Two resistors of resistances ( 2 ) and ( 6 ) are connected in parallel. This combination is then connected to a battery of emf ( 2 ~V ) and internal resistance ( 0.5 ). What is the current flowing through the battery?

Step-by-step solution

Current flowing through circuit is ( I= E R_ e q +r ) Since resistance are connected in parallel, therefore, ( 1 R_ e q = 1 R₁ + 1 R₂ 1 R_ e q = 1 2 + 1 6 = 3+1 6 = 4 6 ) ( R_ 애 = 6 4 = 3 2 =1.5 ) Therefore, ( I= 2 1.5+0.5 = 2 2 =1 ~A ) Thus, current flowing through the battery is ( 1 ~A )