In a series L.C.R circuit, the potential drop across ( L, C ) and ( R ) respectively are ( 40 ~V , 120 ~V ) and ( 60 ~V ). Then the source voltage is

Step-by-step solution

Given, voltage drop across ( L , V_ L =40 ~V ; ) voltage dropacross ( C , V_ c =120 ~V ; ) voltage drop across ( R, V_ R =60 ~V ) Then source voltage ( = V_ R ²+ (V_ L -V_ C )² ) ( = 60²+(120-40)² ) ( = 3600+6400 ) ( = 10000 ) ( =100 ~V ) Therefore, source voltage ( =100 ~V )