It is the 'as if' voltage in an AC circuit. Referred to as Vrms 120 volts in your house is Vrms, the effective voltage, 'as if' it were DC 120V, can do the same work. But 120VACrms is a sine wave with a peak voltage much higher than 120 volts.
When the peak voltage is 311, the RMS voltage is 220. (311 * square root (2))
For a sinusoidal waveorm, RMS (effective, heating) value = 2/pi x (peak voltage). It's not 2/pi for waveforms with other shapes. 2/pi = roughly 63.7%
EFFECTIVE HOW ABOUT AVERAGE .639 of peak.AnswerThe 'effective' value of an a.c. voltage (or current) is the same as its 'root-mean-square' (r.m.s.) voltage which, for a sinusoidal waveform, is 0.707 Umax.The 'average' value of an a.c. voltage (or current) is zero over a complete cycle, or 0.639 Umax, over half a cycle (usually applied to rectified waveforms).
Simply multiply the peak voltage to 2 and you will get the peak to peak voltage.
For a sine wave, the form factor is the square root of 2. Thus, the effective voltage of 56 V (56 Vrms) is 2-1/2 times the peak-to-peak voltage. Thus, the peak-to-peak voltage Vpp = Vrms * sqrt(2)In this example:Vpp = 56V * 1.4142... = 79.2V (rounded to one decimal place)
When the peak voltage is 311, the RMS voltage is 220. (311 * square root (2))
diode
For a sinusoidal waveorm, RMS (effective, heating) value = 2/pi x (peak voltage). It's not 2/pi for waveforms with other shapes. 2/pi = roughly 63.7%
EFFECTIVE HOW ABOUT AVERAGE .639 of peak.AnswerThe 'effective' value of an a.c. voltage (or current) is the same as its 'root-mean-square' (r.m.s.) voltage which, for a sinusoidal waveform, is 0.707 Umax.The 'average' value of an a.c. voltage (or current) is zero over a complete cycle, or 0.639 Umax, over half a cycle (usually applied to rectified waveforms).
I = E/R or Current = Voltage/Resistance (Ohm's Law)
If, by 'effective' voltage, you mean 'root-mean-square' (rms) voltage, then 220 V is already expressed in those terms.Unless otherwise stated, a.c. voltages and currents are always expressed in rms values.
Heat dissipation = (applied voltage)2 / total effective resistance of the circuit
There will be no effect on the voltage. That is the effective voltage will be only 12 volt. But there will be increase of current.
With an AC and a DC voltage source in series, the DC voltage can be added to the RMS value of the AC voltage to give the effective voltage.
That depends on the voltage which the current flows through. I = P / U I = 1500 / U Where I is the current in Amperes and U is the effective voltage in Volts. (P is the power)
Transformers increase and decrease voltage as needed. PLATO
Simply multiply the peak voltage to 2 and you will get the peak to peak voltage.