A sine wave has a peak voltage of 5.8 V . What is the average voltage of this signal expressed in volts ?
a. 4.10 V
b. 11.60 V
c. 5.80 V
d. 3.69 V

Question

A sine wave has a peak voltage of 5.8 V . What is the average voltage of this signal expressed in volts $$ (\mathrm{V}) $$ ? a. 4.10 V b. 11.60 V c. 5.80 V d. 3.69 V

UpStudy AI · Accepted Answer

Given a sine wave with a peak voltage of $$ V_{\text{peak}} = 5.8\,\mathrm{V} $$, the average voltage over one complete cycle is zero because the positive and negative halves cancel out. However, the problem is asking for the average voltage expressed in volts, and the provided options suggest that we are likely looking for the average of the absolute value of the sine wave (also known as the average rectified value).

For a sine wave, the average rectified value is given by

$$
V_{\text{avg}} = \frac{2V_{\text{peak}}}{\pi}.
$$

Substitute $$ V_{\text{peak}} = 5.8\,\mathrm{V} $$:

$$
V_{\text{avg}} = \frac{2 \times 5.8}{\pi} = \frac{11.6}{\pi}.
$$

Calculating,

$$
V_{\text{avg}} \approx \frac{11.6}{3.1416} \approx 3.69\,\mathrm{V}.
$$

Thus, the correct option is (d) $$3.69\,\mathrm{V}$$.
The average voltage of the sine wave is approximately 3.69 V.

A sine wave has a peak voltage of 5.8 V . What is the average voltage of this signal expressed in volts ?
a. 4.10 V
b. 11.60 V
c. 5.80 V
d. 3.69 V

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A sine wave has a peak voltage of 5.8 V . What is the average voltage of this signal expressed in volts ? a. 4.10 V b. 11.60 V c. 5.80 V d. 3.69 V

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Latest Trigonometry Questions

A sine wave has a peak voltage of 5.8 V . What is the average voltage of this signal expressed in volts ?
a. 4.10 V
b. 11.60 V
c. 5.80 V
d. 3.69 V