A transformer is used to light a 100 ~W and 110 ~V lamp from a 220 ~V mains. If the main current is 0.5 ~A , the efficiency of the transformer is approximately

Step-by-step solution

The efficiency of transformer = Energy obtained from the secondary coil Energy given to the primary coil Or = Output power Input power or = V_ s I_ s V_ p I_ p Given, V_ s I_ s =100 ~W , V_ p =220 ~V , I_ p =0.5 ~A Hence, = 100 220 0.5 =0.90=90 % Note The efficiency of an ideal transformer is 1 (or 100 % ). But in practice due to loss in energy, the efficiency of transformer is always less than 1 (or less than 100 % ).