Finding Current after 50s Capacitors

[Googl]

Homework Statement

I have managed to calculate the first part of the question but the second part b I keep getting wrong. I was wondering whether you might give me some guidance. I am trying to find the current after 50s in part b below.

Homework Equations

time constant = RC

The Attempt at a Solution

a) Time constant = RC = (50 x 10-6 F) x (100, 000Ω)
Time constant = (5 x 10-5 F) x (100, 000Ω)
Time constant = 5s Correct

b) T1/2 = 5 x In 2 = 3.4657...
50 seconds corresponds to 50/3.4657 = 14.41 half-times.
Δ changing current 14.41t1/2 = 90µA/2^14.41 = 4.13 x 10^-3 µA = 4.13 x 10^-6A
Current after 50 seconds is 4.13 x 10^-6A

This is wrong. Help. Thanks.

 

[Mike Pemulis]

Your setup looks okay to me, but the arithmetic in your last step is wrong. Try the last calculation again.

A couple of miscellaneous pointers:

1. A handy number to remember is that 10 half-lives corresponds to a decrease of about 1000 (actually 1024). So after more than 14 half-lives, we expect a decrease by a factor of about 10,000. Thus we can see that your answer is much too large; it's about 1/20 of the initial current, not 1/10,000! Quick estimates like this are very useful for finding calculation mistakes.

2. To cut down on calculation, you can use the number e as an exponential base instead of 2. The time-constant RC is actually the "1/e-life" (as opposed to the 1/2-life). So once you calculate RC, you don't need to translate into half-lives, you can just get the answer immediately. Numerically, e = 2.718... but your calculator may have a button labeled "e^"

In this case, RC = 5, so we have,

I (50s) = (90)e^(-50/5) = (90)e^(-10)

And you're done.

 

[gneill]

50/3.4657 is 14.427

 

[Googl]

Thanks a lot.

I cannot believe I missed that.

 

[willem2]

Thanks a lot.

I cannot believe I missed that.

Also, a µA is 10^(-6) A, not 10^(-3) A