The watt rating of a resistor is directly proportional to their physical size. Resistors generate heat just like any resistive load. Small resistors usually are rated at 1/4 watts, larger resistors about 1/4 in diameter have a 2 watt heat dissipation capacity. Ceramic wire wound resistors are wound on a tube form so that there is more surface area and cooling air can pass up through the tube. These types of resistors can dissipate up to 50 watts of unwanted heat.
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None. Resistors do not have watts.
Watts, ohms, amps, volts, farads... are all ways to measure electrical properties, and while there are interrelations, watts refer to the total amount of energy that a device consumes or produces.
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
Find the average of your readings. Divide 220 volts into it and you will have your answer.
The equation for parallel circuit is: 1 / X = 1/R1 + 1/R2 + ... + 1/RN Plug in the known resistances: 1 / 25 = 1 / 220 + 1 / X Solve for X: 1 / 25 - 1 / 220 = 1 / X 8.8 / 220 - 1 / 220 = 1 / X 7.8 / 220 = 1 / X X = 220 / 7.8 X = 28.20512821
A cement resistor is typically used as a power resistor (a resistor whose power rating is greater than 1 W).
A series dropping resistor is a resistor that limits the amount of current flow in a circuit.
No. 2.2K ohm is 2200 Ohms.
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
Find the average of your readings. Divide 220 volts into it and you will have your answer.
LED interface requires typical 220 Ohm resistor in series While in case of relay resistor comes across coil with its value depends on supply voltage
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!
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.
There is insufficient information in the question to answer it. You need to specify something else, such as the resistance of the load.
The equation for parallel circuit is: 1 / X = 1/R1 + 1/R2 + ... + 1/RN Plug in the known resistances: 1 / 25 = 1 / 220 + 1 / X Solve for X: 1 / 25 - 1 / 220 = 1 / X 8.8 / 220 - 1 / 220 = 1 / X 7.8 / 220 = 1 / X X = 220 / 7.8 X = 28.20512821
yes, a variable resistor
220*220 = 48400.
LCM of 110 and 220 is 220.