Question

A series circuit has a capacitor of 0.25*10^-6 farad, a resistor of 5* 10^3 ohms, and an inductor of 1 henry. The initial charge on the capacitor is 0. If a 12volt battery is connected to the circuit and the circuit is closed at t=0, determine the charge on the capacitor at t = 0.001sec, at t= 0.01 sec and at any time t. Also, determine the limiting charge as t --> infinitive
Please, help

Answer 1

@druminjosh

Answer 2

word problem wonderful

Answer 3

No, it killed me, it was my test. I knew nothing about it,

Answer 4

you know its a second order ODE yes?

Answer 5

yes, I had equation of this

Answer 6

and you can solve it?

Answer 7

I just copied from the note and solved it. I never took physics before, know nothing.

Answer 8

\[L \frac{ d^2q }{ dt^2 }+R \frac{ dq }{ dt }+\frac{ 1 }{ C }\frac{ dq }{ dt }=E(t)\]

Answer 9

yes,

Answer 10

ok so we fill in the constants from what's given, yes?

Answer 11

yes,

Answer 12

ok let me do that see what i get... i'm not sure if that 12 volt battery makes E(t) = 12.

Answer 13

hang on hang on

Answer 14

\[Q" + 5*10^3 Q' +\frac{1}{0.25*10^{-6}}Q=12\]

Answer 15

ok 10^-6 in denominator is 10^6 in numerator so i wrote like this:

\[Q''+5000Q'+25000Q=12\]

Answer 16

yes?

Answer 17

wait no hang on grr

Answer 18

should be 4000000Q yes?

Answer 19

\[Q''+5000Q'+4000000Q=12; Q(0)=0\]

Answer 20

yes, but I let the whole thing there to solve

Answer 21

but there's nothing about what happens at Q'(0) hrmm

Answer 22

http://www.wolframalpha.com/input/?i=m%5E2%2B5000m%2B4000000

Answer 23

So
\[y_c = C_1e^{-1000t}+C_2e^{-4000t}\] yes?

Answer 24

:(./......... :( hu huhu... I got it wrong,!!! My bad, I was too nervous to solve that second order differential equation.

Answer 25

but not sure about the constants because i'm not certain about what happens to the first derivative at 0

Answer 26

oh they're both at 0 ... the circuit is closed at 0 and the initial charge on the capacitor is 0.

Answer 27

you can use variation of parameters to solve it LOL

Answer 28

Ok, you help me solve all of them . I need the whole thing to study. I still have a chance to make up the test.

Answer 29

i was kidding about variation of parameters :)

Answer 30

can use coefficients,
\[y_p=At^2+Bt+C\]
yes?

Answer 31

It took me 3 days to pass the depress. Now it is back to me.

Answer 32

nope, the yp is A only, it's a constant

Answer 33

Am I right?

Answer 34

well the second derivative of a quadratic function can be 12, i thought you need the bt and c terms too

Answer 35

there is something wrong at my computer, the letter jumps up and down.

Answer 36

why? My prof taught me that, when the RHS is a constant, the "guessing" of partial solution is an A only

Answer 37

most likely b and c will be 0 ok assume its just At^2

Answer 38

Josh, I don't get!!! whyyyyy??? if it is a polynomial, I can understand yours. This case is not that,

Answer 39

i think you have to use at^2+bt+c but hang on i'm working on it ok bro?

Answer 40

its possible that the derivative of at^2+bt+c is 12

Answer 41

2nd derivative

Answer 42

yes,

Answer 43

actually i think its just A but hang on, i don't do these every day lol

Answer 44

hehehe,, take time, bro

Answer 45

yes the particular solution is of the form y_p = A i am pretty sure

Answer 46

hey, you confused me. Actually, what is the form?

Answer 47

i have a different problem in a book, y"+3y'+2y= 6 and the particular solution is 3 so... i think its just A... the right hand side is a polynomial of degree 0 so the particular solution is a polynomial with degree 0

Answer 48

hahahaa... we meet there.

Answer 49

we could do variation of parameters if we're unsure about the method of coefficients

Answer 50

lets try A, if y_p = A, then both derivatives are 0 ... so we get that 4000000A = 12, so A = 3e-6

Answer 51

yes,

Answer 52

so the particular solutio is just 3e-6 yes?

Answer 53

3E-6 = 3*10^-6

Answer 54

3/1000000

Answer 55

I have question there. In non homogeneous ODE, we have to combine y homo and y partial, then put the initial condition to get the constant, right?

Answer 56

or separate them , solve part by part?

Answer 57

you have to find c1 and c2 with the particular solution

Answer 58

got you

Answer 59

http://www.wolframalpha.com/input/?i=Q%22%2B5000Q%27%2B4000000Q%3D12

Answer 60

ok, thank you very much Josh

Answer 61

so what you get for solution?

Answer 62

I forgot. I was too sad after test, didn't want to think about it. Tomorrow I have class, have to face it on. but I forgot what I was doing.

Answer 63

i meant now... i gave you general solution, you just gotta find c1 and c2

Answer 64

hey, Josh, I need you guide me more about Fourier and Laplace when I study them. Visit this site frequently ,please

Answer 65

send me an e-mail, sometimes i get pretty busy... fourier and laplace transforms are my favorite

Answer 66

you give me the initial condition Q(0) = 0 and Q'(0) =0 I can solve for C1 and C2. don't worry

Answer 67

ok its ugly linear algebra

Answer 68

email? what do you mean?

Answer 69

oh nm when you post my name in here, it e-mails me and it comes up on my phone. if i'm not busy i'll help out

Answer 70

ok, that email not this site pm, right?

Answer 71

pm yes

Answer 72

how can you link them together? this site and your email?

Answer 73

it automatically e-mails me when someone messages me or mentions me in a post i don't know how to be honest

Answer 74

I think its under settings

Answer 75

ah, you set it in your private profile. right?

Answer 76

Under Settings --> Notification Settings

Answer 77

I used to set it until it turned annoying. I reset. so I forgot that part. OK. I will send you if I got stuck
Thanks a lot