Question

A series circuit consists of a 12.0 V source of emf, a 2.00 mF capacitor, a 1000 Ω resistor, and a switch. When the switch is closed, how long does it take for the current to reach one-tenth its maximum value?

Answer 1

(a) 4.61 s
(b) 2.30 s
(c) 18.0 s
(d) 9.00 s
(e) 0.693 s

Answer 2

See this and try by yourself
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

Answer 3

@shivam_bhalla You can answer it?

Answer 4

See the formula is
\[\huge {I = {{v_0} \over {t}}{e^{{-t}/{RC}}}}\]
You know what is V, R ,C , I from the question

Now substitute and find t . If you have doubt in simplification, then post your step one by one and I will help you

Answer 5

So i will then substitute for V_0 with 12 and C with 20 mF and R with 1000 ohms, what about the current and when I simplify the equation, it will be like this : t = V_0 / I * e^-t/RC, so how can i substitute for the power -t and is e for exponential?
Thanks

Answer 6

Yes :)
Make sure you write
C=20 mF = 20 * 10^(-3) F

Answer 7

And what should i substitute for I?

Answer 8

\[I = { 1 \over 10} {v_b \over R}\]

Answer 9

Since the maximimum current in the circuit is
\[I_{\max} = {v_b \over R}\]
and it is given in the question that I = 1/10 of I_max

Answer 10

Yes you are right :).
Well now the power (-t) for the exponential is the problem because that what I need indeed :) so what should I do with?

Answer 11

The best way is to take log_e on both sides and then apply power rule.

Power rule is
\[\large \log_a b^c = c*\log_a b\]

Answer 12

@benanderson5900 , you got the answer because I am getting the answer :)

Answer 13

Hahaha well I just tried but I got irrelevant numbers to the onesI have to choose from :D

Answer 14

I am getting one of your choices, show me your working out so that I can help you out. I am getting option A

Answer 15

Well I think the problem is still in log. After taking log both sides, how the equation should be looks like?

Answer 16

Sorry it should be
\[\large \log_e {1 \over 10} = \log_e e^{-t/2}\]

Answer 17

Now apply power rule and tell me your equation. (For power rule see above in this thread)

Answer 18

Well so log will cancel the exponential, so the equation will be "log(1/10) = -t/2?

Answer 19

Exactly.
Then One more rule which is
\[\large \log_a {p \over q} = \log_a p - \log_a q\]

Answer 20

Well so log(1/10) = log(1) - log(10)?

Answer 21

yes and what is log(1) =??

Answer 22

0

Answer 23

Yes now tell me your final equation after doing all this.

Answer 24

-1 = -t/2?

Answer 25

No.

\[\large \log_e 10 \neq 1 \]

Answer 26

Sorry, equals to 2.303?

Answer 27

Yes :) Now you will get your required answer :)

Answer 28

Haha well yeah I just got it :). Thanks a lot really :)

Answer 29

Anytime :D

Answer 30

Actually it is no the best answer but the best help and support instead :D and that what matters :).

Answer 31

Exactly :D. I will give you a medal for that :)