Voltage does not have a waveform. The waveform is based upon the frequency of the voltage or current. A battery (any voltage) does not waveform, however the voltage coming into your house (US) has a frequency of 60 Hz. The length of the 60 hz waveform
Length (in centimeters) = (3 x (10 ** 10))/ Frequency in hz =
500 000 000 cmUse an oscilloscope. That shows the voltage waveform and you can read the peak value.
Eli the ice man. Voltage (E) before Current (I) in a coil (inductor)(L) Current (I) before Voltage (E) in a Cap. (C) Got it?
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.
no, dc volatage is a type of current direct current, ac is alternating current, average voltage could be any type of voltage ac or dc that maintains a constant rangeAnswerNo. A DC voltage is exactly equivalent to an AC rms-voltage. So, for example, 100 V (DC) is exactly equivalent to 100 V (AC rms). The average value of an AC waveform is zero.
The AC voltage is in the form of a sine wave. Half the wave is above zero volts and half below. In half wave rectification the bottom half of the wave is chopped off. That leaves a series of "humps' interspersed with a half wave time of zero voltage. The capacitor stores charge that decays through the resistor. The more capacitance the longer the charge is held and the voltage smooths out somewhat to approximate a DC voltage with some ripple. A full wave rectifier flips the bottom half of the waveform above zero so that the period of time the voltage is close to zero is reduced and the less ripple in the output voltage and the capacitor will smooth out the voltage even more.
rectangular
You don't. Transformers only work with AC voltage. Their input will be an AC waveform, and their output will be an AC waveform. Other electronics are used to convert the stepped down AC waveform from the transformer to DC.
Use an oscilloscope. That shows the voltage waveform and you can read the peak value.
60 seconds
Frequency.
It would be unity, or 1.0. Since the voltage in a DC circuit does not vary with time, there can be no phase displacement of the current waveform, and therefore the current could not lead or lag the voltage waveform.
bcoz, voltage clipping limits the voltage to a device without affecting the rest of the waveform.
the number u are seeking is 0.639
This would keep the voltage across the inductance a constant, and corrects the non-linearity problem.
Look up sine wave on Google to see a picture.
RMS voltage is the DC equivalent of your AC waveform. Vrms=(Vpeak)/(root two) If your peak voltage is 170V then the RMS voltage would be approx. 120 V (see related link)
Because the resulting voltage waveform is symmetric about the center of the rotor flux, no even harmonics are present in the phase voltage.