Question

Voltage applied to a load is 100√2 sin500t. Current through the load  is 10√2sin(500t + π/3). The power consumed by the load is:
1. 500 W
2. 200 W
3. 1000 W
4. 2000 W

Answer 1

Correct Answer - Option 1 : 500 W

Concept:

Real Power transmitted  to the load can be given as,

P = Vrms × Irms × cos ϕ 

Where,

Vrms = load terminal voltage in volt

Irms = load current in amper

cos ϕ  is the power factor of the load

ϕ is the power factor angle

Calculation:

v = 100√2  sin (500t) V

I = 10√2 sin (500t + π/3) A

Power factor angle \( \phi=\dfrac{\pi }{3} - 0= \dfrac{\pi }{3}\)

Active power consumed by load = V_rms× I_rms × cos ϕ

= \(\dfrac{{100\sqrt2}}{{√ 2 }} × \dfrac{{10\sqrt2}}{{√ 2 }} × \cos \dfrac{\pi }{3} = 100 × 10× \dfrac{1}{2} \)

= 500 W