Question

The current i in a charging circuit is given by i = 10e^(-100t) amperes. Find the charge due to the current after 10 milliseconds, if there is initially no charge in the circuit.

So I did this:
Integration: -[e^(-100t)] / 10

And when I subbed in t = 0.01, I got 0.0036 coulombs. However, the answer is 0.0632 coulombs. Is there any mistake with my integration, or my whole working in itself?

Answer 1

Also, i = dq/dt, where q is the charge in the circuit, if that helps.

Answer 2

http://www.wolframalpha.com/input/?i=%5Cint_0%5E%7B0.01%7D+10e%5E%28-100t%29+dt

Answer 3

If there's no problem with my integration, then why is it that I can't get the answer after subbing it in? :(

Answer 4

i think you're not subtracting the lower bound

Answer 5

check it once

Answer 6

What do you mean by 'lower bound"?

Answer 7

\(\large Q = \int_0^{0.01} 10e^{-100t} dt \)

Answer 8

\(\large Q = \frac{-1}{10} e^{-100t} \Big|_0^{0.01} \)

Answer 9

\(\large Q = \frac{-1}{10} [e^{-1} - e^0] \)

Answer 10

simplify

Answer 11

Oh... I get it. But this question was given in the Indefinite Integral chapter(which is before the Definite Integral chapter), so is there a way to solve it without the bounds?

Answer 12

yeah sure :)
it becomes an IVP

Answer 13

IVP = Initial Value Problem

Answer 14

\(\large Q(t) = \int 10e^{-100t} dt \)

\(\large Q(t) = \frac{-1}{10} e^{-100t} + c \)

Answer 15

find the constant by plugging in the point (0, 0)

Answer 16

Meaning, when t = 0?

Answer 17

yes, when t = 0, Q = 0

using that Initial Values, find the integration constant \(c\)

Answer 18

we got that infor from the question :
... if there is initially no charge in the circuit.

Answer 19

So C = 0.1?

Answer 20

yup

Answer 21

so ur charge equation becomes :

\(\large Q(t) = \frac{-1}{10} e^{-100t} + 0.1 \)

Answer 22

plugin t = 10ms, u should get the same answer..

Answer 23

Oh. Lol. Okay, got it! Thanks a lot!

Answer 24

np.. u wlc :)