In an AC circuit E = 50 (500 t) , I = 600 (500 t + 3 ) mA . What is the power dissipated in the circuit? [ 60^ = 0.5 ]

Step-by-step solution

The average power dissipated in an AC circuit is given by P = E_ rms I_ rms . From the given equations, the peak voltage is E₀ = 50 V and the peak current is I₀ = 600 mA = 0.6 A. The phase difference between the voltage and current is = 3 = 60^ . Using the relations E_ rms = E₀ 2 and I_ rms = I₀ 2 , the power is: P = E₀ I₀ 2 Substituting the values: P = 50 0.6 2 60^ P = 30 2 0.5 = 15 0.5 = 7.5 W