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प्रश्न
(a) An emf e = e0 sin ωt applied to a series L - C - R circuit derives a current I = I0sinωt in the circuit. Deduce the expression for the average power dissipated in the circuit.
(b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
उत्तर
Instantaneous power,
P = ei = (e0 sin ωt) [i0 sin (ωt ± Φ)]
= e0i0 sin ωt (sin ωt cos Φ ± cos ωt sin Φ)
= e0i0 cos Φ sin2 ωt ± e0i0 sin Φ sin ωt cos ωt
Average power over one cycle,
`"P"_"av" = "work done in one cycle"/"time for one cycle"`
`= (int_0^"T" "P dt")/"T"`
`= (int_0^"T" ["e"_0"i"_0 cos phi sin^2 omega"t" +- "e"_0"i"_0 sin phi sin omega "t"]"dt")/"T"`
`= ("e"_0"i"_0)/"T" [cos phi int_0^"T" sin^2 omega "t dt" +- sin phi int_0^"T" sin omega "t" cos omega "t" "dt"]`
Now, `int_0^"T" sin^2 omega "t dt" = int_0^"T" ((1 - cos 2 omega "t")/2)`dt
`= int_0^"T" 1/2 "dt" - int_0^"T" (cos 2 omega "t")/2 "dt" = "T"/2 - 1/2 ((sin 2 omega "t")/(2omega))_0^"T"`
`= "T"/2 - 1/(4omega)(sin 2omega"T" - sin 0)`
`= "T"/2 - 1/(4omega) [sin 2((2pi)/"T")"T" - 0]`
`= "T"/2 - 1/(4omega) [0 - 0] = "T"/2`
Also,
`int_0^"T" sin omega"t" cos omega"t" "dt" = 1/2 int_0^"T" sin 2 omega"t" "dt" = 1/2 [(- cos 2 omega"t")/(2omega)]_0^"T"`
`= - 1/(4omega) [cos 2((2pi)/"T")"T" - cos 0] = - 1/(4omega)[1 - 1] = 0`
Hence, `"P"_"av" = ("e"_0"i"_0)/"T" cos phi xx "T"/2 = ("e"_0"i"_0)/2 cos phi`
`= "e"_0/sqrt2 * "i"_0/sqrt2 cos phi`
`= "e"_"rms""i"_"rms" cos phi = "e"_"rms""i"_"rms" ("R"/"Z")`, where the impedance Z = `sqrt("R"^2 +("X"_"L" - "X"_"C")^2)`
(b) `"P"_"av" = "e"_"rms""i"_"rms" cos phi`
The cos Φ factor is also known as the power factor. A low power factor in a circuit used to transport electric power means that the power available on transportation is much less than `"e"_"rms""i"_"rms"`. It denotes a significant loss of power during transport.
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