Peak value is 1.414 times the RMS voltage. On a 240 volt circuit the peak voltage is 240 x 1.414 = 339.36 volts. The peak to peak value is twice this.
Not much of anything. The 220 volt appliance needs just that ... 220 volts in order to run. If it runs at all, it certainly would not be running at anywhere near peak efficiency.
Once the membrane depolarizes to a peak value of +30 mV, it repolarizes to its negative resting value of -70 mV.
Once the membrane depolarizes to a peak value of +30 mV, it repolarizes to its negative resting value of -70 mV.
peak voltage is the highest level of volts giving from an alternating current. for example, in the UK the mains supply to a house is 230RMS. 230 is the average (simplified so bear with me) volts you will get from the AC supply. but in reality the AC wave is switching between -320 and 320 volts. so 320 is the peak voltage. to find out the peak voltage of an alternating current you must multiply the RMS voltage by root2 or 0.707 (this only works when the AC signal is a sine wave) Totally wrong! Multiply the RMS voltage by 0.707 to find the PEAK voltage? What? The RMS voltage is 230. This times 0.707 is 163 which is far from the peak. Also 0.707 is not the sq. root of 2. I hope people don't take these answers seriously. Peak Voltage is roughly 1.41 times the RMS voltage. (1.41 being the square root of 2) So 230 RMS mains voltage has a peak value of roughly 325V, (1.41x230). The original poster has accidentally multiplied the RMS value by 1 over the square root of 2 (0.707) ... which is what you multiply the peak value by in order to calculate the RMS.
Obtain the molecular mass by determining the m/z value of the molecular ion peak (rightmost in the spectrum).
When you say holdhold supply of 230volts, you are referring to the RMS value, not the peak value.
Adding a DC source to a square wave signal will alter the base line of the wave without changing the peak-to-peak value. For example, if a square wave has a +4V baseline and a +2VDC source is introduced, the resulting square wave will have a +6V baseline. This of course will also affect the high and low peaks of the signal. Assuming that our example has a high peak of +9V and a low peak of -1V (with a total of 10V peak-to-peak), the added +2VDC source would result in a high peak of +11V and a low peak of +1V; however, the total peak-to-peak value remains unchanged at 10V peak-to-peak.
rms. dat means Vp-p will be 325V.
Peak value is the highest value ever reached.
Peak to Peak is the most positive peak to the negative peak value. Or find any peak value and multiply by 2.
You have to convert the peak voltage to an r.m.s. value, before inserting that value into the power equation.
peak - peak.
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'.
30 volts provided zero crossing is at midpoint.
ANSWER: The peak to peak voltage can be found by multiplying 120 v AC x 2.82= 339.41
The input voltage, an AC Sine Wave will have a Peak-to-Peak value equal to 2X its Peak value. Once rectified, all the Peaks will be either above or below the Zero reference line. They'll look like a series of identical bumps. The net value of the unrectified voltage will be Zero. The positive and negative waveforms canceling each other out. The rectified waveform will be all positive or negative and its net value will be non-zero. Its AVERAGE value will be .636 times its Peak value. Its Root Mean Square (RMS) value will be .707 times its Peak value. Its Peak-to-Peak value will equal 1X the Peak value.
The quoted value is usually RMS value, i.e it is lesser than the peak value of the voltage, therefore the peak value is sqrt(2) times the quoted value. (it is a sine wave)