# How do i calculate capicitor in a RC circuit ?

Discussion in 'Homework Help' started by HNC-NEWBIE, May 23, 2016.

1. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
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0
Hi,

I'm just working through my HNC and struggling with this question yet its probably obvious but not to me.

Could some one point me to an explainer video that will help me solve this please.

a) A series electrical circuit features a capacitor (C) charging via a 1MΩ resistor (R) and a 12V dc supply (Vs). The voltage across the capacitor (Vc) may be described by the equation…

vc = vs (1 − e− ⁄)

where t represents time.

Assuming that Vc is 2V after a time of 4 seconds, determine the approximate value of the capacitor.

(b) A series electrical circuit features an inductor (L), a 10Ω resistor (R) and a 10V dc supply (Vs). The current (i) flowing in this circuit is given by the equation…

i = vs R (1 − e−Rt L⁄)

where t represents time.

Assuming that i is 0.632A after a time of 2.2μs, determine the approximate value of the inductor.

Many Thanks

2. ### ericgibbs AAC Fanatic!

Jan 29, 2010
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Last edited: May 24, 2016

May 23, 2016
14
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4. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
14
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I'm struggling , I still do not get it even after reading that.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
After what time the Capacitor voltage reach 0.167*Vs (12V * 0.167 ≈ 2V)? For C = 1F and R = 1Ω?

6. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
What is "e-/" supposed to mean?

Aside from what "e-Rt L/" is supposed to mean, this equation can't produce the correct result. Look at the units. The left hand side is 'i', which has units of amperes, while the right hand side has units of voltage multiplied by resistance, which is voltage-squared per ampere.

7. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
14
0
Sorry it didn't cut / paste correctly

It should be:

Vc=Vs (I-e. -t/Rc)

8. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Assuming you '.' means exponentiation, you still have a units problem. You factor in parentheses consists of a current added to a dimensionless quantity -- you can't add or subtract them.

Please use the tex editor for formulas. It causes a LOT less confusion.

The formula you are probably looking for is

$
V_C \; = \; V_s $$1 \; - \; e^{-\frac{t}{RC}}$$
$

This would be for a capacitor starting out discharged and charging to Vs through a resistance R.

Last edited: May 23, 2016
9. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
14
0
That's it though it should be -t the "." Was an accident

10. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Good catch -- I fixed the typo above.

11. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Can you take that equation and solve it for C in terms of the other variables?

If so, then you are told that Vc = 2 V when t = 4 s. You already know that Vs = 12 V and R = 1 MΩ.

12. ### MrAl Well-Known Member

Jun 17, 2014
2,433
490

Hi,

When you are given an equation for a circuit and you have to find a value, you usually just solve for that value then you have it. For example, if you were given:
y=A+B+C
and you had to solve for the C, you would just use algebra to manipulate it into:
C=y-A-B

and there you have it. Subtract A and B from both sides, and you are left with C on one side. That's it.

Vc=Vs*(1-e^(-t/RC))

you do the same thing, only you have to do a little more work because at some point you have to take the natural log of both sides. I'll start:
Vc=Vs*(1-e^(-t/RC))
The first step is to divide both sides by Vs and we get:
Vc/Vs=1-e^(-t/RC)
Can you take it from here?

Note that i used the shorthand RC=R*C so be careful with that.
Also, ln(e^A)=A just for reference.

13. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
14
0
All. I really appreciate your help on this , thanks for your time.

I just do not understand how I rearrange this part e^(-t/RC), I've looked and looked and looked on the internet for study videos but cannot find anything that explains the way to rearrange a TO THE POWER OFF:

14. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
14
0
HANG ON

so e^(-t/RC) = -t/RC

is that right ?

15. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
14
0
Vc/Vs +e = 1-t/RC

Vc/Vs+e=-t/RC

Vc/Vs+e+t=RC

Vc / Vs +e+t Divide all by R = C

Have I got it ??

16. ### MrAl Well-Known Member

Jun 17, 2014
2,433
490
Hi,

Well, you are close but you need to use the log function too as in y=ln(x).
I'll do an example using the generic e^(-a*t) here...

Vc=Vs*(1-e^(-a*t))

divide both sides by Vs and get:
Vc/Vs=1-e^(-a*t)

subtract 1 and get:
Vc/Vs-1=-e^(-a*t)

multiply by -1 and get:
1-Vc/Vs=e^(-a*t)

and now take the natural log of both sides and get:
ln(1-Vc/Vs)=-a*t

divide both sides by -t and get:
-ln(1-Vc/Vs)/t=a

and flip that around:
a=-ln(1-Vc/Vs)/t

See if you can do it with the actual equation now and finding RC, then just C.

17. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Nope.

One thing that they just seem to not teach much any more is the notion that the exponential function and the logarithm function are reciprocal functions, meaning that one undoes the other. Instead, they seem to just focus on which button on your calculator to punch when you see on of them in an expression.

So let's look at where these come from.

If I have

$
1000 \; = \; 10^3
$

Then I can describe this a couple of ways. One is to say that 1000 is equal to the base 10 raised to the exponent of 3. Another is to say that 3 is the "base-10 logarithm" of 1000.

Similarly, if I have

$
x \; = \; e^y
$

Then the base is 'e', which is just a constant equal to about 2.718. We have this special base because it crops up in mathematical in many different places. Similarly, the "base-e logarithm" is so common that we have a special name for it -- the "natural logarithm" -- and a special function for it -- ln(). In fact, it is so commonly used that frequently when someone just says "logarithm" they mean the natural log.

So instead of saying, "The base-e logarithm of x is y", we can just say, "The natural log of x is y," and we can express this as an equation with

$
\ln(x) \; = \; y
$

Now, from ordinary algebra you should be familiar with the notion that, given

$
a \; = \; b
$

That I can use both sides as the exponent to the same base and maintain the equality. So

$
e^a \; = \; e^b
$

If I do that with the expression above, I have

$
\ln(x) \; = \; y
e^{\ln(x)} \; = \; e^{y}
$

But we already know that the right hand side, e^y, is equal to x. So we can conclude that

$
e^{\ln(x)} \; = \; x
$

We can similarly work though some algebra to conclude that

$
\ln(e^x) \; = \; x
$

Bottom line -- when you need to eliminate an exponential, you take the log; when you need to eliminate a log, you take the exponential. But be careful to first rearrange things so that the thing you are eliminating is all by itself on one side of the equation.

18. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
14
0
Thanks guys ,

Im sure i have it now,

C=-t/RIn(1-Vc/Vs)

19. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
14
0
Thanks guys ,

Im sure i have it now,

C=-t/RIn(1-Vc/Vs)

20. ### HNC-NEWBIE Thread Starter New Member

May 23, 2016
14
0
Vc=Vs(1-e^-t/Rc)

Vc/Vs = 1-e^-t/Rc

Vc/Vs-1=-e^-t/Rc

In(1-Vc/Vs)=-t/Rc

C=-t/RIn(1-Vc/Vs)

????

Have I sussed it ?