- #1

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## Homework Statement

According to Equation 20.7, an ac voltage

*V*is given as a function of time

*t*by

*V*=

*Vo*sin 2

*, where
ft*

*Vo*is the peak voltage and

*f*is the frequency (in hertz). For a frequency of 64.7 Hz, what is the smallest value of the time at which the voltage equals one-half of the peak-value?

## Homework Equations

V = Vo sin2(3.14)ft

## The Attempt at a Solution

1/2Vo = Vo sin(2*3.14*64.7*t)

I know I set this equation wrong with the 1/2Vo but I don't understand how to do that part, the answer I got this way is .0738 V