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Capacitance equation 
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#1
Apr2606, 10:51 AM

P: 41

how would i rearrange this equation to find the capacitance, c?
V=Vo exp(t/RC) 


#2
Apr2606, 11:05 AM

P: 925

Do you know what the inverse of the exponential function is?



#3
Apr2606, 11:16 AM

P: 41

would it go to,
V = Vo ln + (t/RC) then to, V = ln Vo  t x 1/RC 


#4
Apr2606, 11:22 AM

P: 925

Capacitance equation
You're right that you use the natural log function, but your form is incorrect. What would happen if you take the natural log of the exponential function, or ln(e^x) = ?



#5
Apr2606, 11:31 AM

P: 41

ln(e) + x ?????
not really sure you've lost me a bit sorry 


#6
Apr2606, 11:44 AM

Mentor
P: 41,109

log( 10^2 ) = 2 (right?) log( 10^x ) = ? ln( e^x ) is analogous to log( 10^x )... 


#7
Apr2606, 11:50 AM

P: 41

log(10^x) = x ?
analogous?? The same as?? but log e = 1 does that mean log e^x = 1^x ? therefore log e^x = x ? 


#8
Apr2606, 12:12 PM

Mentor
P: 41,109

log( 10^x ) = x ln( e^x ) = ? Try a few numbers on your calculator to help you keep it straight. There's a reason that most calculators overload the log() key with 10^x and overload the ln() key with e^x..... 


#9
Apr2706, 09:59 AM

P: 777

If this is your original equation, then: Furthur evaluation is simple. 


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