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dan greig
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how would i rearrange this equation to find the capacitance, c?
V=Vo exp(-t/RC)
V=Vo exp(-t/RC)
Let's do it first with the more familiar base 10.dan greig said:ln(e) + x ?
not really sure you've lost me a bit sorry
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.dan greig said: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 ?
How come you did not take the natural log on the left hand side?dan greig said:would it go to,
V = Vo ln + (-t/RC)
then to,
V = ln Vo - t x 1/RC
[tex]\ln V = \ln V_0 - {t\over RC}[/tex]V=Vo exp(-t/RC)
The capacitance equation, also known as the charging equation, is given by V=Vo exp(-t/RC), where V is the voltage across a capacitor, Vo is the initial voltage, t is the time, R is the resistance, and C is the capacitance.
You can rearrange the capacitance equation by isolating the capacitance term on one side of the equation. This can be done by dividing both sides by the exponential term and then multiplying by the time and resistance. The resulting equation is C= -t/(Rln(V/Vo)).
The exponential term in the capacitance equation represents the decay of voltage over time in a charging capacitor. It is a characteristic of the capacitor and is determined by the time constant, which is the product of resistance and capacitance (RC).
The time constant (RC) is directly related to the capacitance of a capacitor. A higher capacitance will result in a longer time constant, meaning the capacitor will take longer to charge or discharge. Conversely, a lower capacitance will result in a shorter time constant.
Yes, the capacitance equation can be used for any type of capacitor as long as the other variables (voltage, time, and resistance) are known. However, in some cases, the equation may need to be modified to account for factors such as leakage current or nonlinear behavior of the capacitor.