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egoroff_w [7]
3 years ago
5

Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d

ial is set to 5 msec/div. What is the peak-to-peak amplitude of the displayed signal? What is the frequency in khz?

Physics
1 answer:
denis-greek [22]3 years ago
8 0

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

Finding Frequency:

As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

Time Period = 3 div * 5msec /div\\Time Perod = 15 msec

Since,

Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz

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