In an AC circuit E =200 (50 t ) volt and I=100 (50 t+ 3 ) mA . The power dissipated in the circuit is 30^ = 60^ =0.5 60^ = 30^ = 3 / 2

Step-by-step solution

e=200 (50 t) and I=100 (50 t+ 3 ) Comparing these equations with the standard forms e = e ₀ t and I = I ₀ t , we get, e ₀=200 ~V , I ₀=100 10⁻³ ~A , = phase difference =60^ aligned P=e_ r m s I_ r m s & = e₀ 2 I₀ 2 60^ & = 200 2 100 10⁻³ 2 1 2 =5 ~W aligned