A sine wave has a peak-to-peak value of 88 V. What is the average value?
The correct answer and explanation is :
To find the average value of a sine wave with a peak-to-peak voltage of 88 V, we begin by understanding what “average value” means in this context.
🔹 Step 1: Understand the Peak-to-Peak Value
- The peak-to-peak value (Vpp) is the total voltage from the maximum positive peak to the maximum negative peak.
- For a sine wave: $$
V_{\text{pp}} = 2 \times V_{\text{peak}}
$$ So: $$
V_{\text{peak}} = \frac{88}{2} = 44\text{ V}
$$
🔹 Step 2: Define the Average Value of a Sine Wave
There are two average values depending on what part of the wave you’re considering:
- Full-wave average over a complete cycle: For a pure sine wave, the positive and negative halves cancel each other out. Therefore: $$
\text{Average over full cycle} = 0\text{ V}
$$ - Average of the absolute value (rectified sine wave): This is often what is meant in electrical engineering contexts: $$
V_{\text{avg}} = \frac{2}{\pi} \times V_{\text{peak}} \approx 0.637 \times V_{\text{peak}}
$$
Using the peak value of 44 V:
$$
V_{\text{avg}} = 0.637 \times 44 = 28.028\text{ V}
$$
✅ Final Answer:
Average value = 28.03 V (rounded to two decimal places)
📘 Explanation (300 Words)
The average value of a waveform is a key metric in electrical engineering, especially when analyzing power or signal characteristics. In the case of a sine wave, which alternates symmetrically above and below zero, the arithmetic mean over a complete cycle is zero. This is because the positive and negative halves cancel each other out.
However, in practical applications—such as rectification in power supplies—we are often interested in the average of the absolute value of the sine wave, also called the average rectified value. This is the effective DC voltage you would get if the waveform were rectified (flipped so all values are positive) and then averaged.
To calculate this average, we use the formula:
$$
V_{\text{avg}} = \frac{2}{\pi} \times V_{\text{peak}} \approx 0.637 \times V_{\text{peak}}
$$
Given a peak-to-peak voltage of 88 V, the peak voltage is 44 V. Multiplying by 0.637 yields approximately 28.03 V.
This average value is critical when designing rectifiers, power supplies, and when measuring AC voltages with analog meters. It also provides a way to estimate DC-equivalent levels of AC signals. Note that this is different from RMS (Root Mean Square) value, which measures power content and is equal to $V_{\text{rms}} = 0.707 \times V_{\text{peak}}$. Understanding these distinctions is essential for proper electrical analysis.