For a general cost ,A 1500 w tank heater will be $103.00 per month. Go to morgransou.com/energy calculator, then click on farm equipment.
There are many factors that affect the loss of energy:1 the resistance of the wire2 the kettle will keep most heat in but not all heat within the kettle so more energy is lost3 the water will evaporate out the top and that is also a cause. (they can't close it off otherwise the pressure would build and it would explode).
From What I have read it was rated at 230
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One problem associated with the use of solar energy the cost to purchase the solar panels and convert the energy. Another problem is there needs to be a way to store the energy when the sun is not out.
First of all, bulbs are not normally rated in volts, they are usually recognised by how much energy they use per second (measured in Watts). However, in context of the question, there are two basic properties of an electrical circuit; current and voltage. Current is defined as the rate of flow of charge. That means that if a current of 1 amp flows for 1 second, 1 coulomb of charge is transferred. The equation is: Q=IT Charge transferred (in Coulombs) = Current (in Amps) x Time (in Seconds) Voltage is defined as the energy per unit charge: V=W/Q Voltage (in Voltage) = Energy transferred (in Joules) / Charge transferred (in Coulombs) Theoretically, based on these two equations, if you reduce the voltage but keep the current the same, the energy transferred is halved and thus, the light given out by the bulb will be halved. However, depending on what bulb it may not light at all as an LED (light emitting-diode) will only light if there is a bigger voltage than its threshold voltage. Its threshold voltage is the point where as voltage increases beyond that point, the LED's resitance would be almost zero and the bulb will light. A Filament lamp is a non-ohmic conductor (as is an LED) because the resistance is not constant as the voltage increases. This is because as the voltage increases, the energy it transfers is more and the lamp heats up. This heat vibrates the metal lattice of the wire and inhibits electron flow (charge carriers) through the metal and so its resistance increases. Thus depending on the bulb and its resistance, it may light or it may not. A simple answer to all this is, the bulb may light or not depending on the bulb and current across it. This is assuming that the length and cross-sectional area of the wire as well as the number of electron charge carriers (electron number density of the material) is constant.
Surge protectors are rated in joules because joules measure the energy a surge protector can absorb before it fails to protect your devices. The higher the joule rating, the more energy the surge protector can handle. Choosing a surge protector with a higher joule rating can provide better protection for your electronics.
Because all the electrical energy supplied to the heater is emitted as heat. There is no other way for the energy to escape.
Because all the electrical energy supplied to the heater is emitted as heat. There is no other way for the energy to escape.
The voltage is not relevant, only the power (wattage). At 3 kW = 3000 watts, this means 3000 joules per second (J/s), as 1 watt = 1 J/s.CommentThe voltage is relevant! A 3kW electric fire, rated at 230 V, will only produce 3 kW at that rated voltage.
In joules
If you look at your television set's nameplate, it will specify its power in watts. A watt is simply a joule per second. So, if your television is rated at, say, 300 W, then it is transferring energy at the rate of 300 joules per second.
YAG lasers are rated in Joules because they often operate in pulsed mode, delivering energy in short bursts. The energy output per pulse is more significant for applications like laser surgery or materials processing, where the total energy delivered in a pulse is critical. In contrast, milliwatts (mW) measure continuous output power, which is less relevant for the high-energy, short-duration applications typical of YAG lasers. Thus, Joules provide a more meaningful measure of their performance in these contexts.
If a device is rated at 1000 watts and is used for 1 hour, it will consume 1 kilowatt-hour (kWh) of energy. To calculate the energy consumed for a different duration of use, you can use the formula Energy (kWh) = Power (kW) x Time (hours).
for power factor correction kVAR for lasers Joules if you make them with capacitors and know the voltage and frequency you can figure either from microfarads energy =joules=kg-meters=watt -seconds= (C * e^2)/2 var = X(c) * e
A lamp doesn't produce electrical energy; it utilises it. The power-rating of the lamp will tell you the rate at which that lamp uses expends energy at its rated voltage. For example, a 60-W lamp will expend energy at 60 joules per second.
500kw
A suppressor is rated in joules, and a suppressor is rated as to the amount of joules it can expend before it no longer can work to protect the circuit from the power surge.I think you can find the answer to this question in Dehn-usa.com's "Lightning Protection Guide" or their "ABC's of Lightning". I put the link under the related links below.