Problem with differential equation RLC circuit series

Question

I am trying solve the differential equation of RLC's circuit in series, I have: $C=4\ F, L= 1\ H$, $R=5\ \Omega$, and $V_e=20\ V$.

$1)$ first I got the equation, it is: $i''+5i'+\frac{1}{4}i=0$, what I have to calculate is $v_c$, and I know that $i(0)=-2\ A$ and $v_c(0)=10\ V$

I have calculated the characteristic polynomial and I got that a fundamental system is $\{ e^{-0.051t}; e^{-4.950t}\}$, so $$i=Ae^{-0.051t}+Be^{-4.950t}$$ And now I could calculate $v_c=\int_{0}^{t}idt$

Ok, first problem!... I got $$v_c=-4.950A(e^{-0.051t}-1)-0.051B(e^{-4.950t}-1),$$ SO... when I said that $v_c(0)=10=0-0 $ what happen here?

Ok, now I dont said $\int_0^t$, no. I consider $\int$ only, so I got $$v_c=-4.950A(e^{-0.051t})-0.051B(e^{-4.950t})$$

And when I said that $v_c(0)=0$ I got that $$A=-2.020\ B=0.021,$$ in that, $$ v_c\approx10e^{-0.051t}$$

but in my but said that $v_c=20+0.102e^{-4.950t}-10.102e^{-0.051t}\ [V]$, so what happen, I need help please... please...

PD; To solve can not use the formula that we all know, the problem is solved by mathematical methods. I need help...

Answer 1

I will do the voltage across the capacitor calculations and you can rework the current. Note: You cannot mix the initial conditions for voltage and current and this caused your problem (see below and mimic for current calculation).

We are given the series RLC circuit with component values $R = 5$ Ohms, $L = 1$ Henry and $C = 4$ Farads, with a constant voltage source $V_e = 20$ Volts.

Applying Kirchoff's Voltage Law (KVL), we have:

$$V_R + V_L + V_C = V_E \implies R i(t) + L \dfrac{di(t)}{dt} + V_C(t) = V_E$$

We know the current across the capacitor is given by $i(t) = C \dfrac{dV_C(t)}{dt}$, and we substitute, giving:

$$\dfrac{d^2V_C(t)}{dt^2} + \dfrac{R}{L} \dfrac{dV_C(t)}{dt} + \dfrac{1}{LC} V_C(t) = \dfrac{V_E}{LC}$$

Substituting component values, this leads to:

$$V_C''(t) + 5 V_C'(t) + \dfrac{1}{4} V_C(t) = 5$$

Solving this DEQ:

$$V_C(t) = a e^{(-5/2 + \sqrt{6})t} + be^{(-5/2 - \sqrt{6})t} + 20$$

Now, we have two unknowns, so need two initial conditions. We are given $V_C(0) = 10$ and know that $V_C'(0)= 0$. This leads to:

$$a + b +20 = 10 \\ a (-5/2 + \sqrt{6}) + b (-5/2 + \sqrt{6}) = 0$$

This gives us $a = -10.1031,b= 0.103104$, resulting in:

$$V_C(t) = -10.1031 e^{(-5/2 + \sqrt{6})t} + 0.103104e^{(-5/2 - \sqrt{6})t} + 20$$

This is a handy set of notes and you can even simulate this circuit using the electronic circuit simulator applet.

Update

There is some question regarding the IC $i(0)$.

For $i(0) = 0$, the result is as shown above.

From above, we have the relation $i(t) = C V_C'(t)$ and can use that to find a $V_C'(0) = \dfrac{i(0)}{C}$. For $i(0) = -2, V_C'(0) = -\dfrac{1}{2}$, which leads to a slightly different result of: $$V_C(t) = 20.\, +0.205166 e^{-4.94949 t}-10.2052 e^{-0.0505103 t}$$

You can try doing the problem for $i(t)$ or just the relation above.

Answer 2

Notice (hint):

$$\text{V}_{\text{in}}(t)=\text{V}_{R}(t)+\text{V}_{C}(t)+\text{V}_{L}(t)\Longleftrightarrow$$

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Knowing that:

• $\text{V}_{L}(t)=\text{LI}'_{L}(t)=\text{LI}'_{T}(t)$;
• $\text{I}_{C}(t)=\text{I}_{T}(t)=\text{CV}'_{C}(t)\to\text{I}'_{C}(t)=\text{I}'_{T}(t)=\text{CV}''_{C}(t)$.

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$$\text{V}_{\text{in}}(t)=\text{RI}_{T}(t)+\text{V}_{C}(t)+\text{LI}'_{T}(t)\Longleftrightarrow$$ $$\text{V}_{\text{in}}(t)=\text{R}\cdot\text{CV}'_{C}(t)+\text{V}_{C}(t)+\text{L}\cdot\text{CV}''_{C}(t)\Longleftrightarrow$$ $$\text{V}_{\text{in}}(t)=\text{CRV}'_{C}(t)+\text{V}_{C}(t)+\text{CLV}''_{C}(t)$$

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Given, are the following things:

• $\text{V}_{\text{in}}(t)=\text{V}_{e}=20\space\text{V}$;
• $\text{C}=4\space\text{F}$;
• $\text{L}=1\space\text{H}$;
• $\text{R}=5\space\Omega$;
• $\text{I}_{C}(0)=\text{I}_{T}(0)=\text{CV}'_{C}(0)\to\text{I}_{C}(0)=4\text{V}'_{C}(0)=-2\space\text{A}\Longleftrightarrow\text{V}'_{C}(0)=-\frac{1}{2}\space\text{A}$;
• $\text{V}_{C}(0)=10\space\text{V}$.

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So the problem becomes:

$$ \begin{cases} 20\text{V}'_{C}(t)+\text{V}_{C}(t)+4\text{V}''_{C}(t)=20\\ \text{V}'_{C}(0)=-\frac{1}{2}\\ \text{V}_{C}(0)=10 \end{cases} $$

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The answer will be:

$$\color{red}{\text{V}_{C}(t)=20-\frac{40\cosh\left[t\sqrt{6}\right]+17\sqrt{6}\sinh\left[t\sqrt{6}\right]}{4e^{\frac{5t}{2}}}\space\space\space\text{V}}$$