How to find voltage across capacitor in RLC circuit?

[zenterix]

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)}##.

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

 

[zenterix]

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.

 

[zenterix]

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.

 

[Gavran]

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}} ##.