What is the peak voltage of an AC waveform?
1 Answer
The peak voltage of an alternating current (AC) waveform refers to the highest instantaneous voltage value reached during one complete cycle of the waveform. In mathematical terms, it's the maximum absolute value of the voltage over a single cycle.
For a sinusoidal AC waveform, such as the commonly used sine wave, the peak voltage (V_peak) can be related to the root mean square (RMS) voltage (V_RMS) by the following relationship:
V_peak = √2 × V_RMS
Here, √2 is approximately 1.414. The RMS voltage is the equivalent DC voltage that would produce the same heating effect in a resistive load as the AC voltage does.
For example, if you have an AC voltage waveform with an RMS value of 120 volts, the peak voltage would be approximately:
V_peak = 1.414 × 120 V ≈ 169.7 V
Keep in mind that there are other AC waveforms that might have different peak-to-RMS voltage ratios, but for a standard sinusoidal waveform, the relationship provided above holds true.
For a sinusoidal AC waveform, such as the commonly used sine wave, the peak voltage (V_peak) can be related to the root mean square (RMS) voltage (V_RMS) by the following relationship:
V_peak = √2 × V_RMS
Here, √2 is approximately 1.414. The RMS voltage is the equivalent DC voltage that would produce the same heating effect in a resistive load as the AC voltage does.
For example, if you have an AC voltage waveform with an RMS value of 120 volts, the peak voltage would be approximately:
V_peak = 1.414 × 120 V ≈ 169.7 V
Keep in mind that there are other AC waveforms that might have different peak-to-RMS voltage ratios, but for a standard sinusoidal waveform, the relationship provided above holds true.