Thanks. Well spotted. I have added an annotation to correct the error. That's the problem with making these vids quickly to get them done for this exam season. I'm grateful to you for pointing out the mistake.
I assume you mean a more complex diagram which has a combination of capacitors in series and parallel. The trick is to work outwards. Start with a central grouping (eg a pair in parallel), circle them with a pencil, and work out the effective capacitance. Then move on to the next wider grouping.
How did you do Q4 part b and c ? How did you know that the discharge formula is to be used for the 50% ? After writing 2=e^t/0.6 what is the next step? i don't get it.
so when we are taking RC (in the last question) we are taking indirectly t which we have calculated in part (a). so my question is can we cancel t/t, not in this question but in some other question that is some way similar to this.
These video's are a huge help to me, but on question 4 at 8:18 i keep getting 3.36x10^-9 not 3.36x10^-11 i didnt know if it was a mistake or whether I may be doing something wrong? Again many thanks for your video's they are an amazing resource I enjoy using often :)
Around 2:20, the equation used to find the charge of the capacitor while charging should be Q = Qo [1- e^(-t/RC)]
Thanks. Well spotted. Have added an annotation.
Thanks. Well spotted. I have added an annotation to correct the error. That's the problem with making these vids quickly to get them done for this exam season. I'm grateful to you for pointing out the mistake.
I hope you keep making videos- I really enjoy them.
Keep up the good work!
Can't believe this was on the internet for the past 7 years and I'm stumbling upon this exactly a day before my exam. Sad.
I wish there were actual exam questions, most of these seem very easy.
I assume you mean a more complex diagram which has a combination of capacitors in series and parallel. The trick is to work outwards. Start with a central grouping (eg a pair in parallel), circle them with a pencil, and work out the effective capacitance. Then move on to the next wider grouping.
You've literally just saved my life. Thank you so much!
The only thing that keeps me watching this video for my exam is your accent.
How did you do Q4 part b and c ? How did you know that the discharge formula is to be used for the 50% ? After writing 2=e^t/0.6 what is the next step? i don't get it.
so when we are taking RC (in the last question) we are taking indirectly t which we have calculated in part (a). so my question is can we cancel t/t, not in this question but in some other question that is some way similar to this.
You're welcome.
Thank you for the videos they are great
These video's are a huge help to me, but on question 4 at 8:18 i keep getting 3.36x10^-9 not 3.36x10^-11 i didnt know if it was a mistake or whether I may be doing something wrong? Again many thanks for your video's they are an amazing resource I enjoy using often :)
ln(2) divide by 0.6 equals 1.155 just to point out but very informative vid
Thats ok, many thanks again for the videos :)
Tq dr! Its very useful
@5:15 : 3(300*10^-6) =9*10^-4
... which is the same as 900*10^-6
why are we taking t=RC
That is just a special case. When T equals RC, then the fraction t divided by RC equals one.