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Question
Derive an expression for the average power dissipated in a series LCR circuit.
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Solution
We have seen that a voltage v = vm sin ωt applied to a series RLC circuit drives a current in the circuit given by `"i" = "i"_"m" sin(omega t + ∅)` Where
`"I"_"m" = "v"_"m"/"z" and ∅ = tan^-1 (("X"_"c"-"X"_"L")/("R"))`
Therefore, the instantaneous power p supplied by the source is
`"p" = "vi" = ("v"_"m"sin omega t) xx ["i"_"m" sin(omega t + ∅)]`
= `("v"_"m""i"_"m")/(2)[cos ∅ - cos(2omega t + ∅)]`
The average power over a cycle is given by the average of the two terms in R.H.S. of the above equation. It is only the second term which is time-dependent. Its average is zero (the positive half of the cosine cancels the negative half). Therefore,
`"P" = ("v"_"m""i"_"m")/(2) cos ∅ = ("v"_"m")/sqrt2 ("i"_"m")/sqrt2 cos ∅`
= `"V" "I" cos ∅`
This can also be written as
`"P" = "I"^2 "Z"cos ∅`
So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle ∅ between them. The quantity cos ∅ is called the power factor.
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