In a L R circuit of 3 mH inductance and 4 resistance, emf E=4 1000 t volt is applied. The amplitude of current is

Step-by-step solution

E=E₀ t Given E=4 1000 t From Eqs. (i) and (ii), we get Peak value of emf, E₀=4 ~V Augular fiequency, =1000 ~Hz Now pcak valuc of currcnt is aligned i₀ &= E₀ Z = E₀ R²+X_ L ² &= E₀ R²+ ² L² aligned Putting E₀=4 ~V , R=4 , array l =1000 ~Hz , L=3 mH =3 10⁻³ H array we get i₀=0.8 ~A