Exam Help

Discussion in 'Homework Help' started by hitmen, Dec 5, 2008.

1. hitmen Thread Starter Active Member

Sep 21, 2008
159
0
Well, my exams are over and here I am going to post the questions I have difficulty with.

In the first question, I am able to derive that Ix(0+) = (4-Vc(0+))/6 and that Ix(t) = (1+2e-1000t) / 3 for t>0. However, I am not sure how to calculate the energy absorbed by capacitor over 0<t<∞ and the amount of charge deposited into the capacitor.

Can someone explain how the first 2 equations relate to the 3th part because I only know how to solve them blindly.

Do kindly ignore my markings on the exam paper.

File size:
305.5 KB
Views:
34
• DSCF0050.jpg
File size:
304.5 KB
Views:
27
Last edited: Dec 5, 2008
2. hgmjr Moderator

Jan 28, 2005
9,030
214
The image is a bit fuzzy. Can you post a clearer picture?

hgmjr

3. studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Both answers require calculus can you do this?

4. hitmen Thread Starter Active Member

Sep 21, 2008
159
0
Ummm... I am only interested in the Energy absorbed and the charge deposited in the capacitor. I cant do it in the exam. I can do calculus but I am not sure of how to even START doing the problem. how about giving me the first step or a rough concept?

5. hitmen Thread Starter Active Member

Sep 21, 2008
159
0
Do you mind giving the first step?
Thanks.

6. studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Having some trouble posting here.

OK now we have done the diode problem, we can go on with this one. Sorry I couldn't post anything earlier.

I would guess from this that you have been given the formulae energy = 1/2CV^2 and charge q = CV for a capacitor.

So noting that initially the capacitor appears as a short circuit and finally as an open circuit calculate the voltages using normal theorems (egKVL) for these two conditions and apply the formula to the voltage change across the capacitor.

I note by inspection that the capacitor is at -2 volts initially.

The charge is easier as it is just the voltage rise times the capacitance.

An alternative method is to use you currents to calculate the current through the capacitor using the differential equation

i = C dVc/dt where Vc is the capacitor voltage, seen as a node voltage in KVL

Then you integrate this to find the same solution

Oh and well done for getting as far as you have. Most courses skate over this stuff quickly because there are better more comprehensive mathods you will come on to.

7. Bigcountry Active Member

Jul 4, 2008
71
0
In reference of the clearer picture. Since it looks like you took a picture of it You could put it in paint and trace over the numbers and schematic. Just idea though.

8. hitmen Thread Starter Active Member

Sep 21, 2008
159
0
I know the formula i = C dVc/dt but how I calculate Δt when it is from 0 to infinity?