Question
Download Solution PDFWhat is the peak to peak value of a sine waveform whose average value is 12.74 volts?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The average value of sinusoidal waveform over one complete cycle is zero as two halves cancel each other, so the average value is taken over half a cycle.
\({V_{avg}} = \frac{1}{\pi }\mathop \smallint \nolimits_0^\pi {V_p}\sin \theta \;d\theta \)
\({V_{avg}} = \frac{{{V_p}}}{\pi }\left( { - \cos \theta } \right)_0^\pi \)
\({V_{avg}} = \frac{{2{V_p}}}{\pi } = 0.637 × {V_p}\)
Therefore, the average value of a sinusoidal wave is 0.637 times the peak value.
Calculation:
Given,
Vavg = 12.74 V
\({V_{avg}} = \frac{{2{V_p}}}{\pi } = 0.637 × {V_p}\)
\(V_p=\frac{12.47}{0.637}~ V\)
= 19.57 V
∵ We are asked to find the peak to peak voltage.
∴ Vp-p = 2 × 19.57 ≈ 40 V
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