If a sine wave has an average value of 90 V, the peak value is approximately what?

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Multiple Choice

If a sine wave has an average value of 90 V, the peak value is approximately what?

Explanation:
For a sine wave, the average value taken over a half cycle relates to the peak value by a factor of 2/π. If the instantaneous peak is Vpeak, the average over the positive half-cycle is (2/π) × Vpeak. Setting this average equal to 90 V gives Vpeak = 90 × (π/2) ≈ 141.37 V, which rounds to about 141.5 V. Context: the RMS value would be Vpeak/√2, so with Vpeak ≈ 141.5 V, Vrms ≈ 100 V. The full-cycle average is zero, so 90 V wouldn’t be the average over a full cycle.

For a sine wave, the average value taken over a half cycle relates to the peak value by a factor of 2/π. If the instantaneous peak is Vpeak, the average over the positive half-cycle is (2/π) × Vpeak. Setting this average equal to 90 V gives Vpeak = 90 × (π/2) ≈ 141.37 V, which rounds to about 141.5 V.

Context: the RMS value would be Vpeak/√2, so with Vpeak ≈ 141.5 V, Vrms ≈ 100 V. The full-cycle average is zero, so 90 V wouldn’t be the average over a full cycle.

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