The formula you are looking for is W = I x E. Watts = Amps x Volts. W = 18 x 240 = 4320. Your dryer is rated at 4320 watts.
However, just because your 240 volt outlet is rated at 18 amps does not mean that is what your appliance will draw.
If the appliance is listed (by the manufacturer) as 1500 watts, then 1500 / 240 = 6.25, would mean that it will draw 6.25 amps at 240 volts.
Note:
Within the United States, most regular size electric dryers require a 240 volt, 30 amp power source. In addition to the electric heating element, there is the drum electric motor that must be factored in determining the wattage rating. That is the reason that it is best to use the manufacturer's nameplate rating for the answer to this question.
An electric clothes dryer demands 22A from a 240V outlet at approximately 90% Power Factor. The power demand on the outlet should be about 240V x 22A x 0.9 = 4.75 kW. The active components in an electric clothes dryer are the heating element (100% PF) and the electric motor that turns the tumbler (70-80% PF). The formula you are looking for is W = I x E. (W is watts, I is current in amps, E is volts)
To find the current drawn, use the formula: Current (I) = Power (P) / Voltage (V). Plugging in the values, I = 795 W / 120 V = 6.625 A. Therefore, the appliance draws approximately 6.625 A of current when plugged into a 120 V outlet.
It will increase the current since the water heater is made of a heating element and which is resistive in nature. Ohms law states that V=IR where V is the voltage, I the current and R the resistance. Now the resistance will always remain constant. Thus, when the voltage is increased, the current will also increase.
A clothes dryer typically draws the most amps among common home appliances, with an average of around 20 amps.
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.
4800 watts because watts = volts x amps.
watts = volts X amps, so: 15A X 240V = 3600 watts
1.7amp
We know that Voltage = Current x Resistance, so if E = I x R, then E = 20 x 12 = 240 volts, and the dryer must be plugged into a 240 volt outlet.
An electric clothes dryer demands 22A from a 240V outlet at approximately 90% Power Factor. The power demand on the outlet should be about 240V x 22A x 0.9 = 4.75 kW. The active components in an electric clothes dryer are the heating element (100% PF) and the electric motor that turns the tumbler (70-80% PF). The formula you are looking for is W = I x E. (W is watts, I is current in amps, E is volts)
The formula you are looking for is W = I x E.
The electric heater draws some of the electrical current in the circuit, which reduces the current available for the electric bulb. This leads to a decrease in the brightness of the bulb as it is receiving less power.
By Ohm's Law Voltage = Current x Resistance R = V / I = 120 / 12 = 10 Ohms
11.5 x 240 = 2760W motor.
Power (Watts) = Current (Amps) * VoltagePower = 22Amps * 240 VoltsPower = 5,280 Watts5280
Ohm's Law says Voltage = Current x Resistance V = 5 x 24 = 120 V.
P=VI so I=P/V I= 60/230 I=0.261 A