V = 100 sin100t, I = 100 sin (100t + π/6). Then find the watt less power:

The correct answer is option 4) i.e. 2.5 × 103 W.

CONCEPT:

• Watt less power: Watt less power corresponds to the power that cannot perform any work in a given AC circuit. This happens when the average power consumed in the circuit is zero. 

The Watt-less power (P) in a circuit is given by:

P = V_rms × I_rms × cos ϕ

​Where I is the current, V is the voltage, and θ is the angular phase difference between the current and the voltage.

• The voltage applied across an AC circuit if given by V = V₀sin(ωt) and the current flowing across it is given by I = I₀sin(ωt - ϕ)

​Where V₀ is the peak voltage, ω is the frequency, ϕ is the phase lag of current with respect to voltage, and I₀ is the peak current.

CALCULATION:

Given that:

V = 100 sin100t ⇒ V₀ = 100 V

I = 100 sin (100t + π/6) ⇒ I0 = 100 A and ϕ = π/6

Watt less power, P = V_rms × I_rms × cos ϕ

=  \(\frac{V_0}{√2}×\frac{I_0}{√2} × cos \frac{\pi}{6}\)          (∵ V_rms = \(\frac{V_0}{√2}\), I_rms = \(\frac{I_0}{√2}\))

= \(\frac{100}{√2}×\frac{100}{√2} × \frac{√{3}}{2}\) = \(\frac{100 × 100\times √ {3}}{4} \) = 2500 = √3× 2.5 × 103  W