I need help understanding!!

Aug 21, 2014

Help, I found a calculation I am having trouble with. I have a poor background in math and would like to understand in plain English how to solve for Vc.
Vc = V * (1- e^(-t / R*C))
This where I got it...OH! I can't include a screen grab, so here goes...

Vc is the capacitor voltage, V is the supply voltage,
R is the resistance, C is the Capacitance, t is the time,
and e = 2.71828

The numbers are V= 13.8V
R= 33K ohms
C= 66mF
t= 2000 milliseconds
e= 2.71828
I think I've got the ending part R*C =2178 and that gets divided by ??? -t
I get, e^ means e to the power of the last part (-t / R*C)
My biggest problem is I don't understand the minus signs 1- and the last thing is to multiply by V and that should equal Vc.
Thanks for looking and any help anyone could give will be appreciated.
Jerry

RobertMacy · Aug 22, 2014

On Thu, 21 Aug 2014 11:40:50 -0700, <jweigh@gmail.com> wrote:

Help, I found a calculation I am having trouble with. I have a poor
background in math and would like to understand in plain English how to
solve for Vc.
Vc = V * (1- e^(-t / R*C))
This where I got it...OH! I can't include a screen grab, so here goes...

Vc is the capacitor voltage, V is the supply voltage,
R is the resistance, C is the Capacitance, t is the time,
and e = 2.71828

The numbers are V= 13.8V
R= 33K ohms
C= 66mF
t= 2000 milliseconds
e= 2.71828
I think I've got the ending part R*C =2178 and that gets divided by ???
-t
I get, e^ means e to the power of the last part (-t / R*C)
My biggest problem is I don't understand the minus signs 1- and the last
thing is to multiply by V and that should equal Vc.
Thanks for looking and any help anyone could give will be appreciated.
Jerry

homework? is ok, been there, too

The best way to solve math problems is to simply start 'poking' at them.

When confronted with an exponential, e, or a power of 10,you've first got
to get everything to be 'like' with 'like'

How to do that? First divide both sides by V, changing to
Vc/V=(1-e^(-t/R*C))

oh, oh, still not got 'like' with 'like', so add the exponential to both
sides AND subtract Vc/V from both sides, leaving:
e^(-t/R*C)=1-Vc/V
now like is with like, and you can take the natural log of BOTH sides and
keep going.

Jim Thompson · Aug 22, 2014

On Thu, 21 Aug 2014 12:05:31 -0700, RobertMacy
<robert.a.macy@gmail.com> wrote:

On Thu, 21 Aug 2014 11:40:50 -0700, <jweigh@gmail.com> wrote:

Help, I found a calculation I am having trouble with. I have a poor
background in math and would like to understand in plain English how to
solve for Vc.
Vc = V * (1- e^(-t / R*C))
This where I got it...OH! I can't include a screen grab, so here goes...

Vc is the capacitor voltage, V is the supply voltage,
R is the resistance, C is the Capacitance, t is the time,
and e = 2.71828

The numbers are V= 13.8V
R= 33K ohms
C= 66mF
t= 2000 milliseconds
e= 2.71828
I think I've got the ending part R*C =2178 and that gets divided by ???
-t
I get, e^ means e to the power of the last part (-t / R*C)
My biggest problem is I don't understand the minus signs 1- and the last
thing is to multiply by V and that should equal Vc.
Thanks for looking and any help anyone could give will be appreciated.
Jerry

homework? is ok, been there, too

The best way to solve math problems is to simply start 'poking' at them.

When confronted with an exponential, e, or a power of 10,you've first got
to get everything to be 'like' with 'like'

How to do that? First divide both sides by V, changing to
Vc/V=(1-e^(-t/R*C))

oh, oh, still not got 'like' with 'like', so add the exponential to both
sides AND subtract Vc/V from both sides, leaving:
e^(-t/R*C)=1-Vc/V
now like is with like, and you can take the natural log of BOTH sides and
keep going.

Forget the Algebra... go golfing >:-}

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: skypeanalog | |
| Voice

Tom Biasi · Aug 22, 2014

On 8/21/2014 2:40 PM, jweigh@gmail.com wrote:

Help, I found a calculation I am having trouble with. I have a poor background in math and would like to understand in plain English how to solve for Vc.
SNIP.
Jerry

It's not an English problem it's a math problem.

Maynard A. Philbrook Jr. · Aug 22, 2014

In article <7d4d3143-589e-4cb2-a941-f83a2564fbb4@googlegroups.com>,
jweigh@gmail.com says...

Help, I found a calculation I am having trouble with. I have a poor background in math and would like to understand in plain English how to solve for Vc.
Vc = V * (1- e^(-t / R*C))
This where I got it...OH! I can't include a screen grab, so here goes...

Vc is the capacitor voltage, V is the supply voltage,
R is the resistance, C is the Capacitance, t is the time,
and e = 2.71828

The numbers are V= 13.8V
R= 33K ohms
C= 66mF
t= 2000 milliseconds
e= 2.71828
I think I've got the ending part R*C =2178 and that gets divided by ??? -t
I get, e^ means e to the power of the last part (-t / R*C)
My biggest problem is I don't understand the minus signs 1- and the last thing is to multiply by V and that should equal Vc.
Thanks for looking and any help anyone could give will be appreciated.
Jerry

Normally I don't do this however.. Let me take a crack at it.
e^1 = 2.71828 as you have found.
e^-1= 0.3678 so what does this mean?
If you were to do this;
1 / 2.71828, it would equal 0.3678

do you see the correlation here? When e^ has a "- sign" it returns
the reciprocal (1/result) of the answer. That is, what ever math is
going on for the ^??? part, that returns a -sign number, will return
the inverse of the results.

P.S.
use 2 secs instead of 2000 msec in the calculations.

with out giving you the answer, I can tell you there isn't
much of a charge in the cap after 2 secs. So little that you'd
need a high end DMM to see it.

Jamie

George Herold · Aug 22, 2014

On Thursday, August 21, 2014 2:40:50 PM UTC-4, jwe...@gmail.com wrote:

Help, I found a calculation I am having trouble with. I have a poor background in math and would like to understand in plain English how to solve for Vc.

Vc = V * (1- e^(-t / R*C))
This where I got it...OH! I can't include a screen grab, so here goes...

Vc is the capacitor voltage, V is the supply voltage,
R is the resistance, C is the Capacitance, t is the time,
and e = 2.71828

The numbers are V= 13.8V

R= 33K ohms

C= 66mF

t= 2000 milliseconds

e= 2.71828

I think I've got the ending part R*C =2178 and that gets divided by ??? -t
I get, e^ means e to the power of the last part (-t / R*C)
My biggest problem is I don't understand the minus signs 1- and the last thing is to multiply by V and that should equal Vc.
Thanks for looking and any help anyone could give will be appreciated.

Jerry

Hi Jerry, Well maybe you need to review some basic algebra.
The equation describes the charging of a capacitor through a resistor. So at time = 0, Vc = 0 (no voltage on C) and at time = infinity the cap should be charge fully. Vc=V.

One of the first things I like to do is check these initial and final conditions and make sure the equation works correctly. So as the first part to answering your question I'll ask you to find Vc(t=0) and Vc(t=infinity). Can you repeat my results?

George H.

Cydrome Leader · Aug 22, 2014

George Herold <gherold@teachspin.com> wrote:

On Thursday, August 21, 2014 2:40:50 PM UTC-4, jwe...@gmail.com wrote:
Help, I found a calculation I am having trouble with. I have a poor background in math and would like to understand in plain English how to solve for Vc.

Vc = V * (1- e^(-t / R*C))
This where I got it...OH! I can't include a screen grab, so here goes...

Vc is the capacitor voltage, V is the supply voltage,
R is the resistance, C is the Capacitance, t is the time,
and e = 2.71828

The numbers are V= 13.8V

R= 33K ohms

C= 66mF

t= 2000 milliseconds

e= 2.71828

I think I've got the ending part R*C =2178 and that gets divided by ??? -t
I get, e^ means e to the power of the last part (-t / R*C)
My biggest problem is I don't understand the minus signs 1- and the last thing is to multiply by V and that should equal Vc.
Thanks for looking and any help anyone could give will be appreciated.

Jerry

Hi Jerry, Well maybe you need to review some basic algebra.
The equation describes the charging of a capacitor through a resistor. So at time = 0, Vc = 0 (no voltage on C) and at time = infinity the cap should be charge fully. Vc=V.

One of the first things I like to do is check these initial and final conditions and make sure the equation works correctly. So as the first part to answering your question I'll ask you to find Vc(t=0) and Vc(t=infinity). Can you repeat my results?

George H.

Are you folks really reading the "mF" as "millifarads"? I think even the
East Germans stopped using nonsense units like that, decades ago.

For the OP, take a look at a RC time constant chart. All that matters is 0
to 5 time constants. My math shows that in this case, in 2.178 seconds
(one time constant) that cap will reach about 63% of 13.8 volts. The rate
of charge decreases over time.

Helmut Wabnig · Aug 22, 2014

On Thu, 21 Aug 2014 11:40:50 -0700 (PDT), jweigh@gmail.com wrote:

Help, I found a calculation I am having trouble with. I have a poor background in math and would like to understand in plain English how to solve for Vc.
Vc = V * (1- e^(-t / R*C))
This where I got it...OH! I can't include a screen grab, so here goes...

Vc is the capacitor voltage, V is the supply voltage,
R is the resistance, C is the Capacitance, t is the time,
and e = 2.71828

The numbers are V= 13.8V
R= 33K ohms
C= 66mF
t= 2000 milliseconds
e= 2.71828
I think I've got the ending part R*C =2178 and that gets divided by ??? -t
I get, e^ means e to the power of the last part (-t / R*C)
My biggest problem is I don't understand the minus signs 1- and the last thing is to multiply by V and that should equal Vc.
Thanks for looking and any help anyone could give will be appreciated.
Jerry

That's what your function looks like,
I inserted some numerical values.

http://img540.imageshack.us/img540/6416/R0B17U.gif

w.

Jim Thompson · Aug 23, 2014

On Thu, 21 Aug 2014 12:05:31 -0700, RobertMacy
<robert.a.macy@gmail.com> wrote:

On Thu, 21 Aug 2014 11:40:50 -0700, <jweigh@gmail.com> wrote:

Help, I found a calculation I am having trouble with. I have a poor
background in math and would like to understand in plain English how to
solve for Vc.
Vc = V * (1- e^(-t / R*C))
This where I got it...OH! I can't include a screen grab, so here goes...

Vc is the capacitor voltage, V is the supply voltage,
R is the resistance, C is the Capacitance, t is the time,
and e = 2.71828

The numbers are V= 13.8V
R= 33K ohms
C= 66mF
t= 2000 milliseconds
e= 2.71828
I think I've got the ending part R*C =2178 and that gets divided by ???
-t
I get, e^ means e to the power of the last part (-t / R*C)
My biggest problem is I don't understand the minus signs 1- and the last
thing is to multiply by V and that should equal Vc.
Thanks for looking and any help anyone could give will be appreciated.
Jerry

homework? is ok, been there, too

The best way to solve math problems is to simply start 'poking' at them.

When confronted with an exponential, e, or a power of 10,you've first got
to get everything to be 'like' with 'like'

How to do that? First divide both sides by V, changing to
Vc/V=(1-e^(-t/R*C))

oh, oh, still not got 'like' with 'like', so add the exponential to both
sides AND subtract Vc/V from both sides, leaving:
e^(-t/R*C)=1-Vc/V
now like is with like, and you can take the natural log of BOTH sides and
keep going.

Maynard A. Philbrook Jr. · Aug 23, 2014

In article <lt7opj$ep2$1@reader1.panix.com>, presence@MUNGEpanix.com
says...

The equation describes the charging of a capacitor through a resistor. So at time = 0, Vc = 0 (no voltage on C) and at time = infinity the cap should be charge fully. Vc=V.

One of the first things I like to do is check these initial and final conditions and make sure the equation works correctly. So as the first part to answering your question I'll ask you to find Vc(t=0) and Vc(t=infinity). Can you repeat my results?

George H.

Are you folks really reading the "mF" as "millifarads"? I think even the
East Germans stopped using nonsense units like that, decades ago.

For the OP, take a look at a RC time constant chart. All that matters is 0
to 5 time constants. My math shows that in this case, in 2.178 seconds
(one time constant) that cap will reach about 63% of 13.8 volts. The rate
of charge decreases over time.

really? I still use it... yes, I read as millifarards. If he meant
something else then he better properly specify it.

Maybe you're thinking of the "mmF"?

Jamie

Tom Biasi · Aug 23, 2014

On 8/22/2014 6:01 PM, Maynard A. Philbrook Jr. wrote:

In article <lt7opj$ep2$1@reader1.panix.com>, presence@MUNGEpanix.com
says...
The equation describes the charging of a capacitor through a resistor. So at time = 0, Vc = 0 (no voltage on C) and at time = infinity the cap should be charge fully. Vc=V.

One of the first things I like to do is check these initial and final conditions and make sure the equation works correctly. So as the first part to answering your question I'll ask you to find Vc(t=0) and Vc(t=infinity). Can you repeat my results?

George H.

Are you folks really reading the "mF" as "millifarads"? I think even the
East Germans stopped using nonsense units like that, decades ago.

For the OP, take a look at a RC time constant chart. All that matters is 0
to 5 time constants. My math shows that in this case, in 2.178 seconds
(one time constant) that cap will reach about 63% of 13.8 volts. The rate
of charge decreases over time.

really? I still use it... yes, I read as millifarards. If he meant
something else then he better properly specify it.

Maybe you're thinking of the "mmF"?

Jamie

In the old days it would be micro farad and mmF would be micro micro
Farad or picoFarad in today's terms. By his calculation I assumed he
meant milli Farad. I agree also that the term milli Farad is obscure.

John Fields · Aug 23, 2014

On Thu, 21 Aug 2014 11:40:50 -0700 (PDT), jweigh@gmail.com wrote:

Help, I found a calculation I am having trouble with. I have a poor background in math and would like to understand in plain English how to solve for Vc.
Vc = V * (1- e^(-t / R*C))
This where I got it...OH! I can't include a screen grab, so here goes...

Vc is the capacitor voltage, V is the supply voltage,
R is the resistance, C is the Capacitance, t is the time,
and e = 2.71828

The numbers are V= 13.8V
R= 33K ohms
C= 66mF
t= 2000 milliseconds
e= 2.71828
I think I've got the ending part R*C =2178 and that gets divided by ??? -t
I get, e^ means e to the power of the last part (-t / R*C)
My biggest problem is I don't understand the minus signs 1- and the last thing is to multiply by V and that should equal Vc.
Thanks for looking and any help anyone could give will be appreciated.
Jerry

---
Here's the easy way:

https://www.dropbox.com/s/e1dpv0h2nvcr8j4/scan0003.png?dl=0

John Fields

RobertMacy · Aug 23, 2014

On Fri, 22 Aug 2014 18:58:38 -0700, Tom Biasi <tombiasi@optonline.net>
wrote:

...snip...
In the old days it would be micro farad and mmF would be micro micro
Farad or picoFarad in today's terms. By his calculation I assumed he
meant milli Farad. I agree also that the term milli Farad is obscure.

the terms F, mF, uF, nF, pF are all alive and well, being used in PSpice
and LTspice.

I think fF is interpreted as Farad, though.

Cydrome Leader · Aug 23, 2014

Maynard A. Philbrook Jr. <jamie_ka1lpa@charter.net> wrote:

In article <lt7opj$ep2$1@reader1.panix.com>, presence@MUNGEpanix.com
says...
The equation describes the charging of a capacitor through a resistor. So at time = 0, Vc = 0 (no voltage on C) and at time = infinity the cap should be charge fully. Vc=V.

One of the first things I like to do is check these initial and final conditions and make sure the equation works correctly. So as the first part to answering your question I'll ask you to find Vc(t=0) and Vc(t=infinity). Can you repeat my results?

George H.

Are you folks really reading the "mF" as "millifarads"? I think even the
East Germans stopped using nonsense units like that, decades ago.

For the OP, take a look at a RC time constant chart. All that matters is 0
to 5 time constants. My math shows that in this case, in 2.178 seconds
(one time constant) that cap will reach about 63% of 13.8 volts. The rate
of charge decreases over time.

really? I still use it... yes, I read as millifarards. If he meant
something else then he better properly specify it.

Do you use kilofeet and decimeters too? Goofy units should be avoided.

Jim Thompson · Aug 23, 2014

On Sat, 23 Aug 2014 07:11:16 -0700, RobertMacy
<robert.a.macy@gmail.com> wrote:

On Fri, 22 Aug 2014 18:58:38 -0700, Tom Biasi <tombiasi@optonline.net
wrote:

...snip...
In the old days it would be micro farad and mmF would be micro micro
Farad or picoFarad in today's terms. By his calculation I assumed he
meant milli Farad. I agree also that the term milli Farad is obscure.

the terms F, mF, uF, nF, pF are all alive and well, being used in PSpice
and LTspice.

I think fF is interpreted as Farad, though.

Maynard A. Philbrook Jr. · Aug 24, 2014

In article <ltap3e$h4g$1@reader1.panix.com>, presence@MUNGEpanix.com
says...

Maynard A. Philbrook Jr. <jamie_ka1lpa@charter.net> wrote:
In article <lt7opj$ep2$1@reader1.panix.com>, presence@MUNGEpanix.com
says...
The equation describes the charging of a capacitor through a resistor. So at time = 0, Vc = 0 (no voltage on C) and at time = infinity the cap should be charge fully. Vc=V.

One of the first things I like to do is check these initial and final conditions and make sure the equation works correctly. So as the first part to answering your question I'll ask you to find Vc(t=0) and Vc(t=infinity). Can you repeat my results?

George H.

Are you folks really reading the "mF" as "millifarads"? I think even the
East Germans stopped using nonsense units like that, decades ago.

For the OP, take a look at a RC time constant chart. All that matters is 0
to 5 time constants. My math shows that in this case, in 2.178 seconds
(one time constant) that cap will reach about 63% of 13.8 volts. The rate
of charge decreases over time.

really? I still use it... yes, I read as millifarards. If he meant
something else then he better properly specify it.

Do you use kilofeet and decimeters too? Goofy units should be avoided.

Oh gee, brand me a odd ball, I still use Webers..

I bet you still use stones!

Jamie

Aug 25, 2014

High again guys, Sorry now I realized I made an error in the info I posted after reading some of the replies...... the correct info would be: The numbers are
V= 13.8V
R= 33K ohms
***C= 66uF***
**t= 2 seconds**
e= 2.71828
What I'm expecting is the cap will reach 63% of full charge(13.8V)at around 2 seconds this will be used within a circuit to turn on a relay so there is at least 2-3 seconds delay of the relay closing turning on a circuit to charge a battery.
So I guess I need to know if this R/C combo will give me the result I hope for

Thanks for the replies and help, It is greatly appreciated.

Jerry

Maynard A. Philbrook Jr. · Aug 26, 2014

In article <30badaa9-1837-411a-a4bf-dd434b1f429d@googlegroups.com>,
jweigh@gmail.com says...

High again guys, Sorry now I realized I made an error in the info I posted after reading some of the replies...... the correct info would be: The numbers are
V= 13.8V
R= 33K ohms
***C= 66uF***
**t= 2 seconds**
e= 2.71828
What I'm expecting is the cap will reach 63% of full charge(13.8V)at around 2 seconds this will be used within a circuit to turn on a relay so there is at least 2-3 seconds delay of the relay closing turning on a circuit to charge a battery.
So I guess I need to know if this R/C combo will give me the result I hope for

Thanks for the replies and help, It is greatly appreciated.

Jerry

Your first mistake was indicating that you were looking for
63% which is a text book figure.

May I suggest that you work a little harder on your home work.

Jamie

George Herold · Aug 26, 2014

On Sunday, August 24, 2014 8:37:35 PM UTC-4, jwe...@gmail.com wrote:

High again guys, Sorry now I realized I made an error in the info I posted after reading some of the replies...... the correct info would be: The numbers are

V= 13.8V

R= 33K ohms

***C= 66uF***

**t= 2 seconds**

e= 2.71828

What I'm expecting is the cap will reach 63% of full charge(13.8V)at around 2 seconds this will be used within a circuit to turn on a relay so there is at least 2-3 seconds delay of the relay closing turning on a circuit to charge a battery.

So I guess I need to know if this R/C combo will give me the result I hope for

Thanks for the replies and help, It is greatly appreciated.

Jerry

Yup, tau is the RC time constant, an Ohm times a Farad is a second.
George H.

I need help understanding!!

Guest

RobertMacy

Guest

Jim Thompson

Guest

Tom Biasi

Guest

Maynard A. Philbrook Jr.

Guest

George Herold

Guest

Cydrome Leader

Guest

Helmut Wabnig

Guest

Jim Thompson

Guest

Maynard A. Philbrook Jr.

Guest

Tom Biasi

Guest

John Fields

Guest

RobertMacy

Guest

Cydrome Leader

Guest

Jim Thompson

Guest

Maynard A. Philbrook Jr.

Guest

Guest

Maynard A. Philbrook Jr.

Guest

George Herold

Guest

Log in

Welcome to EDABoard.com

Sponsor