Final question from me!

Hi there I'm still plugging away at this nightmare. Ill attach a picture of the question, I've highlighted it as I've scribbled around it and I can't figure it out. I understand the time constant is the time taken for charge/voltage to fall to 37% of initial figures. But it doesn't seem to work here...where am I going wrong!

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Reply 1

uberteknik

21

Original post by Hexaneandheels

Hi there I'm still plugging away at this nightmare. Ill attach a picture of the question, I've highlighted it as I've scribbled around it and I can't figure it out. I understand the time constant is the time taken for charge/voltage to fall to 37% of initial figures. But it doesn't seem to work here...where am I going wrong!

The voltage across the capacitor is falling hence discharging.

V_{(t)} = V_{(t=0)}(e^{(\frac {-t}{CR})})

Your task is to make CR the subject and substitute the known values where:

V_{(t)} = 5V

V_{(t=0)} = 10V

t = 48 \mathrm x 10^{-3}S

HINT: use logarithms.

(edited 9 years ago)

Reply 2

lerjj

13

Original post by Hexaneandheels

Hi there I'm still plugging away at this nightmare. Ill attach a picture of the question, I've highlighted it as I've scribbled around it and I can't figure it out. I understand the time constant is the time taken for charge/voltage to fall to 37% of initial figures. But it doesn't seem to work here...where am I going wrong!

I know this isn't especially helpful in general, but in an exam like this my thinking would go:

time taken to halve = 48 ms
therefore time taken to quarter = 96ms
the time taken to get somewhere in between a half and a quarter (37% or whatever), must be between 48 and 96 ms.

You only have one option that's around there, so in an exam this would be quicker than calculating it. Of course, you should still learn how to work it out (although if you're doing AQA the formula is given to you).

(edited 9 years ago)

Reply 3

MSB47

2

You can re arrange V=Voe^-t/RC

To make RC the subject. Using logarithms RC can be made the subject by...

•

(taking natural logarithms of both sides...)

•

ln(V/Vo)=-t/RC

•

[size="6"]RC=-t/ln(V/Vo)

RC is equivalent to the time constant therefore, sub in the values the question provides and you should get answer C (69ms).

Hope thaat helps :smile:

Reply 4

Hexaneandheels

OP

4

After a little brief revision I am annoyed that I couldn't figure this earlier, thank you for your help!!!

Original post by MSB47

You can re arrange V=Voe^-t/RC

To make RC the subject. Using logarithms RC can be made the subject by...

•

(taking natural logarithms of both sides...)

•

ln(V/Vo)=-t/RC

•

RC=-t/ln(V/Vo)

RC is equivalent to the time constant therefore, sub in the values the question provides and you should get answer C (69ms).

Hope thaat helps :smile:

Reply 5

MSB47

2

Original post by Hexaneandheels

After a little brief revision I am annoyed that I couldn't figure this earlier, thank you for your help!!!