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Obtain an expression for average power of AC over a cycle. Discuss its special cases. - Physics

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प्रश्न

Obtain an expression for average power of AC over a cycle. Discuss its special cases.

दीर्घउत्तर

उत्तर

It is given by the product of the voltage and current. In an AC circuit, the voltage and current vary continuously with time. Let us first calculate the power at an instant and then it is averaged over a complete cycle.

The alternating voltage and alternating current in the series RLC circuit at an instant are given by

υ = Vm sin ωt and i = Im sin (ωt + 4>)

where φ is the phase angle between υ and i. The instantaneous power is then written as

P = υi = VIm sin ωt sin(ωt + φ)

= VIm sin ωt (sin ωt cos φ – cos ωt sin φ)

P = VIm (cos φ sin2 ωt – sin ωt cos ωt sin φ) …… (1)

Here the average of sin2 ωt over a cycle is `1/2` and that of sin ωt cos ωt is zero. Substituting these values, we obtain average power over a cycle.

Pav = VIm cos φ `xx 1/2 = "V"_"m"/sqrt2 "I"_"m"/sqrt2` cos φ

Pav = VRMS IRMS cos φ …… (2)

where VRMS IRMS is called apparent power and cos φ is power factor. The average power of an AC circuit is also known as the true power of the circuit.

Special Cases:

  1. For a purely resistive circuit, the phase angle between voltage and current is zero and cos
    φ = 1.
    ∴ Pav = VRMS IRMS
  2. For a purely inductive or capacitive circuit, the phase angle is 
    `+- pi/2 and cos (+- pi/2) = 0`
    ∴ Pav = 0
  3. For series RLC circuit, the phase angle φ = tan-1 `(("x"_"L" - "x"_"c")/"R")`

    ∴ Pav = VRMS IRMS cos φ

  4. For series RLC circuit at resonance, the phase angle is zero and cos φ = 1.
    ∴ Pav = VRMS IRMS

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अध्याय 4: Electromagnetic Induction And Alternating Current - Evaluation [पृष्ठ २६१]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 4 Electromagnetic Induction And Alternating Current
Evaluation | Q III. 23. | पृष्ठ २६१

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