A voltage of 100 V is applied to an impedance of Z = (3 + j4) Ω. What are the values of active power, reactive power and volt-amperes respectively?

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ESE Electrical 2015 Paper 1: Official Paper
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  1. 1200 W, 1200 VAR and 2200 VA
  2. 1600 W, 1600 VAR and 2200 VA
  3. 1200 W, 1600 VAR and 2000 VA
  4. 1600 W, 1200 VAR and 2200 VA

Answer (Detailed Solution Below)

Option 3 : 1200 W, 1600 VAR and 2000 VA
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Detailed Solution

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Concept:

The power triangle is as shown below.

F1 U.B Madhu 23.03.20 D10

P = Active power (or) Real power in W = Vrms Irms cos ϕ

Q = Reactive power in VAR = Vrms Irms sin ϕ

S = Apparent power in VA = Vrms Irms

S = P + jQ

\(S = \sqrt {{P^2} + {Q^2}}\)

ϕ is the phase difference between the voltage and current

Power factor \(\cos \phi = \frac{P}{S}\)

Calculation:

V = 100 V

Impedance, Z = 3 + j4

Power factor, \(\cos \phi = \frac{R}{Z} = \frac{3}{{\sqrt {{3^2} + {4^2}} }} = 0.6\)

Current (I) = V/Z = 100/5 = 20 A

Real power (P) = 100 × 20 × 0.6 = 1200 W

Reactive power (Q) = 100 × 20 × 0.8 = 1600 W

Apparent power (S) = 100 × 20 = 2000 W

The corresponding power triangle is as shown below.

F1 U.B Madhu 23.03.20 D3

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