First find the equivalent resistance between points #a #and # b# :
= #(6\times 3)/ (6+3)# ------ as they are in parallel.
= #18/9#
= #2 Omega#
Now the equivalent resistor of the total circuit will be #4 Omega# in series with #2 Omega#, i.e. #4+2= 6 Omega#
So the current #i# flowing through the circuit will be
# i = (12V)/ (6Omega)# = #2# Ampere
Voltage drop across the 4 ohm resistor will be :
#V= i R = 2A\times4Omega = 8V#
Now 12V - 8V = 4V is the voltage drop across the points #a and b#
It will be same in both the #6 Omega# resistor and the #3 Omega# resistor as they are parallel.So
# v_0 = 4V#
Now, as we know the voltage across #3Omega# resistor is #4V#, we can find the current #i_0# flowing through it;
#i_0 = (4V)/(3Omega) = 1.333 A#
The power dissipated in the #3Omega# resistor will be :
#P = VI#
#P = 4 \times 1.333#
#therefore P = 5.333W#
