A voltage of V = 150 < 60° V is applied to a load which carries a current of I = 50 < 30° A. The complex power supplied to the load is:

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MPPGCL JE Electrical 19 March 2019 Shift 2 Official Paper
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  1. S = 7.5 < 30° kVA
  2. S = 7.5 < 90° KVA
  3. S = 3 < 30° KVA
  4. S = 3 ≤ 90° kVA

Answer (Detailed Solution Below)

Option 1 : S = 7.5 < 30° kVA
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Detailed Solution

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Power in an AC circuit

The complex power in an AC circuit is defined as the product of phasor voltage and conjugate of phasor current.

S = V × I

where, S = Complex power

V = Voltage

I= Conjugate of current

Calculation

Given, V = 150∠60° 

I = 50∠30°

S = (150∠60°) (50∠30°)*

S = (150∠60°) (50∠-30°)

S = 7.5∠30° kVA

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