An AC source given by E = (12 V ) ! ((250 s ⁻¹) t ) is connected across a resistor of resistance 100 . Determine the energy dissipated as heat from t = 0 to t = 1.0 ms .

Step-by-step solution

The instantaneous power dissipated in the resistor is given by P = E ^2 R . The energy dissipated as heat from t = 0 to t = 1.0 ms is: H = ₀^ t P dt = ₀^ 10⁻³ (12 (250 t))^2 100 dt H = 1.44 ₀^ 10⁻³ ^2(250 t) dt Using the trigonometric identity ^2 = 1 - (2 ) 2 : H = 1.44 ₀^ 10⁻³ 1 - (500 t) 2 dt H = 0.72 [ t - (500 t) 500 ]₀^ 10⁻³ Substituting the limits: H = 0.72 ( 10⁻³ - (500 10⁻³) 500 ) H = 0.72 ( 10⁻³ - ( /2) 500 ) H = 0.72 10⁻³ ( 1 - 2 ) Taking 3.1416 : H 0.72 10⁻³ ( 1 - 0.6366 ) H 0.72 10⁻³ 0.3634 H 2.61 10⁻⁴