A resistor of resistance of 100 is connected to an AC source =10 (250 s ⁻¹ ) t . The energy dissipated as heat during t =0 to t =1 ~ms is approximately.

Step-by-step solution

Given that =10 (250 s ⁻ ) t Compair with = ₀ ( t ) =250 Rad / s The energy dissipated as aligned & H = 1 R ₀^ 10⁻³ ₀^2 ^2 tdt & = ₀^2 2 R ₀^ 10⁻³ (1- 2 t ) dt & = ₀^2 2 R [ t - 2 t 2 ]^ 10⁻³ & = ₀^2 2 R [10⁻³- 2 250 2 250 10⁻³ ] & = 100 2 100 [10⁻³- 1 500 ] & = 1 2 [ 1 1000 - 1 500 ]= 1 2 [ -2 1000 ] & = 0.57 mJ aligned