Horrible Homework question on alternating currents

[Sofie1990]

Homework Statement

An alternating current source is connected to a resistor to form a complete circuit. The oscilloscope settings are: Y sensitivity 4.0v per division
Time base 1.0ms per division
(i)determine peak voltage of the a.c source
(ii)Hence calculate the r.m.s voltage
(iii)Determine the time period of the a.c signal
(iv)Hence calculate the frequency of the a.c signal

The oscilliscope trace was 4 divisions from peak to trough and 3 divisions for 1 complete wave.

Homework Equations

The Attempt at a Solution

The answers i got was
(i) 8.0v
(ii) 5.7v
(iii) 3.0ms
(iv) could not work out

 

[collinsmark]

Homework Statement

An alternating current source is connected to a resistor to form a complete circuit. The oscilloscope settings are: Y sensitivity 4.0v per division
Time base 1.0ms per division
(i)determine peak voltage of the a.c source
(ii)Hence calculate the r.m.s voltage
(iii)Determine the time period of the a.c signal
(iv)Hence calculate the frequency of the a.c signal

The oscilliscope trace was 4 divisions from peak to trough and 3 divisions for 1 complete wave.

Homework Equations

The Attempt at a Solution

The answers i got was
(i) 8.0v

So far so good.

But just to be nit-picky, I've never liked the use of the phrase "peak voltage" to describe the amplitude of an AC signal. It's too easily confused with "peak-to-peak." But I guess you have no choice in that, given your coursework. The term "peak voltage" is standard terminology even if I don't like it. Your answer is correct as it is.

(ii) 5.7v

I'm not sure how many significant figures you are supposed to use, but with only two significant figures, your answer is correct.

(iii) 3.0ms

Correct.

(iv) could not work out

The frequency, f, of a signal is the inverse of its period, T. In other words, f = 1/T.

If T is measured in seconds, than f is measured in Hertz (Hz).