How to find voltage across capacitor in RLC circuit?

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To find the voltage across a capacitor in an RLC circuit, the discussion centers on applying Faraday's law and forming differential equations based on charge (q) and current (I). The derived equations allow for solutions that express charge and current as sinusoidal functions, incorporating phase relationships. The voltage across the capacitor is calculated as V_c(t) = q(t)/C, with considerations of initial conditions and phase lag between current and voltage. The analysis reveals that current leads charge by π/2, and the phase lag between current and voltage varies depending on the frequency relative to the natural frequency of the oscillator. The discussion highlights the complexity of accurately determining capacitor voltage under varying conditions.
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Homework Statement
This question is based on a problem in MIT OCW's 8.02 course. There is automated grading and I just cannot get it right.

Given a series RLC circuit (pictured below), find the voltage across the capacitor.
Relevant Equations
Assume that ##V(t)=V_0\sin{(\omega t)}##.
1715485540407.png

By Faraday's law

$$-V(t)+I(t)R+\frac{q(t)}{C}=-L\dot{I}(t)\tag{1}$$

$$\dot{I}+\frac{R}{L}I+\frac{1}{LC}q=\frac{V(t)}{L}\tag{2}$$

Here we can either form a differential equation in ##q(t)## or we can differentiate and form one in ##I(t)##.

These equations are

$$\ddot{q}+\frac{R}{L}\dot{q}+\frac{1}{LC}q=\frac{V(t)}{L}\tag{3}$$

$$\ddot{I}+\frac{R}{L}\dot{I}+\frac{1}{LC}I=\frac{\dot{V}(t)}{L}\tag{4}$$

The solution to (3) is

$$q(t)=\frac{V_0\sin{(\omega t-\phi)}}{\omega\sqrt{R^2+\left ( \frac{1}{\omega C}-\omega L \right )^2}}$$

where

$$\tan{\phi}=\frac{\omega RC}{1-\omega^2 LC}$$

The solution to (4) is

$$I(t)=\frac{V_0\sin{\left (\omega t+\frac{\pi}{2}-\phi\right )}}{\sqrt{R^2+\left ( \frac{1}{\omega C}-\omega L \right )^2}}$$

In MIT OCW's 8.02 there is the following question

Calculate ##V_{cap}(t)=\frac{q}{C}##, the voltage across the capacitor. Hint: do this calculation assuming that at ##t=0## there is no charge on the capacitor and consider the time right after that where charge on the capacitor is increasing.

I am really not sure how to take this hint into account.

Let

$$I_0=\frac{V_0}{\sqrt{R^2+\left ( \frac{1}{\omega C}-\omega L \right )^2}}$$

Then

$$q(t)=\frac{I_0}{\omega}\sin{(\omega t-\phi)}$$

and

$$V_c(t)=\frac{q(t)}{C}=\frac{I_0}{\omega C}\sin{(\omega t-\phi)}$$

As you can see below, the automated grading system for this question tells me I am wrong

1715528083192.png
 
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For the record, this problem also asked a few questions about the phase lag between current, ac voltage and voltage across the capacitor. I answered these successfully, and here is my analysis

Let ##\omega_0=\frac{1}{\sqrt{LC}}##, the natural frequency of the oscillator.

Note that from the equations

$$V(t)=V_0\sin{(\omega t)}$$

$$q(t)=\frac{V_0\sin{(\omega t-\phi)}}{\omega\sqrt{R^2+\left ( \frac{1}{\omega C}-\omega L \right )^2}}$$

$$I(t)=\frac{V_0\sin{\left (\omega t+\frac{\pi}{2}-\phi\right )}}{\sqrt{R^2+\left ( \frac{1}{\omega C}-\omega L \right )^2}}$$

$$\tan{\phi}=\frac{\omega RC}{1-\omega^2 LC}$$

We can see that current ##I(t)## always leads charge on the capacitor ##q(t)## by ##\pi/2##.

The phase lag of current relative to voltage is ##\beta=\frac{\pi}{2}-\phi##.

$$\tan{\beta}=(...)=\frac{1}{\tan{\phi}}=\frac{L(\omega_0^2-\omega^2)}{R\omega}$$

Suppose ##\omega>\omega_0##. Then, ##\beta<0## which means that current lags voltage.

Conversely, if ##\omega<\omega_0## then ##\beta>0## and current leads voltage.
 
One other thing I tried to do to find the capacitor voltage (while taking into account the hint) was to try to force ##q(0)=0##.

After all, if the AC voltage is ##V_0\sin{\omega t}## then at time zero this voltage is zero but the charge on the capacitor is not since there is a phase lag and so the sine in the expression for ##q(t)## is not zero.

If the AC voltage is instead ##V_0\sin{(\omega t+\theta)}## then

$$q(t)=\frac{I_0}{\omega}\sin{(\omega t+\theta -\phi)}$$

and

$$q(0)=\frac{I_0}{\omega}\sin{(\theta-\phi)}=0$$

$$\implies \theta=\phi$$

so that now

$$q(t)=\frac{I_0}{\omega}\sin{(\omega t)}$$

and

$$V_c(t)=\frac{I_0}{\omega C}\sin{(\omega t)}$$

But this is also incorrect in the automated grader.
 
Something is missed.

## I(t) = \frac{d}{dt} (\frac{V_0 \sin (\omega t-\phi)}{\omega \sqrt{R^2+(\frac{1}{\omega C}-\omega L)^2} }+\text{constant}) ##

where

## \text{constant} = \frac{V_0 \sin \phi}{\omega \sqrt{R^2+(\frac{1}{\omega C}-\omega L)^2}} ##.
 
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