# Homework Help: How do I find out the capacitor?

1. Mar 28, 2004

### 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

2. Mar 28, 2004

### 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

3. Mar 28, 2004

### 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.

4. Mar 28, 2004

### 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()...

5. Mar 29, 2004

### tigrot

Thank you for correcting me Chen.You are correct.

6. Mar 29, 2004

### cseet

re

Hi there,

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

thanks again!
Cseet

7. Mar 29, 2004

### expscv

yeah ppl are so helpful here

8. Mar 29, 2004

### 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

9. Mar 29, 2004

### 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

10. Mar 29, 2004

### cseet

hi guys,

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

11. Mar 29, 2004

### 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:
$$ln(1 - \frac{V_{(t)}}{V_f}) = -\frac{t}{RC}}$$
$$C = -\frac{t}{R ln(1 - \frac{V_{(t)}}{V_f})}}$$

12. Mar 29, 2004

### 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

13. Mar 29, 2004

### 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 $-\frac{t}{RC}$.