Let's say you increase the diameter of the hose and all of the fittings to the tank. The same is true of an electrical system: Increasing the voltage will make more current flow. What happens if you increase the pressure in the tank? You probably can guess that this makes more water come out of the hose. Let's say you have a tank of pressurized water connected to a hose that you are using to water the garden. Let's see how this relation applies to the plumbing system. It says that the current is equal to the voltage divided by the resistance or I = V/R. There is a basic equation in electrical engineering that states how the three terms relate. The voltage is equivalent to the water pressure, the current is equivalent to the flow rate, and the resistance is like the pipe size. capacitor) and to draw a significant current (and thus need to rate the cables correctly for this current) but to consume no energy (zero watts).Īs far as kWh are concerned, Earl Wilson remarked: Benjamin Franklin may have discovered electricity, but it was the man who invented the meter who made the money.A neat analogy to help understand these terms is a system of plumbing pipes. As you electrical types know, it is possible to have a voltage across a device (e.g. Power factor is always between zero and 1 because watts (real power) is always less than or equal to volt amperes. This is because the currents for each device are not necessarily in phase with each other.īut – importantly – you can add up the individual VA ratings to get a conservative figure since the actual total (calculated correctly) will always be less than or equal to this value. Unfortunately (apart from direct current circuits) you can’t simply add the VA rating of devices to come up with the total VA rating. You must ensure your wires and associated circuits can cope with 2.2amps (rms). If it is supplied by a 230Vrms ac line you would calculate maximum current as 500 VA/230Vrm = 2.2amps (rms). So to work out the current draw of a device you simply take the VA and divide by rms voltage.įor example, you have a device which is rated at 500VA (maximum VA the device will draw).
Volt amps are very important for calculating current draw (and is essential to know in sizing cables). A standard multimeter is not much help here. The measurement of watts does require specialized equipment where both voltage and current needs to be measured over a specific time. When calculating the real power for multiple devices you simply add the watts for each appliance. Rms volts refers to root mean square voltage (which is peak volts divided by square root of 2).įor dc circuits, this simply becomes W = V (dc) x I (dc). Watts are calculated by W = volts (rms) x amps (rms) x cos (phi) where phi is the angle between the current and voltage for ac circuits (cos (phi) is often referred to as the power factor). People building aluminium smelters would consider these numbers as they would make a big difference to a business case. Other interesting ones are Australia (29c/kWh), UK (20c/kWh) and South Africa and Canada (10c/kWh). For example (for 2011 in US cents), in India it is 8c/kWh against Denmark where it is 41c/kWh (perhaps – and I am speculating here – they are paying for the enormous investment in wind energy).
We have a surprisingly wide range of charges throughout the world for kilowatt-hour charges. This is what you as a consumer generally pay your electrical utility in kilowatt-hours (a 60W light bulb left on for 10 hours consumes 0.6 kWh). One watt is the consumption or generation of energy at the rate of one joule per second. So herewith a quick summary of the differences with some interesting alternative calculations to work out total VA (I am waiting for the deluge of critiques from my electrical brotherhood).
As you would know, electrical products generally indicate both to show how much energy and current they draw.
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