Determine the active - reactive and apparent power:

In the given circuit, determine the active, reactive and apparent power.

Figure

Solution

 The inductive reactance, X_L = 2π fL

= 2π × 60 × 0.2

= 75.408 Ω

Impedance, Z = 15 + j 75.408 Ω

In polar form = 76.885 ∠ 78.749^o Ω

Current in the Circuit

 i =  v/ Z   = 220 ∠ 0^o / 76.885 ∠78.749^o

= 2.861 ∠ - 78.749^o Amp (lagging)

Power factor cos φ = cos 78.749^o

= 0.195 (lag)

and      sin φ = sin 78.749^o = 0.98

Active Power

P = VI cos φ

= 220 × 2.861 × 0.195

= 122.7 Watts

or         P = I ² R

= 122.7 Watts

Reactive Power

Q = VI sin φ

= 220 × 2.861 × 0.98

= 616.83 volt-Amp reactive or           

Q = I ² X

= 617.1 VA

Apparent Power

S = VI

= 220 × 2.861

= 629.42 Volt-amp

or        

= 629.2 Volt-amp.

Posted Date: 2/5/2013 1:00:29 AM | Location : United States

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