Three resistors `8 Omega , 12 Omega and 6 Omega ` are connected to a 12 V battery as shown in Fig. below.

the current through the 8 `Omega ` resistor,

Explanation

If `R_p` is the equivalent resistance of resistors 12 `Omega `and `6 Omega ` connected in parallel, then
` (1)/(R_p) = 1/12 + 1/6 = (1+2 )/( 12) = 3/(12) or R_p = (12)/(3) = 4 Omega `
Now total resistance of the circuit will be the equivalent resistance of series combination of resistors 8 `Omega ` and `R_p (= 4Omega )` i.e., total resistance `R_s = 8 + 4 = 12 Omega `.
` therefore I=(" e.m.f of battery ")/(" total resistance ") = (12V)/( 12 Omega ) = 1A`