Capacitor

[cseet]

Hi all

I've a question to the following and would be greatly appreciated if anyone could lend a hand....

Question:
A series combination of a 12 kilo ohm and an unknown capacitor is connected to a 12 V battery. One second after the circuit is completed the voltage across the resistor is 2 V. How do I find out the capacitor?

pls help me to understand and problem...

thanks
cseet

[tigrot]

V on capacitor = 12V - 2V = 10V

now,charging formula for a capacitor is: Vs=Vi(1-[e to the power of t/RC])
where Vs is the supply voltage
Vi is the voltage on capacitor
t is the time
R is the resistor
and C is the capacitance

therefor your formula will becoma 12=10(1- [e to the power of 1/12k * C)

all you have to do is just subject of the formula

[flexten]

Also remember that the value of capacitance for a parallel plate capacitor is:

C= Ae/t where e is the dielectric constant, A is the area of the plates and t is the distance between the plates. Usually values of e are given relative, so multiply them by 8.854E-12 farad/meter to get actual material dielectric (permitivity) constant.

[Chen]

tigrot made a little mistake, the power of e is negative:
V_{(t)} = V_f(1 - e^{-\frac{t}{RC}})
1 - \frac{V_{(t)}}{V_f} = e^{-\frac{t}{RC}}
And then just use ln()...

[tigrot]

Thank you for correcting me Chen.You are correct.

[cseet]

Hi there,

thanks for all your replies. they're really helpful and takes the pain away from learning Physics!

thanks again!
Cseet

[expscv]

yeah ppl are so helpful here

[cseet]

Hi,
I came to an answer of 15C.

the following is how I worked out... correct me if I'm wrong... thanks

12 = 10 (1-exo(-1/12C))
12 = 10 * 0.08 * C
C = 12 / (10 * 0.08)
C = 15 C

am I correct?
thanks
cseet

[cseet]

Hi,
I came to an answer of 15C.

the following is how I worked out... correct me if I'm wrong... thanks

12 = 10 (1-exo(-1/12C))
12 = 10 * 0.08 * C
C = 12 / (10 * 0.08)
C = 15 C

am I correct?
thanks
cseet

[cseet]

hi guys,

pls ignore my above equation.... I got it all wrong... fond out the answer... thanks anyway...
cseet

[Chen]

I don't understand how you got from here:
12 = 10(1 - e^{-\frac{1}{12C}})
To here:
12 = 10 (0.08C)
I don't think you read the equations above right. The capacity C is part of the power of e, you cannot leave it out and only raise e to the power of 1/12. Do as I said - from this:
1 - \frac{V_{(t)}}{V_f} = e^{-\frac{t}{RC}}
Take the ln() of both sides to get:
ln(1 - \frac{V_{(t)}}{V_f}) = -\frac{t}{RC}}
C = -\frac{t}{R ln(1 - \frac{V_{(t)}}{V_f})}}

The answer I get is C = 46.5 micro-farads.

[cseet]

sorry this may be a very stupid question... but what does In() mean? and where's exp in this equation.

I do apologise for my ignorance....
cseet

[Chen]

ln is Logarithm Natural, i.e log with the natural base e. If:
b = a^x
Then:
x = \log _ab = \log _aa^x
With ln, the base of the log is simply e. So if:
b = e^x
Then:
x = \ln b = \ln e^x

So when you have this equation:
1 - \frac{V_{(t)}}{V_f} = e^{-\frac{t}{RC}}
You are allowed to take the ln of both sides:
\ln (1 - \frac{V_{(t)}}{V_f}) = \ln e^{-\frac{t}{RC}}
The left side stays the same, and the right side just becomes -\frac{t}{RC}.