How do I find out the capacitor?

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The discussion revolves around calculating the capacitance of an unknown capacitor in a circuit with a known resistor and voltage. The charging formula for the capacitor is provided, and participants clarify the correct application of logarithmic functions in the calculations. One user initially miscalculates the capacitance but later corrects themselves, while another participant arrives at a capacitance value of 46.5 microfarads using the correct logarithmic approach. The conversation emphasizes the importance of understanding the mathematical principles behind capacitor charging in physics. Overall, the thread highlights collaborative problem-solving in learning electrical concepts.
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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
 
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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
 
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.
 
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()...
 
Thank you for correcting me Chen.You are correct.
 
re

Hi there,

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

thanks again!
Cseet
 
yeah ppl are so helpful here
 
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
 
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
hi guys,

pls ignore my above equation... I got it all wrong... fond out the answer... thanks anyway...
cseet
 
  • #11
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.
 
  • #12
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
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}.
 
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