What is the relationship between the RMS value and the peak value of a sine wave?

What is the relationship between the RMS value and the peak value of a sine wave?

• A. V_rms = V_peak / 2
• B. V_rms = V_peak / √2
• C. V_rms = 2 V_peak
• D. V_rms = V_peak

Answer: (B)

For a sine wave, the RMS value represents the equivalent DC voltage that would deliver the same heating effect in a resistor. If the voltage is v(t) = V_peak sin(ωt), squaring gives V_peak^2 sin^2(ωt). Averaging sin^2 over a full cycle yields 1/2, so the mean of v^2 is V_peak^2/2. Taking the square root gives V_rms = V_peak / √2 ≈ 0.7071 × V_peak. This shows why the RMS value is lower than the peak value for a sinusoid and is the standard measure of effective voltage in AC circuits. The other expressions don’t match this squared-averaging relationship: dividing by 2 understates, doubling overstates, and equating to the peak would ignore the alternating nature of the waveform.

For a sine wave, the RMS value represents the equivalent DC voltage that would deliver the same heating effect in a resistor. If the voltage is v(t) = V_peak sin(ωt), squaring gives V_peak^2 sin^2(ωt). Averaging sin^2 over a full cycle yields 1/2, so the mean of v^2 is V_peak^2/2. Taking the square root gives V_rms = V_peak / √2 ≈ 0.7071 × V_peak. This shows why the RMS value is lower than the peak value for a sinusoid and is the standard measure of effective voltage in AC circuits. The other expressions don’t match this squared-averaging relationship: dividing by 2 understates, doubling overstates, and equating to the peak would ignore the alternating nature of the waveform.