Study the following electric circuit and calculate the energy drawn from the battery in 10 s.

Explanation

From ohm.s law V =IR
`(1)/(R) = (1)/(R_(1))+(1)/(R_(2))=(1)/(30)+(1)/(20)=(20+30)/(600) implies (50)/(600)= (1)/(12) Omega`
`:. R = 12 Omega`
As `V= IR implies 3 = Ixx12`
`I=(3)/(12)=(1)/(4) A`
Electrical energy W = Vt
`3xx(1)/(4) xx10`
`= (30)/(4) = (15)/(2) = 7.5 J`