How does a capacitor in an RLC circuit lose charge?

deedee5
Messages
1
Reaction score
0
1. First you fully charge a capacitor (C = 6.4 µF) with a 12V battery. Then you disconnect the battery and discharge the capacitor through a resistor (R = 3 Ohms) and an inductor (L = 2.0 x 10-3 H) in series.

2. What is the charge on the capacitor when you begin discharging at t= 0?

How long does it take for the charge amplitude to fall to one tenth of initial value?

At what time does the charge first fall to one tenth of its initial value?



The Attempt at a Solution


 
Physics news on Phys.org
Well what have you thought about doing? What dictates how fast a capacitor in an RLC (Resistor - Inductor - Capacitor) circuit will lose charge?
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top