Why do you suddenly go from 1MOhm to 10MOhm??

Looks like a transcription error in the set up equation -- which is why it is so important to verify that the set up is correct before proceeding, because that is where the physics lies, everything after that is math and the math neither knows nor cares whether we got the physics right. I clearly didn't do that properly.

I also didn't do something else that I normally do and often recommend. Once we have an answer, we can almost always determine the correctness of that answer from the answer itself by simply asking whether it actually solves the problem.

The original problem says that the capacitor has a voltage of 2 V after 4 s when charging towards a 12 V final value.

So

\(v_c(t) \; = \; 12 \, V \, \left( 1 \; - \; e^{-\frac{t}{RC}}\right)\)

Using the answer I got, we have

\(v_c(4 \, s) \; = \; 12 \, V \, \left( 1 \; - \; e^{-\frac{4 \, s}{(1 \, M \Omega)(2.19 \, \mu F)}}\right)\)

\(v_c(4 \, s) \; = \; 12 \, V \, \left( 1 \; - \; e^{-1.8265}\right)\)

\(v_c(4 \, s) \; = \; 12 \, V \, \left( 1 \; - \; 0.16098 \right)\)

\(v_c(4 \, s) \; = \; 10.07 \, V\)

Which does not agree with the problem statement that the voltage at this time is 2 V. So we know our answer is wrong.

Note that one very important thing we need to do when verifying our answer is to go back to the original problem statement. If we just start from somewhere in our work (like my original set up equation) we risk incorporating the very mistake we are trying to detect into our verification work.

It's quite possible that in setting up the verification work we would catch the mistake (very likely in this particular case).

To summarize, I was in such a hurry to answer the question (don't remember why) that I failed to do two things that I always preach about: Set up your equations and then verify them before proceeding, and checking whether the answer is correct from the answer itself. Had I done either one of those, the mistake would have been caught.

Thanks for catching it and pointing it out.

 

Looks like a transcription error in the set up equation -- which is why it is so important to verify that the set up is correct before proceeding, because that is where the physics lies, everything after that is math and the math neither knows nor cares whether we got the physics right. I clearly didn't do that properly.

I also didn't do something else that I normally do and often recommend. Once we have an answer, we can almost always determine the correctness of that answer from the answer itself by simply asking whether it actually solves the problem.

The original problem says that the capacitor has a voltage of 2 V after 4 s when charging towards a 12 V final value.

So

\(v_c(t) \; = \; 12 \, V \, \left( 1 \; - \; e^{-\frac{t}{RC}}\right)\)

Using the answer I got, we have

\(v_c(4 \, s) \; = \; 12 \, V \, \left( 1 \; - \; e^{-\frac{4 \, s}{(1 \, M \Omega)(2.19 \, \mu F)}}\right)\)

\(v_c(4 \, s) \; = \; 12 \, V \, \left( 1 \; - \; e^{-1.8265}\right)\)

\(v_c(4 \, s) \; = \; 12 \, V \, \left( 1 \; - \; 0.16098 \right)\)

\(v_c(4 \, s) \; = \; 10.07 \, V\)

Which does not agree with the problem statement that the voltage at this time is 2 V. So we know our answer is wrong.

Note that one very important thing we need to do when verifying our answer is to go back to the original problem statement. If we just start from somewhere in our work (like my original set up equation) we risk incorporating the very mistake we are trying to detect into our verification work.

It's quite possible that in setting up the verification work we would catch the mistake (very likely in this particular case).

To summarize, I was in such a hurry to answer the question (don't remember why) that I failed to do two things that I always preach about: Set up your equations and then verify them before proceeding, and checking whether the answer is correct from the answer itself. Had I done either one of those, the mistake would have been caught.

Thanks for catching it and pointing it out.

i only noticed because i too am trying to do the HNC and i have the very same question - but when working it out i am trying to use 10 to the power of 6 x ln (1 - 2/12) (i dont know how to do formulas and functions on here) which gives me -182321.5568 - which is obviously a few decimal places away from your workings.

would i be correct in doing it that way or should i do 1 ln (1-2/12) ?

 

i only noticed because i too am trying to do the HNC and i have the very same question - but when working it out i am trying to use 10 to the power of 6 x ln (1 - 2/12) (i dont know how to do formulas and functions on here) which gives me -182321.5568 - which is obviously a few decimal places away from your workings.

would i be correct in doing it that way or should i do 1 ln (1-2/12) ?

Do it the way I showed, simply correct the error in the setup equation.

\( C \; = \; \frac{-4 \, s}{10 \, M \Omega \, \ln{ \left( 1 \; - \; \frac{2 \, V}{12 \, V} \right)}} \)

Just needs to be

\( C \; = \; \frac{-4 \, s}{1 \, M \Omega \, \ln{ \left( 1 \; - \; \frac{2 \, V}{12 \, V} \right)}} \)

Now just evaluate it and you get C = 21.9 μF

Then verify the answer by seeing if it solves the problem:

\(v_c(4 \, s) \; = \; 12 \, V \, \left( 1 \; - \; e^{-\frac{4 \, s}{(1 \, M \Omega)(21.9 \, \mu F)}}\right)\)

\(v_c(4 \, s) \; = \; 12 \, V \, \left( 1 \; - \; e^{-0.18265}\right)\)

\(v_c(4 \, s) \; = \; 12 \, V \, \left( 1 \; - \; 0.8331 \right)\)

\(v_c(4 \, s) \; = \; 2.002 \, V\)

And it does.