sin-wave

[stealthelectric]

Hi:
Please see the question attached.

If I find the peak value - √2 x 12.0V = 17.0V (2s.f.)
We have 6 divisions peak-peak, so divide 17.0 by 6 to get 2.8V (2s.f.)

Could somebody please tell me if I have the right solution here?
Thanks

 

[Hero999]

The peak voltage calculation is correct.

The peak-to-peak calculation is wrong, it should be double the peak voltage.

 

[stealthelectric]

peak value - √2 x 12.0V = 17.0V (2s.f.)
So, peak to peak = 34.0V
We have 6 divisions peak-peak, so divide 34.0 by 6 to get 5.67V (3s.f.)

Considering the figures given in the question, should I choose 2 or 3 significant figures for the answer?

 

[Hero999]

Yes, the peak is three divisions and the peak-to-peak is six divisions.

I don't know how you should round the answer, I used to hate that at school. I generally gave a figure more than what was given to me in the question.

 

[stealthelectric]

Thank you Hero

 

[stealthelectric]

so for the x-direction. I take 1/250Hz=0.004, for the period. Then 0.004/5squares=0.8x10^-3s for the time of one division?

 

[Hero999]

Yes, it's 800×10^6 seconds.

 

[stealthelectric]

thank you hero
But do you mean 10^-6?