RMS is used to determine the average power in an alternating current. Since the voltage in an A/C system oscillates between + and -, the actual average is zero. The RMS or "nominal" voltage is defined as the square root of the average value of the square of the current, and is about 70.7% of the peak value.
************************************************************
The r.m.s. value of an alternating current or voltage is the value of direct current or voltage which produces the same heating effect.
Fo a sine wave, the r.m.s. value is 0.707 x the peak value.
The average value is different; for a sine wave it is 0.636 x the peak value.
Assuming "quoted value" to be RMS value, or average, [what you would see on a meter], the peak would be that value times 1.414. Going backward, peak times .707 is RMS.
To get the average:Volts avg=0.637 X Vp (peak)0.637 X 80 Vp = 50.96 VavgTo get rms (root mean square):Volts rms = 0.707 X Vp (peak)0.707 X 80 Vp = 56.56 VrmsCommentIt should be pointed out that the average value, described above, is for half a cycle. The average for a complete cycle is zero.
rms. dat means Vp-p will be 325V.
The peak of a waveform that is purely sinusoidal (no DC offset) will be RMS * sqrt(2). This is the peak to neutral value. If you are looking for peak to peak, multiply by 2.
You can work this out yourself. For a sinusoidal waveform the rms value is 0.707 times the peak value. As you quote a peak-to-peak value, this must be halved, first. Incidentally, the symbol for volt is 'V', not 'v'.
Assuming "quoted value" to be RMS value, or average, [what you would see on a meter], the peak would be that value times 1.414. Going backward, peak times .707 is RMS.
Average Current = 0.636 * (Peak Current)so Peak Current = (Average Current)/0.636RMSCurrent = 0.707 * (Peak Current)so Peak Current = (RMS Current)/0.707Because both equations are in terms of Peak Current, we can set them equal to each other.(Average Current)/0.636 = (RMS Current)/0.707(42.5)/0.636 = (RMS Current)/0.707thenRMS Current = (0.707)(42.5)/0.636 = 47.24 ampsAnother AnswerSince the average value of a single sine wave is zero, you cannot calculate its r.m.s. value!
To get the average:Volts avg=0.637 X Vp (peak)0.637 X 80 Vp = 50.96 VavgTo get rms (root mean square):Volts rms = 0.707 X Vp (peak)0.707 X 80 Vp = 56.56 VrmsCommentIt should be pointed out that the average value, described above, is for half a cycle. The average for a complete cycle is zero.
For a 12V peak voltage (V_peak), the peak-to-peak value (V_pp) is 24V, as it is twice the peak voltage (V_pp = 2 * V_peak). The root mean square (RMS) value is approximately 8.49V, calculated as V_rms = V_peak / √2. The half-cycle average voltage is about 7.64V, calculated as V_avg = (V_peak / π).
rms. dat means Vp-p will be 325V.
The peak of a waveform that is purely sinusoidal (no DC offset) will be RMS * sqrt(2). This is the peak to neutral value. If you are looking for peak to peak, multiply by 2.
RMS is an average. If you have a 50% duty square wave, the average will be 1/2 the peak. for a 33.3% duty cycle, the average will be 1/3 the peak, etc. VRMS = Vpeak x duty cycle
Not sure what you mean by Class A current. Normally, when measuring AC voltage or current you either measure the peak to peak value or the Root Mean Squared (RMS) value. Since RMS is essentially an average measured over time, it would always be less than Peak to Peak value.
You can work this out yourself. For a sinusoidal waveform the rms value is 0.707 times the peak value. As you quote a peak-to-peak value, this must be halved, first. Incidentally, the symbol for volt is 'V', not 'v'.
When you say holdhold supply of 230volts, you are referring to the RMS value, not the peak value.
To find the root mean square (rms) value for a voltage given in peak-to-peak (Vpp), you need to divide the Vpp value by 2√2. In this case, the Vpp is 300mV, which is equivalent to 0.3V. Dividing 0.3V by 2√2 ≈ 2.828, the rms value is approximately 0.106 V.
rms value of ac power = dc power in reference to heat production in pure resistive load So ac power of some rms value will produce the same heat in resistive load as dc power will of same value