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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:

$$\frac{d^2I}{dt^2}+20\frac{dI}{dt}+500I=100000\cos(100t)$$

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:

$$I(t)=e^{-10t}(10.080 \cos(100t) -5.570 \sin(20t))-10.080\cos(20t)+2.122\sin(20t)$$

But $I(\frac{\pi}{4})=0.080111$, and the answer the professor give was 10.076. Am I doing something wrong?

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  • $\begingroup$ $0.001I_h''+0.02I_h'+500I_h \equiv I_h''+20I_h'+500000I_h=0$ $\endgroup$
    – Cesareo
    Commented Nov 23, 2018 at 1:22
  • $\begingroup$ @Cesareo, it wa a mistake. Now it's fixed. $\endgroup$ Commented Nov 23, 2018 at 1:28
  • $\begingroup$ @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). $\endgroup$ Commented Nov 23, 2018 at 2:37
  • $\begingroup$ 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? $\endgroup$
    – Dylan
    Commented Nov 24, 2018 at 13:46

1 Answer 1

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Hint.

We have

$$ E(t) = \frac 1C \int_0^t i(\tau)d\tau+Ri(t) + L\frac{di}{dt} $$

or calling $\int_0^t i(\tau)d\tau = q(t)$

$$ E(t) = \frac{q}{C}+R\frac{dq}{dt}+L\frac{d^2q}{dt^2} $$

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

$$ q_h = (C_1\sin\beta t+C_2\cos\beta t)e^{-\alpha t} $$

and calling $E(t) = E_0\cos\omega t$

$$ q_p = \frac{C E_0 \left(\left(1-C L \omega ^2\right) \cos (\omega t)+C \omega R \sin (\omega t)\right)}{C^2 \omega ^2 R^2+\left(CL \omega ^2-1\right)^2} $$

then

$$ q = q_h + q_p $$

and now knowing that $i(t) = q'$ we can establish the $C_1, C_2$ constants.

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