Ohms law is: R= V x I (resistance = voltage / current) ...where an led typically likes 20mA (0.02A) of current running through it and 2V of voltage across it, therefore:
R= V x I = (220-2) / 0.02 = 218 / 0.02 = 10,900 ohms = 10.6kohms (10kohms will do fine).
hope that helps
You will need to know the amount of current flowing through the coil when 220 volts is applied across it. A resistor in series with the coil will limit the current so that the coil only sees 220 volts. The resistor will need to drop 57 volts. So, 57 volts divided by the current in amps will give you your required resistance. You will need a resistor with a high power dissipation rating with 57 volts across it. Your resistor will probaly need to dissipate several watts. For example: A 220 volt coil with 300 milliamps (.3 amps) will require a resistor of 733 ohms. The power dissipation of the resistor would need to be 17.1 Watts! You might try using a light bulb as a series resistor. Ensure that it can handle 57 volts. To complicate matters, is that AC or DC you are using? AC relays have inductance build in, in order to increase the specific "ac resistance", thus the same coil could use as little as 0,001A so you will need a very low value resistor. Anyway, if any 220V relay uses as much as 300mA, I doubt if you will be able to pick it up with one hand! Such a relay coil will draw about 66W of power! I have a 16A rated contact 230V relay. Its current is 0,0015A that is equivelant to 0.33W at 220V!
Actually, if you connect the above described LED to 220 V, it will immediately burn out then the circuit will be "open" and the resistance will be "infinite".
U = RxI so 11x20=220 volts
No. 2.2K ohm is 2200 Ohms.
1.25 A
3.0 or threeAnswerIt depends how they are connected.In series, ther total resistance will be 220 ohms and, so, the current will be 120/220 = 0.545 A.In parallel, ther total resistance will be 20 ohms and, so, the current will be 120/20 = 6 A.
As the question didn't say whether the resistors were in series of parallel, perhaps both eventualities should be looked at. In series, the total resistance is the sum of the series resistors. The 30-ohm and 60-ohm resistors sum to 90 ohms. Total current will be voltage divided by resistance. The 220 volts applied divided by the 90 ohms will give 2.44 amps (2 4/9ths amps). In parallel, each resistor is connected indendently across the 220-volt source. Each one will "feel" the 220 volts and draw current accordingly. Since current equals voltage divided by resistance (just like always), we have to make the calculation for each resistor. The 220 divided by 30 equals 7.33 amps (7 1/3 amps). The 220 divided by 60 equals 3.66 amps (3 2/3 amps). The two branch currents each represent a part of the total circuit current, we have to find the sum of the branch currents to find the total current. Our 7.33 plus 3.66 amps sums to 11 amps.
I(current) = V(voltage)/R(resistence) Example : 220 V / 5000 Ohm = 0.044 A (Ampère) = 44mA
The formula you are looking for is R = E/I
The formula you are looking for is R = E/I.
A 4000-Ω resistor is connected across 220 V will have a current flow of 0.055 A.Ohm's law: Voltage equals Current times Resistance
At 110 volts it is 0.8 amps. At 220 it is 0.4 amps. I=E/R. I=amps.E=volts R=resistance.