Question

You place an AC voltage signal of amplitude 1.4 V and with a frequency of 1400.0 Hz, into the input of an oscilloscope.

a.) Find the distance displayed on the screen of the oscilloscope for a full wave, if the sweep speed is adjusted to 119 �sec/DIV. (in divisions)



b.) If the vertical gain is set to 0.40 V/DIV, how many divisions will the wave span from peak-to-peak of the wave? (in divisions)

Answer 1

(a). The horizontal distance on the screen is 6 Div.

(b). The number of division from peak to peak is 7 Div.

Step-by-step explanation:

Given that,

Amplitude = 1.4 V

Frequency = 1400.0 Hz

Vertical gain = 0.40 V/Div

Sweep speed
`v= 119 *10^(-6)\ sec/Div`

We need to calculate the period

Using formula of period of input voltage

`T=(1)/(f)`

Put the value into the formula

`T=(1)/(1400.0)`

`T=0.714*10^(-3)\ sec`

(a). We need to calculate the distance displayed on the screen of the oscilloscope for a full wave,

`x=(T)/(v)`

Here, v = sweep speed

Put the value into the formula

`x=(0.714*10^(-3))/(119*10^(-6))`

`x=6\ Div`

The horizontal distance on the screen is 6 Div.

(b). We need to calculate the peak to peak vertical amplitude of the wave

Using formula of amplitude

`v_(pp)=2* Amplitude`

Put the value into the formula

`v_(pp)=2*1.4`

`v_(pp)=2.8\ V`

We need to calculate the number of division from peak to peak

Using formula of division

`y=(v_(pp))/(verical\ gain)`

Put the value into the formula

`y=(2.8)/(0.40)`

`y=7\ Div`

The number of division from peak to peak is 7 Div.

Hence, (a). The horizontal distance on the screen is 6 Div.

(b). The number of division from peak to peak is 7 Div.