What is the average value of an AC voltage whose peak value is 120 volts?

Study for the NEIEP Electrical Fundamentals Test. Use flashcards and multiple choice questions, each with hints and explanations. Get ready for your exam!

Multiple Choice

What is the average value of an AC voltage whose peak value is 120 volts?

Explanation:
When a voltage is sinusoidal, the average value of its magnitude over a full cycle is 2Vm/π, where Vm is the peak value. This comes from averaging the absolute value of Vm sin(ωt) over one period: (1/2π) ∫0^{2π} |Vm sin θ| dθ = (Vm/π) ∫0^{π} sin θ dθ = 2Vm/π. With Vm = 120 V, the average magnitude is 2×120/π ≈ 240/3.1416 ≈ 76.4 V. This matches the option 76.32 V. So the correct concept is the average value of a sinusoid’s magnitude over a cycle, not the signed average (which would be zero).

When a voltage is sinusoidal, the average value of its magnitude over a full cycle is 2Vm/π, where Vm is the peak value. This comes from averaging the absolute value of Vm sin(ωt) over one period: (1/2π) ∫0^{2π} |Vm sin θ| dθ = (Vm/π) ∫0^{π} sin θ dθ = 2Vm/π.

With Vm = 120 V, the average magnitude is 2×120/π ≈ 240/3.1416 ≈ 76.4 V. This matches the option 76.32 V.

So the correct concept is the average value of a sinusoid’s magnitude over a cycle, not the signed average (which would be zero).

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