial conditions of $I$? We know the initial charge and the initial current, so isn't it better so find $q$ first?
– Dylan
Nov 24 at 13:46
add a comment |
1 Answer
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Hint.
We have
$$
E(t) = frac 1C int_0^t i(tau)dtau+Ri(t) + Lfrac{di}{dt}
$$
or calling $int_0^t i(tau)dtau = q(t)$
$$
E(t) = frac{q}{C}+Rfrac{dq}{dt}+Lfrac{d^2q}{dt^2}
$$
Solving this DE with the initial conditions
$$
q(0) = 0\
frac{dq}{dt}(0) = 0
$$
we will obtain the desired result.
NOTE
Calling $alpha = frac{R}{2L}$ and $beta = sqrt{frac{C^2R-4L}{4L^2C}}$ we have
$$
q_h = (C_1sinbeta t+C_2cosbeta t)e^{-alpha t}
$$
and calling $E(t) = E_0cosomega t$
$$
q_p = frac{C E_0 left(left(1-C L omega ^2right) cos (omega t)+C omega R sin (omega
t)right)}{C^2 omega ^2 R^2+left(CL omega ^2-1right)^2}
$$
then
$$
q = q_h + q_p
$$
and now knowing that $i(t) = q'$ we can establish the $C_1, C_2$ constants.
answered Nov 23 at 13:56
Cesareo
7,8953516
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
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down vote
Hint.
We have
$$
E(t) = frac 1C int_0^t i(tau)dtau+Ri(t) + Lfrac{di}{dt}
$$
or calling $int_0^t i(tau)dtau = q(t)$
$$
E(t) = frac{q}{C}+Rfrac{dq}{dt}+Lfrac{d^2q}{dt^2}
$$
Solving this DE with the initial conditions
$$
q(0) = 0\
frac{dq}{dt}(0) = 0
$$
we will obtain the desired result.
NOTE
Calling $alpha = frac{R}{2L}$ and $beta = sqrt{frac{C^2R-4L}{4L^2C}}$ we have
$$
q_h = (C_1sinbeta t+C_2cosbeta t)e^{-alpha t}
$$
and calling $E(t) = E_0cosomega t$
$$
q_p = frac{C E_0 left(left(1-C L omega ^2right) cos (omega t)+C omega R sin (omega
t)right)}{C^2 omega ^2 R^2+left(CL omega ^2-1right)^2}
$$
then
$$
q = q_h + q_p
$$
and now knowing that $i(t) = q'$ we can establish the $C_1, C_2$ constants.
answered Nov 23 at 13:56
Cesareo
7,8953516
add a comment |
up vote
0
down vote
Hint.
We have
$$
E(t) = frac 1C int_0^t i(tau)dtau+Ri(t) + Lfrac{di}{dt}
$$
or calling $int_0^t i(tau)dtau = q(t)$
$$
E(t) = frac{q}{C}+Rfrac{dq}{dt}+Lfrac{d^2q}{dt^2}
$$
Solving this DE with the initial conditions
$$
q(0) = 0\
frac{dq}{dt}(0) = 0
$$
we will obtain the desired result.
NOTE
Calling $alpha = frac{R}{2L}$ and $beta = sqrt{frac{C^2R-4L}{4L^2C}}$ we have
$$
q_h = (C_1sinbeta t+C_2cosbeta t)e^{-alpha t}
$$
and calling $E(t) = E_0cosomega t$
$$
q_p = frac{C E_0 left(left(1-C L omega ^2right) cos (omega t)+C omega R sin (omega
t)right)}{C^2 omega ^2 R^2+left(CL omega ^2-1right)^2}
$$
then
$$
q = q_h + q_p
$$
and now knowing that $i(t) = q'$ we can establish the $C_1, C_2$ constants.
answered Nov 23 at 13:56
Cesareo
7,8953516
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint.
We have
$$
E(t) = frac 1C int_0^t i(tau)dtau+Ri(t) + Lfrac{di}{dt}
$$
or calling $int_0^t i(tau)dtau = q(t)$
$$
E(t) = frac{q}{C}+Rfrac{dq}{dt}+Lfrac{d^2q}{dt^2}
$$
Solving this DE with the initial conditions
$$
q(0) = 0\
frac{dq}{dt}(0) = 0
$$
we will obtain the desired result.
NOTE
Calling $alpha = frac{R}{2L}$ and $beta = sqrt{frac{C^2R-4L}{4L^2C}}$ we have
$$
q_h = (C_1sinbeta t+C_2cosbeta t)e^{-alpha t}
$$
and calling $E(t) = E_0cosomega t$
$$
q_p = frac{C E_0 left(left(1-C L omega ^2right) cos (omega t)+C omega R sin (omega
t)right)}{C^2 omega ^2 R^2+left(CL omega ^2-1right)^2}
$$
then
$$
q = q_h + q_p
$$
and now knowing that $i(t) = q'$ we can establish the $C_1, C_2$ constants.
answered Nov 23 at 13:56
Cesareo
7,8953516
Hint.
We have
$$
E(t) = frac 1C int_0^t i(tau)dtau+Ri(t) + Lfrac{di}{dt}
$$
or calling $int_0^t i(tau)dtau = q(t)$
$$
E(t) = frac{q}{C}+Rfrac{dq}{dt}+Lfrac{d^2q}{dt^2}
$$
Solving this DE with the initial conditions
$$
q(0) = 0\
frac{dq}{dt}(0) = 0
$$
we will obtain the desired result.
NOTE
Calling $alpha = frac{R}{2L}$ and $beta = sqrt{frac{C^2R-4L}{4L^2C}}$ we have
$$
q_h = (C_1sinbeta t+C_2cosbeta t)e^{-alpha t}
$$
and calling $E(t) = E_0cosomega t$
$$
q_p = frac{C E_0 left(left(1-C L omega ^2right) cos (omega t)+C omega R sin (omega
t)right)}{C^2 omega ^2 R^2+left(CL omega ^2-1right)^2}
$$
then
$$
q = q_h + q_p
$$
and now knowing that $i(t) = q'$ we can establish the $C_1, C_2$ constants.
answered Nov 23 at 13:56
Cesareo
7,8953516
edited Nov 24 at 14:35
answered Nov 23 at 13:56
Cesareo
7,8953516
answered Nov 23 at 13:56
Cesareo
7,8953516
answered Nov 23 at 13:56
Cesareo
7,8953516
7,8953516
add a comment |
add a comment |
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$0.001I_h''+0.02I_h'+500I_h equiv I_h''+20I_h'+500000I_h=0$
– Cesareo
Nov 23 at 1:22
@Cesareo, it wa a mistake. Now it's fixed.
– Lexie Walker
Nov 23 at 1:28
@Moo I didn't know if writing everything was crucial. I wrote cosine becuase with that form of equation is the derivate of E(t).
– Lexie Walker
Nov 23 at 2:37
What did you use for the initial conditions of $I$? We know the initial charge and the initial current, so isn't it better so find $q$ first?
– Dylan
Nov 24 at 13:46