From one site: "Though I'm not a worthy engineer ,but my common sense tells that ,the load on any circuit determines the voltage and power of the path of the current.
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the circuit ( very badly drawn,my apologies ) shows a circuit coming with a power supply (a.c in this case) flows through a network ( 3 phase network or the socket or what not) that ends with the terminal 'a' and 'b' ,which could be assumed to be the plug point.a load of variable resistance is depicted onefich the power supply is acting.
now when the switch is not pressed ON
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it has no defined path ,thus no current passes through it .although a voltage is still present. this state is called open circuit.thus the circuit path itself is not complete ,o no current flows .
power: i²r (load) W
since i=0 ,P=0 W.
Now ,when the switch is turned ON it acts as a key and defines the path for the load . But, suppose we remove the load from the circuit i.e R=0 ohms. the path now becomes short circuited.Where there is no drop of voltage although the same current in the branch is present.
P=0W (theoretically),
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but remember the instruments have non ideal components, along with faulty parts.Thus,due to heating effects and other losses there are some power losses ,so exactly the same power doesn't returns to the original source.so YES, there are some power consumed even when the switch is ON with no appliances connected to it.The quantity is uncertain but negligible.
Thanks for asking,thanks for making me realise how important it is to switch off things.
