To figure this out, you need to know the expected forward voltage and current of the LED. Lets assume 2ma and 2V. (Actually, 2ma is small, but I intend to make a point.)
By Kirchoff's Voltage Law, you know that the signed sum of the voltage drops going around a series circuit must add up to zero. This means that the voltage across the resistor must be 228 volts. (-230 + 228 + 2 = 0)
By Kirchoff's Current Law, you know that the signed sum of the currents entering a node is zero. As a consequence, you also know that the current at every point in a series circuit is the same. Therefore, the current through the resistor is also 2ma.
By Ohm's Law, you know that resistance is voltage divided by current, so you know that the resistor is 228V divided by 2ma, which is 114K. The nearest standard value in the E12 scale is 100K. Recalculate the current for 100K, and you get about 2.25ma. (You could also use 120K, and I'll let you run the calculations yourself.)
Don't stop here. There are some issues...
By the power law, you know that power is voltage times current, so you know that the power dissipated by the resistor is 228V times 2.25ma, which is 513mW. I would put a one watt resistor in there.
However, consider this. 2ma is a low current LED. Some of them pull 25ma. The power in the resistor in that case is about 6.5W, which is getting pretty high.
Secondly, you need to consider the reverse breakdown voltage on the LED. I assume that when you said 230V, you meant AC, not DC, which means that there is going to be 230V (actually, a peak value of 325) across that LED for one half the line cycle. You need to check the datasheet and make sure the LED can handle that. If not, you need to put an ordinary signal diode, such as a 1N4148, in parallel with the LED, in the reverse direction, so that it clamps the reverse voltage at about 0.7 volts. (Don't worry about the reverse breakdown on the 1N4148, because the LED will protect it, on opposite half line cycles.)
Last, but not least, you need to consider the safety of the operator. 230V is a high voltage, and LED's are not the most rugged thing around. If the LED breaks, you need to consider if its internal wiring could come into contact with the operator. I would certainly demand a UL listed device in this application.
You are working with two different values. Watts is the product of amps times volts. W = A x V, as you can see if there is a value missing then there can be no answer given.
The 440 volts listed on the cap is the maximum allowable voltage the capacitor can handle. You could actually use a 370 volt cap on 230 volts. ANSWER; 230 volts AC can it actually be 644 volts peak to peak . It is 44ov because it must be rectified and sees only 324 volt peak which is withing the 440 volt capacitor handling voltage
It is a component or substance which reduces the flow of electricity in a circuit. Their will be one in your mobile phone charger to stop 240/230 volts being routed into your phone( which would probably charge it very quickly but then it would blow up!) Also a resistor is the opposite of a conductor i.e something that resists conducting electricity.
The only way to do that is with a transformer. They make inline transformers that convert 230 to 115 and they simply plug in inline.
Philippines follows 60 hz.
You would not connect a current transformer to a 230 v supply. To get 5.6 v 12 mA you could get a 230 to 6 volt transformer, then drop the supply from 6 to 5.6 using a 33-ohm resistor.
The Netherlands uses 220 volts at 50hz and the UK uses 230 volts at 50 hz. So as long as you use a 230 volt light bulb in your UK home the fixture will work perfectly. It can handle the extra 10 volts.
the volts are 230
230 Volts
230 volts
230 volts.
Yes, 220 volts is in the same voltage classification as 230 volts.
No, you will need to obtain an adapter to change the 120 volts to 230 volts. The adapter must be sized to the load wattage of the 230 volt appliance. To find the load wattage multiply the amperage times the voltage of the appliance. Once this is assessed, the size of the adapter or transformer will be of equal or greater value than that of the connected appliance. The adapter will be rated in VA or KVA depending on the needed load amperage.
You are working with two different values. Watts is the product of amps times volts. W = A x V, as you can see if there is a value missing then there can be no answer given.
0.0005
230 volts
That depends on what the suply is if your power suply is 230 then you run it on 230.