This depends on the duty of the square wave - if it is 50%, then it will be 1/2 the peak. If it is 33.3%, then it will be 1/3 the peak.
square root of area enclosed by a half wave to the base
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
rms voltage will give the ability to predict how much work will be done by an ac voltage. the rms value of a pure sine wave is 0.707 times it's maximum amplitude.
The wave with the maximum RMS value, in comparision with the peak value, is the square wave.
No, the peak-to-peak voltage is 2sqrt(2) times as much as the rms for a pure sine-wave.
rms value of voltage
RMS stands for Root Mean Square. Power is calculated as V2/R where V is the voltage and R is the resistive component of a load, This is easy toi calculate for a DC voltage, but how to calculate it for a sinusoidal voltage? The answer is to take all the instantaneous voltages in the sine wave, square them, take the mean of the squares, then take the square root of the result. This is defined as the "heating effect voltage". For a sine wave, this is 0.707 of the peak voltage.
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
For an alternating voltage, the simple mean over a cycle would be zero. 'RMS' means 'root mean square', and is defined as the square root of the mean value of the square of the voltage, taken over a cycle. Thus whether the voltage is + or - , as it is in alternate half cycles, the value of its square is always positive, giving a real number for the square root. In fact the RMS value of voltage produces an RMS current which dissipates power at the same rate as a DC current of the same value. To find the RMS value of a sine wave with no DC offset, divide the peak value of the sine wave by square root of 2. **************************************************** Since the r.m.s. value of a sine wave is 1.414Vpk, and the mean voltage of a sine wave is 1.57Vpk, then, starting with the r.m.s. value: Vmean = (Vr.m.s. x 1.414) ÷ 1.57
rms voltage will give the ability to predict how much work will be done by an ac voltage. the rms value of a pure sine wave is 0.707 times it's maximum amplitude.
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.
The wave with the maximum RMS value, in comparision with the peak value, is the square wave.
rms values refer to "root mean square" mathematical values of the sine wave of electricity. This is essentially an "average" value of the voltage being measured as voltage in any circuit varies constantly.
A square wave has the highest RMS value. RMS value is simply root-mean-square, and since the square wave spends all of its time at one or the other peak value, then the RMS value is simply the peak value. If you want to quantify the RMS value of other waveforms, then you need to take the RMS of a series of equally spaced samples. You can use calculus to do this, or, for certain waveforms, you can use Cartwright, Kenneth V. 2007. In summary, the RMS value of a square wave of peak value a is a; the RMS value of a sine wave of peak value a is a divided by square root of 2; and the RMS value of a sawtooth wave of peak value a is a divided by cube root of 3; so, in order of decreasing RMS value, you have the square wave, the sine wave, and the sawtooth wave. For more information, please see the Related Link below.
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)
RMS stands for "Root of the Means Squared", and is a mathematical method of defining the "operating" voltage of a sine wave power source. Typical home lighting and outlet voltage presently is 120 VAC (volts alternating current), 60 Hz. (Hertz, formerly referred to as "cycles per second".) But the PEAK voltage is the absolute maximum voltage at the "peak" of each sine wave of voltage. Mathematically, the "Peak" voltage is 1.414 (which is the square root of the number 2) times the RMS voltage, and conversely, the RMS voltage is 0.707 times the PEAK voltage.
The rms of 10V is 6.02V. Take the peak voltage of the sine wave and multiply it by 0.707.
RMS and peak voltage for a square waveform are the same. There is a small caveat, and that is that you'd have to have a "perfect" square wave with a rise time of zero. Let's have a look. If we have a perfect square wave, it has a positive peak and a negative peak (naturally). And if the transition from one peak to the other can be made in zero time, then the voltage of the waveform will always be at the positive or the negative peak. That means it will always be at its maximum, and the effective value (which is what RMS or root mean square is - it's the DC equivalent or the "area under the curve of the waveform") will be exactly what the peak value is. It's a slam dunk. If we have a (perfect) square wave of 100 volts peak, it will always be at positive or negative 100 volts. As RMS is the DC equivalent, or is the "heating value for a purely resistive load" on the voltage source, the voltage will always be 100 volts (either + or -), and the resistive load will always be driven by 100 volts. Piece of cake.