Engineering maths help

Original post by richjpark

Hi,

I've recently started studying engineering maths after being away from formal education for some time! struggling my way through but I'm stumped here. Any help would be fantastic!

A series electrical circuit for a computer game which you are testing, features a capacitor (C) discharging via a 270 kΩ resistor (R). The capacitor is initially charged to a voltage of 15 V. The voltage across the capacitor (Vc) may be described by the following equation, where t represents time.

V_c= 15(e ^-t/RC)

The circuit produces an output based upon the time it takes for the voltage on the capacitor to discharge to a value of 2 V. The value of the capacitance is based upon the year of your birth, which is input at the start of the game. For example, if you were born in 1990, then C = 1 + 9 + 9 + 0 = 20 μF.

Determine how long it takes for the circuit to produce an output.

You must have a similar example in your notes/textbook or google for something like
RC circuit
gives a lot of help/examples. So what have you done/what are you stuck with.

(edited 5 months ago)

Reply 2

OP

Original post by mqb2766

You must have a similar example in your notes/textbook or google for something like
RC circuit
gives a lot of help/examples. So what have you done/what are you stuck with.

Hi, thanks for the reply.

yes using a similar worked example in the workbook I have ended up with...

Vc/15 = e^-t/RC

Loge (Vc/15) = loge (e(^-t/Rc))

Loge (Vc/15) = -t/Rc

Loge (Vc/15)Rc = (-t/Rc)Rc

RcLoge (Vc/15) = -t

That's where I'm up to. Not sure if that looks anything like correct so far! I assume i would then input R=270 and C=19 but unsure where to get Vc?

I assume im looking for a figure to express as some unit of time at the end but unsure if it would be seconds or miliseconds or anything else?

t = RcLoge (Vc/15)/-1

Reply 3

Original post by richjpark

Hi, thanks for the reply.

yes using a similar worked example in the workbook I have ended up with...

Vc/15 = e^-t/RC

Loge (Vc/15) = loge (e(^-t/Rc))

Loge (Vc/15) = -t/Rc

Loge (Vc/15)Rc = (-t/Rc)Rc

RcLoge (Vc/15) = -t

That's where I'm up to. Not sure if that looks anything like correct so far! I assume i would then input R=270 and C=19 but unsure where to get Vc?

I assume im looking for a figure to express as some unit of time at the end but unsure if it would be seconds or miliseconds or anything else?

t = RcLoge (Vc/15)/-1

Sort of, but a reasonable number of issues.
* Youre calculating the time until the voltage on the capacitor is 2. Youve not used the value anywhere, but its use should be obvious?
* Check your units for the actual values of R and C. You must have covered this already?
* Your last equation is about right, though dividing by -1 is the same as ... If you use SI units throughout (see second point), the unit of time will be s.

(edited 5 months ago)

Reply 4

OP

I should have explained that its a distance learning course I'm doing and have had no contact with the a tutor yet (I've booked to see them in a few weeks but trying to get as much done as possible this week whilst off work!)

•

Would Vc be 13? the difference between the initial charge (15V) and 2V?

•

You've lost me a bit on the second point I'm afraid

•

dividing by -1 same as multiply by one so...

t = RcLoge (Vc/15)

thanks for your help with this!

Reply 5

Original post by richjpark

I should have explained that its a distance learning course I'm doing and have had no contact with the a tutor yet (I've booked to see them in a few weeks but trying to get as much done as possible this week whilst off work!)

•

Would Vc be 13? the difference between the initial charge (15V) and 2V?

•

You've lost me a bit on the second point I'm afraid

•

dividing by -1 same as multiply by one so...

t = RcLoge (Vc/15)

thanks for your help with this!

Vc is defined to be the capacitor voltage. It has a value of 15 at t=0 and you want to find the time that it takes for the capacitor to discharge until it gets a value of 2. You may be overthinking it.

The units are kilo ohms and micro farads. Youre missing the bold parts.

What does 5/-1 evaulate to? Do the same to your formula.

THere are some description / worked examples in
https://www.electronics-tutorials.ws/rc/rc_2.html

(edited 5 months ago)

Reply 6

OP

so is it as simple as Vc = 2?
270kilo ohms = 270000 ohms = 2.7^5?
25 micro farads = 0.000025 = 2.5^-5 ?
t = (RcLoge (Vc/15))-1?

I really appreciate this! thanks

(edited 5 months ago)

Reply 7

Original post by richjpark

so is it as simple as Vc = 2?
270kilo ohms = 270000 ohms = 2.7^5?
25 micro farads = 0.000025 = 2.5^-5 ?
t = (RcLoge (Vc/15))^-1?

I really appreciate this! thanks

1) Yes for vc
2) Bold isnt correct, you could type it into your calculator to check. Maybe 2.7 * 10^5 for instance?
3) Dividing by -1 is the same as multiplying by -1. Again you could simply type it into your calculator to check.

Reply 8

OP

Original post by mqb2766

1) Yes for vc
2) Bold isnt correct, you could type it into your calculator to check. Maybe 2.7 * 10^5 for instance?
3) Dividing by -1 is the same as multiplying by -1. Again you could simply type it into your calculator to check.

vc=2
R = 2.7x10^5
c = 2.5x10^-5

t = (2.7x10^5 ( 2.5x10^-5) loge (2/15))-1

Am i getting any closer?

Reply 9

Original post by richjpark

vc=2
R = 2.7x10^5
c = 2.5x10^-5

t = (2.7x10^5 ( 2.5x10^-5) loge (2/15))-1

Am i getting any closer?

Looks about right if the last -1 is a multiplication. To discharge to 2 from 15 should be around a couple of time constants (see previous link) so you can check your numbers if its not approximately that.

Reply 10

OP

yes the -1 was a multiplication. When entering it in to my calculator I get -14.60059539.
so is this -14.6 seconds? back to the brief does mean it takes 14.6seconds to produce an output? I've read over the page on the link a few times, will keep going over it until it makes more sense to me!

Reply 11

Original post by richjpark

yes the -1 was a multiplication. When entering it in to my calculator I get -14.60059539.
so is this -14.6 seconds? back to the brief does mean it takes 14.6seconds to produce an output? I've read over the page on the link a few times, will keep going over it until it makes more sense to me!

The time should be positive, but as youve used SI units, then time is in seconds.
-14.6 looks like youre still subtracting -1 rather than multiplying it by -1.

Id really google a few basic tutorials/youtube videos and pick a couple of appropriate ones, otherwise its pretty much guesswork how youre interpreting this discharging equation.

Reply 12

OP

ah yes I think it was me being fairly unfamiliar with my calculator. I've done it again and came up with 13.60059539
I really appreciate all your time with this, its been a huge help!

Reply 13

OP

Any help with my next question would be appreciated too!

one of your labrator instantaneous test signal voltages (Vs) is described by the equation....
Vs = 8sin (2 π f t - ( π / 4)
where f=500kHz and t represents time.
Make t the subject and hence determine the first point in time when the instantaneous signal voltage has a magnitude of +5v.

I have ended up with...

t = (sin^-1(5/8)+ π/4) / 2 π f

this looks remarkably similar to the answer of one of the worked examples in my workbook so I've got high hopes for that one?!

Would I be right in saying f = 5 x 10^5?

I then get 4.649005 x10^-7 which doesnt seem right

Reply 14