if you know the wave shape
the average voltage is a synonym for phase voltage
a triangle wave would have a peak that is 4 x the average
a square wave the peak is 2 x the average
a sine wave has a peak that is sqrt(2) times the average
RMS root mean squared is a way of saying that you are working with curves rather than lines and it a bit harder to find the area
AnswerUnless otherwise stated, a.c. voltages and currents are always expressed as root-mean-square (r.m.s.) values. For a sinusoidal voltage, the amplitude is 1.414 times the r.m.s. value. For example, the amplitude of a 120-V supply is 170 V.
The average value of any alternating voltage or current, over one complete cycle, is zero.
This applies whether you are describing a phase voltageor a line voltage. The terms, 'phase' and 'line' simply describe what is being measured, not its value.
It is the highest value of the amplitude, called the peak value. Scroll down to related links and look at "RMS voltage, peak voltage and peak-to-peak voltage". Look at the figure in the middle below the headline "RMS voltage, peak voltage and peak-to-peak voltage".
Unless otherwise stated, the value of an a.c. current or voltage is expressed in r.m.s. (root mean square) values which, for a sinusoidal waveform, is 0.707 times their peak value. The output of a voltage (or potential) transformer is no different, its measured voltage will be its r.m.s value which is lower than its peak value.
Frequency modulation Phase modulation
Need to know the voltage and phase for this question.
To calculate the peak voltage of an RMS voltage in a sine wave simply multiply the RMS voltage with the square root of 2 (aprox. 1,414) like this: 240 x 1,414 = 339,4 V RMS x sqr.root of 2 = peak voltage
To measure peak voltage using an oscilloscope, adjust the voltage scale setting on the vertical axis until the entire waveform is visible on the screen. Then, use the cursors or measurement tools on the oscilloscope to determine the maximum amplitude of the waveform, which represents the peak voltage. You can directly read the peak-to-peak value if it's available, or calculate it by multiplying the peak voltage by 2 if only the peak amplitude is displayed.
It is the highest value of the amplitude, called the peak value. Scroll down to related links and look at "RMS voltage, peak voltage and peak-to-peak voltage". Look at the figure in the middle below the headline "RMS voltage, peak voltage and peak-to-peak voltage".
Yes.
The same as in single phase with the same RMS voltage.
Conversions of RMS voltage, peak voltage and peak-to-peak voltage. That are the used voltages. The expression "average" voltage is used for RMS voltage.Scroll down to related links and seach for "RMS voltage, peak voltage and peak-to-peak voltage".Answer'Average' is not the same as 'root mean square'. As the average value of a sinusoidal voltage is zero, you cannot convert it to a peak-to-peak value.
first we connect the oscilloscope with the function generator or whatever the source of the input voltage , there will be a wave ,we try to adjust its amplitude using oscilloscope ..and this amplitude will be the peak to peak voltage..putting into consideration how volt/ div while measuring the amplitude
ANSWER: The peak to peak voltage can be found by multiplying 120 v AC x 2.82= 339.41
Unless otherwise stated, the value of an a.c. current or voltage is expressed in r.m.s. (root mean square) values which, for a sinusoidal waveform, is 0.707 times their peak value. The output of a voltage (or potential) transformer is no different, its measured voltage will be its r.m.s value which is lower than its peak value.
You do not specify, in your question, what the 'input' device is.
Frequency modulation Phase modulation
For a sine wave with maximum amplitude at time zero, there is no phase shift. The wave starts at its peak at time zero, and therefore, its phase angle is zero.
In a pure inductive circuit, the relationship between current and voltage is such that the current lags behind the voltage by a phase angle of 90 degrees. This means that the current and voltage are out of phase with each other, with the current reaching its peak value after the voltage has reached its peak value.