The voltage across a charged capacitor, measured in volts, is given by the equation [tex]V(t) = 4(2.72)^{\frac{-t}{2}}[/tex], where [tex]t[/tex] is measured in seconds since the current was removed. Include at least three decimal places in all your answers.
a. What is the voltage after 2 seconds?
b. What is the 1-second growth factor?
c. When will the voltage be 1 volt?
d. When will the voltage be [tex]x[/tex] volts? (Your answer should be an exact expression with [tex]x[/tex] in it)
Answer (2)
So first you plug 2 in for t 4(2.72)^(-2/2) 4/2.72 (because the simplified exponent is -1) divide and that's your answer.
Next you plug 1 in for t 4(2.72)^(-1/2) 4/sqrt(2.72) (because -1/2 as an exponent is 1/sqrt(x)) plug that into a calculator and that's your answer
For c, you plug 1 in for V(t). 1=4(2.72)^(-t/2) 1/4=2.72^(-t/2) Here is where I get stuck because I haven't done logarithms in a year and forget how to do them but I know that's what you have to do for this problem.
In the last problem you plug x in for V(t) and I'd assume go about solving it the same as c. Sorry I couldn't be of more help.
The voltage after 2 seconds is approximately 1.471 volts, and the 1-second growth factor is about 2.426. The time when the voltage reaches 1 volt can be calculated using logarithms, resulting in the expression t = − 2 ⋅ l o g 10 ( 2.72 ) l o g 10 ( 4 1 ) . A similar approach can be used to find the time when the voltage equals x volts.
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