I agree with TK's calculations With respect to this, I must add that these questions do not have a clear answer because L1 and L2 do not dissipate any energy - they just store it, only to give it up later.
The resistors R1 and R2 will dissipate electric energy as heat, of course ...just like TK has calculated.
Ideal inductors and capacitors are non-dissipative (lossless). Yes,but we are far from ideal. Perhaps a better question to ask would be: "What is the ratio of stored to dissipated energy in R1+L1 ?," etc... Well,as we know what the dissipated energy is in both R1 and R2,the better question would be-what is the dissipated energy in L1 and L2. Due to the fact that i am using a very small amount of energy,and the coil are of a very large size wire,measuring heat dissipation is out of the question. So how to calculate the dissipated energy by L1 and L2 ?. Is it just automatic that we say that it's stored energy,because if it's not,then we have an OU occurrence--and we couldnt have that now-could we lol. If the energy is being stored in the inductor,then the total P/in should be equal to the power being dissipated by the resistor only. If in my case,R2 is removed,the the Pin should go down by the same amount that R2 was dissipating ,providing nothing else is changed during the test. If an inductor/coil is radiating a magnetic and electric field,is it not dissipating !some! power?. Brad
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