When an inductor ' L ' and a resistor ' R ' in series are connected across a 15 ~V , 50 ~Hz a.c. supply, a current of 0.3 ~A flows in the circuit. The current differs in phase from applied voltage by ( 3 )^c . The value of ' R ' is ( 6 = 3 = 1 2 , 3 = 6 = 3 2 )

Step-by-step solution

array ll & Given: E_v=15 ~V , f =50 ~Hz , I =0.3 ~A , & = 3 rad & Impedance Z= E _ v I = 15 0.2 =50 & = X _ L R & 3 = X _ L R & 3 = X _ L R & X _ L = 3 R & Impedance Z= R ^2+ X _ L ^2 & Z = R ^2+( 3 R )^2 & Z = 4 R ^2 2 R = Z & R = Z 2 = 50 2 =25 & array