- #1
nelsyeung
- 11
- 0
I've started an experiment lately with charging a capacitor through a resistor, I wanted to prove that the equation of charging capacitor by using linear law, but it didn't quite work and needed help.
This is what I've done:
Equation of charging a capacitor:
V = [tex]V_{max}[/tex](1-[tex]e^{\frac{-t}{RC}}[/tex])
Multiply out of bracket:
V = [tex]V_{max}[/tex] - [tex]V_{max}[/tex] [tex]e^{\frac{-t}{RC}}[/tex]
Apply natural log to remove the e:
ln(V) = ln([tex]V_{max}[/tex]) - ln([tex]V_{max}[/tex] [tex]e^{\frac{-t}{RC}}[/tex])
Use laws of logs and ln(e) will cancels out:
ln(V) = ln([tex]V_{max}[/tex]) - ln([tex]V_{max}[/tex]) + [tex]\frac{-t}{RC}[/tex]
ln([tex]V_{max}[/tex]) - ln([tex]V_{max}[/tex]) = 0, therefore:
ln(V) = [tex]\frac{-t}{rc}[/tex]
So this tells me if I plot a graph with ln(V) against t, I should get a straight line, but it doesn't, the new graph looks identical to the first graph I drew only with different values, and also I found out that using that formula created by linear law, I can seem to get it back to the original equation.. :( Please help.
Thanks, Nels
This is what I've done:
Equation of charging a capacitor:
V = [tex]V_{max}[/tex](1-[tex]e^{\frac{-t}{RC}}[/tex])
Multiply out of bracket:
V = [tex]V_{max}[/tex] - [tex]V_{max}[/tex] [tex]e^{\frac{-t}{RC}}[/tex]
Apply natural log to remove the e:
ln(V) = ln([tex]V_{max}[/tex]) - ln([tex]V_{max}[/tex] [tex]e^{\frac{-t}{RC}}[/tex])
Use laws of logs and ln(e) will cancels out:
ln(V) = ln([tex]V_{max}[/tex]) - ln([tex]V_{max}[/tex]) + [tex]\frac{-t}{RC}[/tex]
ln([tex]V_{max}[/tex]) - ln([tex]V_{max}[/tex]) = 0, therefore:
ln(V) = [tex]\frac{-t}{rc}[/tex]
So this tells me if I plot a graph with ln(V) against t, I should get a straight line, but it doesn't, the new graph looks identical to the first graph I drew only with different values, and also I found out that using that formula created by linear law, I can seem to get it back to the original equation.. :( Please help.
Thanks, Nels