it's highest swing positive or negative. 90 and 270 degrees. the peak for a 10vAC generator would be 10vAC
The wave with the maximum RMS value, in comparision with the peak value, is the square wave.
Do you mean a "saw-tooth" wave? Also, a "peak value" is not an RMS value (by definition), it is simply the highest value attained (positive or negative) over the period of analysis.
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)
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.
From your description, this sounds like it is a sine wave offset to 10A, so the peak is at 20A, and the min is at 0? For this case, you have 10A DC (RMS) wave and a 10A Peak - neutral AC wave; The RMS value of the AC wave is: 10/2*sqrt(2) = 3.54A. So the RMS amplitude of this wave is 13.54A.
30 volts provided zero crossing is at midpoint.
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.
A square wave will have the highest value since it has a peak, positive or negative, all of the time. Other wave shapes such as triangular and sine, have a lower value than this.
The wave with the maximum RMS value, in comparision with the peak value, is the square wave.
Do you mean a "saw-tooth" wave? Also, a "peak value" is not an RMS value (by definition), it is simply the highest value attained (positive or negative) over the period of analysis.
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'.
The rms value of a sine wave current is 0.707 Imax. So the answer to your quesion is 0.707 x 4 = 2.83 A.
For a sine wave ONLY - and assuming you are talking plus and minus 100V (200V peak to peak) - the RMS voltage is about 71V. (One half square root of 2 * single sided 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)
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.
100v divided by 1.41
From your description, this sounds like it is a sine wave offset to 10A, so the peak is at 20A, and the min is at 0? For this case, you have 10A DC (RMS) wave and a 10A Peak - neutral AC wave; The RMS value of the AC wave is: 10/2*sqrt(2) = 3.54A. So the RMS amplitude of this wave is 13.54A.