Resistors in series: # R= R_1+R_2+R_3+.....#

Resistors in parallel: # 1/R= 1/R_1+1/R_2+1/R_3+.....#

Start by combining the resistances so that we can work out the current flowing in the various paths.

The # 8Omega# resistor is in parallel with #14Omega# (#3+5+6#) so the combination (let's call it #R_a#) is

#1/R=(1/8 +1/14) = 11/28#

#R_a = 28/11 " "(=2.5454 Omega)#

#R_a# is in series with #4Omega# and the combination is in parallel with #10Omega#, so

#1/R_b = (1/10 + 1/( 4+28/11)) = 0.1 +1/(72/11) = 0.1+11/72=0.2528#

**#R_b = 3.9560 Omega#**

#R_b # is in series with #2Omega# so

#R_(Total) = 2 + 3.9560 = 5.9560 Omega#

#I = V/R = (12)/(5.9560) =2.0148A # (total current flowing from battery)

This current flowing through the #2Omega# resistor divides into 2 paths, the #10Omega# resistor and #R_b#

It is possible to proportion the currents through #R_b# and then #R_a#, but easier to subtract the voltage drops across the #2Omega# and #4Omega# resistors.

#V_(R_a) = 12- (2*2.0148)= 7.9705V#

so current through the #4Omega# resistor

#= (7.9705/(4+R_a)) = 7.9705/(4+(28/11)) = 1.2177A#

so #V_(4Omega)= 4* 1.2177 = 4.8708V#

Voltage across #8Omega# is #7.9705 - 4.8708 = 3.0997V#

So current through the #8Omega# resistor is #3.0997/8 = 0.387A#