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At what phase angle is the voltage momentarily constant in a sine wave?
We often see the peak and trough (maximum positive and maximum negative excursions) of the sine wave considered as points of momentarily constant voltage. Those points are at phase angles of 90 degrees and at 270 degrees.
What is the approximate peak-to-peak voltage of a 2 VRMS sine wave?
the answer is 5.6vp-p
What is the relationship between peak to peak and peak voltage for a sine wave?
The RMS (root mean square) of the peak voltage of a sine wave is about 0.707 times the peak voltage. Recall that the sine wave represents a changing voltage, and it varies from zero to some positive peak, back to zero, and then down to some negative peak to complete the waveform. The root mean square (RMS) is the so-called "DC equivalent voltage" of the sine wave. The voltage of a sine wave varies as described, while the voltage of a DC source can be held at a constant. The "constant voltage" here, the DC equivalent, is the DC voltage that would have to be applied to a purely resistive load (like the heating element in a toaster, iron or a clothes dryer) to get the same effective heating as the AC voltage (the sine wave). Here's the equation: VoltsRMS = VoltsPeak x 0.707 The 0.707 is half the square root of 2. It's actually about 0.70710678 or so.
What is the rms voltage for 144 volts modified sine wave?
if that 144 is the peak voltage if its a sine wave the rms voltage is that voltage divided by sqrt(2) if not a sine wave (modified) you must find the area under the curve by integrating a cycle of that wave shape (root mean squared)
What is the peak-to peak voltage of a 56 Vrms ac voltage?
For a sine wave, the form factor is the square root of 2. Thus, the effective voltage of 56 V (56 Vrms) is 2-1/2 times the peak-to-peak voltage. Thus, the peak-to-peak voltage Vpp = Vrms * sqrt(2)In this example:Vpp = 56V * 1.4142... = 79.2V (rounded to one decimal place)
If the peak voltage of a sine wave is 100 volts then the instantaneous voltage at 150 and deg is volts?
To find the instantaneous voltage of a sine wave at a given angle, you can use the formula ( V(t) = V_{peak} \cdot \sin(\theta) ). For a peak voltage of 100 volts and at 150 degrees, convert 150 degrees to radians if necessary or use the sine function directly. The sine of 150 degrees is 0.5, so the instantaneous voltage is ( 100 \cdot \sin(150^\circ) = 100 \cdot 0.5 = 50 ) volts.
What is the instantaneous voltage of 90 degrees?
The instantaneous voltage at 90 degrees in a sinusoidal waveform is at its peak value, as this angle corresponds to the maximum point of the sine function. Mathematically, if the voltage is represented as ( V(t) = V_{\text{max}} \sin(\omega t + \phi) ), at 90 degrees (or ( \frac{\pi}{2} ) radians), the voltage is ( V(t) = V_{\text{max}} ). Thus, the instantaneous voltage is equal to the maximum amplitude of the waveform.
If an AC voltage begins rising or going positive at 0 degrees when will it reach its maximum negative value?
It's a sine wave (if there is no distortion). Voltage is zero at 0 degrees, at its positive peak at 90 degrees, back to zero at 180 degrees, at its negative peak at 270 degrees, and back to zero at 360 degrees.
How does time relate to degrees in an AC sine wave?
One cycle of the sine wave is equal to 360 degrees. In US the frequency of power is typically 60 Hz and hence one cycle is 1/60 of a second. Therefore you can calculate the degrees at any instant of time. If at zero degrees the voltage amplitude is zero, then at 90 degrees,which is 1/4 cycle, wave is at peak voltage. At 180 degrees it is at 1/2 cycle and zero voltage and then at 270 degrees it is 3/4 of the cycle and a peak negative voltage. Finally at 360 degrees the cycle is complete and the voltage is again zero.
Are peak-to-peak and rms voltage measurements the same?
No, the peak-to-peak voltage is 2sqrt(2) times as much as the rms for a pure sine-wave.
At what phase angle is the voltage momentarily constant in a sine wave?
We often see the peak and trough (maximum positive and maximum negative excursions) of the sine wave considered as points of momentarily constant voltage. Those points are at phase angles of 90 degrees and at 270 degrees.
What is the approximate peak-to-peak voltage of a 2 VRMS sine wave?
the answer is 5.6vp-p
What is the conversion of rms voltage to Peak to Peak voltage?
Assuming sine wave (it is different if not): Vp-p = 2.828 * Vrms
What is the peak voltage of a 240 VRMS sine wave?
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
What is the approximate instantaneous voltage at 37 degrees on a 169 Vp sine wave?
169sin(37*) = 101.7067389 (round to 101.7) *=degrees (function found on TI Calculators under "Angle") you can not do like that generally VpSIN(Wt
4.0Vrms sine wave equals peak to peak voltage?
4volts x 2.8 =9.6 v
What is the relationship between peak to peak and peak voltage for a sine wave?
The RMS (root mean square) of the peak voltage of a sine wave is about 0.707 times the peak voltage. Recall that the sine wave represents a changing voltage, and it varies from zero to some positive peak, back to zero, and then down to some negative peak to complete the waveform. The root mean square (RMS) is the so-called "DC equivalent voltage" of the sine wave. The voltage of a sine wave varies as described, while the voltage of a DC source can be held at a constant. The "constant voltage" here, the DC equivalent, is the DC voltage that would have to be applied to a purely resistive load (like the heating element in a toaster, iron or a clothes dryer) to get the same effective heating as the AC voltage (the sine wave). Here's the equation: VoltsRMS = VoltsPeak x 0.707 The 0.707 is half the square root of 2. It's actually about 0.70710678 or so.
