Which equation correctly expresses the peak value in terms of the average value?

Which equation correctly expresses the peak value in terms of the average value?

• A. Peak = Average ÷ 0.636
• B. Peak = Average × 0.636
• C. Peak = Average × 2
• D. Peak = Average ÷ 2

Answer: (A)

When a sinusoidal voltage is considered, the average value of its magnitude over a full cycle is a fixed fraction of its peak value. Specifically, the average of the absolute value of a sine wave equals (2/π) times the peak. So, to express the peak in terms of that average, you divide the average by (2/π), which is the same as multiplying by π/2. In other words, Peak = Average ÷ 0.636 equals Peak = Average × (π/2). The other simple multipliers don’t match this fixed ratio, so they don’t give the correct peak from the average.

When a sinusoidal voltage is considered, the average value of its magnitude over a full cycle is a fixed fraction of its peak value. Specifically, the average of the absolute value of a sine wave equals (2/π) times the peak. So, to express the peak in terms of that average, you divide the average by (2/π), which is the same as multiplying by π/2. In other words, Peak = Average ÷ 0.636 equals Peak = Average × (π/2). The other simple multipliers don’t match this fixed ratio, so they don’t give the correct peak from the average.