If a sine wave has a peak voltage of 100 V, what is its average value?

If a sine wave has a peak voltage of 100 V, what is its average value?

• A. 63.6 V
• B. 100 V
• C. 50 V
• D. 0 V

Answer: (A)

The important idea is how to find the average value of a sine wave when you’re looking at the DC level after rectification. A sine wave with peak Vm can be written as v(t) = Vm sin(ωt). If you average over a full cycle without rectifying, the positive and negative halves cancel, giving an average of zero. But when we consider the average value of the waveform after rectification (the average of the magnitude, or the DC component of a rectified sine), the result is 2Vm/π. Applying this with a peak of 100 V gives 2 × 100 / π ≈ 63.66 V, which rounds to 63.6 V. So the correct average value is about 63.6 V. For context, the RMS value would be Vm/√2 ≈ 70.7 V, and the half-wave rectified average would be Vm/π ≈ 31.8 V.

The important idea is how to find the average value of a sine wave when you’re looking at the DC level after rectification. A sine wave with peak Vm can be written as v(t) = Vm sin(ωt). If you average over a full cycle without rectifying, the positive and negative halves cancel, giving an average of zero. But when we consider the average value of the waveform after rectification (the average of the magnitude, or the DC component of a rectified sine), the result is 2Vm/π.

Applying this with a peak of 100 V gives 2 × 100 / π ≈ 63.66 V, which rounds to 63.6 V. So the correct average value is about 63.6 V. For context, the RMS value would be Vm/√2 ≈ 70.7 V, and the half-wave rectified average would be Vm/π ≈ 31.8 V.