You place an AC voltage signal of amplitude 2.0 V and with a frequency of 1100.0 Hz, into the input of an oscilloscope. (Note, you will not include units in your answers.) (a) Find the distance displayed on the screen of the oscilloscope for a full wave, if the sweep speed is adjusted to 455 µsec/DIV. divisions

Respuesta :

Answer:

distance displayed on the screen of the oscilloscope for a full wave is 2 DIV

Explanation:

given data

amplitude  = 2.0 V

frequency = 1100.0 Hz

solution

we get here time period of input voltage that is express as

T = [tex]\frac{1}{f}[/tex]    ...........1

put here value

T = [tex]\frac{1}{1100}[/tex]    

T = 0.909 × [tex]10^{-3}[/tex] s

and now we get here

distance x horizontal distance that is

x = [tex]\frac{T}{sweep\ speed}[/tex]    

x = [tex]\frac{0.909\times 10^{-3}}{455\times 10^{-6}}[/tex]  

x = 2 DIV