Find Capacitance: Rearranging the Capacitance Equation V=Vo exp(-t/RC)

[dan greig]

how would i rearrange this equation to find the capacitance, c?

V=Vo exp(-t/RC)

 

[daveb]

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

 

[dan greig]

would it go to,

V = Vo ln + (-t/RC)

then to,

V = ln Vo - t x 1/RC

 

[daveb]

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) = ?

 

[dan greig]

ln(e) + x ?

not really sure you've lost me a bit sorry

 

[berkeman]

ln(e) + x ?

not really sure you've lost me a bit sorry

Let's do it first with the more familiar base 10.

log( 10^2 ) = 2 (right?)

log( 10^x ) = ?

ln( e^x ) is analogous to log( 10^x )...

 

[dan greig]

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 ?

 

[berkeman]

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 ?

No, you need to keep your log() and ln() straight. log() is used with base 10 math, and ln() is used with base e math.

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...

 

[Reshma]

would it go to,

V = Vo ln + (-t/RC)

then to,

V = ln Vo - t x 1/RC

How come you did not take the natural log on the left hand side?
If this is your original equation, then:

V=Vo exp(-t/RC)

$$\ln V = \ln V_0 - {t\over RC}$$

Furthur evaluation is simple.