When the peak voltage is 311, the RMS voltage is 220. (311 * square root (2))
Ohm's Law: Resistance is voltage divided by current Power Law: Power is current times voltage Combining them gives: Resistance is voltage squared divided by power 220 volts squared divided by 100 watts = 484 ohms. Note that this is hot resistance. If you measure the bulb in the cold state, you will get an entirely different, smaller, value, due to the extreme temperature coefficient of the filament. Independently of that, since you ask for peak voltage, that means you are talking about an AC voltage source. We have to assume a sinusoidal waveform, and that the 220 volts was the RMS value. In this case, the peak value is simply the RMS value multiplied by the square root of 2, i.e. 0.707..., making the peak value 311 volts.
The effective current is the square-root of the average value of the current-squared, root-mean-square or rms for short. That is because the power in a resistive load like a water heater is proportional to the square of the current. The rms voltage of an ac supply gives the same heating effect as a dc supply of the same voltage. The peak current is sqrt(2) times higher. The rms voltage and current are always quoted when referring to an ac power supply.
Answer Yes you can use a AC rated bulb on DC, When using an AC bulb on 110volt DC you would need to double the wattage of the bulb to get the same amount of light output. So if you need 50 watts of light you would need to use a 100 watt bulb on 110 volt DC. One benfit of using DC is the bulbs last longer! Hope this helps. It is important to understand that 110 volts AC is also referred to as "effective voltage" or "DC equivalent voltage". In reality, the true voltage at the wall outlet is anywhere from 311 to 340 volts peak to peak alternating current in the U.S. at 60 cycles per second. A cycle is a sine wave starting at zero volts at the short slot on the outlet swinging negative to approximately 160 volts negative in respect to neutral or ground then going to approximately 160 volts above ground or neutral 120 times per second. The 110 volts is derived by dividing the peak to peak voltage by two, then multiplying .707 root mean square, which equals the term 110 to 120 volts alternating current.AnswerThe original answer is incorrect. An incandescent lamp will work on a.c. or d.c. and will provide exactly the same power at the same voltage. This is because 110 V d.c. provides exactly the same heating effect as 110 V a.c.
Not all capacitors can be used for 220V; There are capacitors that are specially designed to withstand such high voltages.AnswerThe voltage rating of a capacitor is normally expressed as a d.c. value. If you want to use it on a 220-V a.c. system, then you must take into account that 220 V is an rms value, so you must determine its peak value. The peak value of 220 V (rms) is 311 V. So your capacitor must have a rated value in excess of 311 V d.c., or its insulation will fail.
220 V is rms in europe if that is what you are getting at. Peak is at about 311 V.AnswerUnless otherwise stated, all a.c. voltages and currents are expressed in r.m.s. values.
311, 311!
12% of 311= 12% * 311= 0.12 * 311= 37.32
10% of 311 = 10% * 311 = 0.1 * 311 = 31.1
The positive integer factors of 311 are: 1, 311
1 x 311, 311 x 1
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1 and 311.