- #1

zenterix

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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)}##.

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

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