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A sine wave of 17 volts peak is equivalent to how many volts RMS?
Wiki User
∙ 13y ago
ERMS = 0.707 times Epeak = .707 times 17 volts = 12 volts RMS
Erms = Epeak / Sqrt (2)
(Note I'll be using different notation so, E = V)
The above is derived from using.
v = VMsin ωt ...... Eq1
Where ω = 2πf (angular frequency) & VM is amplitude & π (pi radians)
and
VRMS = √[1/2π ʃ2π v2(ωt) d(ωt)] , Eq2
and the standard trig relationship from your maths text book
Sin2A = ½(1 - cos 2A), which is rewritten by replacing A with ωt, thus
sin2ωt = ½(1 - cos 2ωt) Eq3
Armed with these three equations, lets begin...
substitute v in Eq1 into Eq2
VRMS = √[1/2π ʃ2π vM2 sin2ωt d(ωt)] Eq4
Now apply trig relationship Eq3 to Eq 4 above & rearrange.
VRMS = √[vM2/4π ʃ02π (1 - cos 2ωt) d(ωt)] Eq5
Now take the integral using limits 0 to 2π as shown by ʃ02π .
√{vM2/4π [ωt - ½ sin 2ωt]02π}
continuing...
√{vM2/4π [(2π - 0) - (0 - 0)]}
VM / √2 which is the same as your first responders Epeak / Sqrt (2) , just using different notation.
0.707VM
or
0.707Epeak
Wiki User
∙ 13y ago
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If a sine wave has a peak value of 220 volts what is the route means square value?
If the Peak to neutral voltage is 220 volts, the root mean square voltage is 155.6 volts (sqrt(220)).
Vt 17 sin 314 t load 68 Ohms find peak value?
You don't say whether you're looking for the peak value of voltage or current.-- The peak value of the sine is ' 1 ', so the peak voltage is 17 volts.-- You haven't mentioned whether the load is complex or all real,so naturally I'll assume it to be all real. Then the peak current is 17 volts/68 ohms = 0.25 amp.
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.
What is the peak value of the input to a half wave rectifier if the half wave rectifier is 10V?
In this case, the peak voltage, which is half the peak to peak voltage, is 100 volts. Additionally, the half-wave rectifier will only provide an output for half the input cycle. In the case of a full wave rectifier, the RMS output voltage would be about 0.707 times the value of the peak voltage (100 volts), which would be about 70.7 volts. But with the output operating only half the time (because of the half wave rectification), the average output voltage will be half the 70.7 volts, or about 35.35 volts RMS.
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4volts x 2.8 =9.6 v
What is the peak value of a sine wave that has a peak to peak value of 60v?
30 volts provided zero crossing is at midpoint.
If a sine wave has a peak value of 220 volts what is the route means square value?
If the Peak to neutral voltage is 220 volts, the root mean square voltage is 155.6 volts (sqrt(220)).
What is DC offset in signal generator?
A sine wave centered at zero will have a positive peak that is the same magnitude as the negative peak. This can be offset so the negative peak magnitude does not match the positive peak magnitude. For example a 1volt peak - neutral sine wave could be DC offset by 1 volt so the positive peak is at 2 volts and the negative peak is at 0.
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)
The rms voltage is obtained by multiplying peak by a factor?
Multiply the peak to neutral by .707 for a sine wave. The RMS value is the equivalent level available if it were DC, so it will always be lower than the peak value.
What is effective voltage?
It is the 'as if' voltage in an AC circuit. Referred to as Vrms 120 volts in your house is Vrms, the effective voltage, 'as if' it were DC 120V, can do the same work. But 120VACrms is a sine wave with a peak voltage much higher than 120 volts.
Vt 17 sin 314 t load 68 Ohms find peak value?
You don't say whether you're looking for the peak value of voltage or current.-- The peak value of the sine is ' 1 ', so the peak voltage is 17 volts.-- You haven't mentioned whether the load is complex or all real,so naturally I'll assume it to be all real. Then the peak current is 17 volts/68 ohms = 0.25 amp.
What is the approximate peak-to-peak voltage of a 2 VRMS sine wave?
the answer is 5.6vp-p
What is harmonics in electrical system?
in a 60hz system anything that is not a 60Hz sine wave is a harmonic this is caused by non linear devices with 120 volts if a device draws more current at 170V than at 0 volts it will cause current at peak and -peak for a 120hz harmonic in the current If the wiring is not 0 ohms you get funny ringing through the whole system
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.
What is the peak value of the input to a half wave rectifier if the half wave rectifier is 10V?
In this case, the peak voltage, which is half the peak to peak voltage, is 100 volts. Additionally, the half-wave rectifier will only provide an output for half the input cycle. In the case of a full wave rectifier, the RMS output voltage would be about 0.707 times the value of the peak voltage (100 volts), which would be about 70.7 volts. But with the output operating only half the time (because of the half wave rectification), the average output voltage will be half the 70.7 volts, or about 35.35 volts RMS.
