The power used, assuming Unity Power Factor (resistive load), is the product of resistance and the square of the current -- or 1210 Watts.
Use the formula: P=IR (power = current x resistance).
Power = Potential Difference (Voltage) x Current So in this case, Power = 6 x 0.5 = 3 watts
P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.
There is no direct relation of electric current and power. In order to knowhow much power (or energy) the current gives up, you must know what thecurrent is flowing through.The easiest way to describe anything through which the current is flowing isto measure and state its electrical resistance.Once you know the resistance through which the current is flowing . . .Power delivered by the current = (magnitude of the current, amperes)2 x (resistance of the path, ohms)The power is delivered continuously. Its unit is watts.Each watt of power means 1 joule of energy every second.
The amount of electric energy that is converted into thermal energy increases as the resistance of wire increases. As the resistance in the current increases, the current in the circuit decreases.
Ohm's law: Voltage = Current times Resistance Solve: Resistance = Voltage divided by Current So, a device drawing 50ma with 150V has a resistance of 150 / 0.05, or 3000 ohms. p.s. Since power is volts times amps, that device is dissipating 7.5 watts.
Power is inversely proportional to resistance. Ohm's law: Current is voltage divided by resistance Power law: Power is voltage times current, therefore power is voltage squared divided by resistance.
All resistances will emit heat energy when a current flows. The heat production rate (or power) can be found by any of these formulas: Power = Current * Voltage Power = Current2 * Resistance Power = Voltage2 / Resistance. Power is given in Watts when Current is in Amps, Voltage in Volts, and Resistance in Ohms.
If the resistance is 1.2k and the current is 0.024 ma, then the voltage is 0.0288 volts. (Voltage = resistance times current) If the voltage is 0.0288V and the current is 0.024 ma, then the power is 0.6912 microwatts. (Power = voltage times current)
A: If the transformer is connected to a power input of course it will draw current. The primary is a long wire it has own resistance wrap around an iron core. Of course there will be primary current whether there is a load on the secondary or not.
Power=current squared times resistance
Power = (current) times (voltage)Current = (Power) divided by (voltage)Voltage = (Power) divided by (current)
Use the formula: P=IR (power = current x resistance).
The power dissipated across a resistor, or any device for that matter, is watts, or voltage times current. If you don't know one of voltage or current, you can calculate it from Ohm's law: voltage equals resistance times current. So; if you know voltage and current, power is voltage times current; if you know voltage and resistance, watts is voltage squared divided by resistance; and if you know current and resistance, watts is current squared times resistance.
To find the resistance necessary, one would need to know how much current the bulb draws. If one knows the current the bulb draws, then one would subtract the 14 volts from 120 volts then divide that by the current the bulb draws and one will find the resistance needed. Once this has been done, one would need to multiply the current drawn by the voltage drop to get the wattage rating necessary. Another important detail to note is that the power dissipated by the resistor will be much greater than the power consumed by the bulb itself. Finally if the bulb burns out the voltage across the contacts will be 120V. I would not recommend using this method to drop the voltage for the bulb.
if the resistance is decreased and the current stays the same, then the power decreases.
P=I^2*R where P=power I=Current R=Resistance