. Consider this waveform equation vx=80sin(1000t) <?xml:namespace prefix = o ns
ID: 1938193 • Letter: #
Question
. Consider this waveform equation
vx=80sin(1000t)
<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /?>
<?xml:namespace prefix = v ns = "urn:schemas-microsoft-com:vml" /?>
: a. What is the frequency of the waveform (in Hz)?
b. What is the voltage Vx at a time t = 1.5 ms?
c. What is the voltage Vx at a time t = 3 ms?
d. What is the voltage Vx at a time t = 4.5 ms? 7.
What is the rms voltage of the waveform from question 6?
8. An AC signal has an rms voltage of 125 mV. What is the peak voltage of the signal?
Explanation / Answer
Vx=80sin(1000t) which looks like Vx=Asin(wt) where w=2*pi*f comparing above eq'ns 2*pi*f=1000 which gives f=159.23 Hz at t=1.5ms Vx=80sin(1000*1.5*10^-3)=2.09 V at t=3 ms Vx=80sin(1000*3*10^-3)=4.18 V at t=4.5ms Vx=80sin(1000*4.5*10^-3)=6.27 V rms voltage is given by Vpeak/sqrt(2) = 56.56 v 8. Vrms=Vpeak/sqrt(2) so I peak = 176.77 mV
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