Here resistors of value `3 Omega and 6 Omega` are in parallel. They together are in series with resistor `2 Omega`. They all the three together are in parallel with `4 Omega` resistors. Thus the equivalent circuit.
Effective resistance between points A and D
`R' = 2 + (3 xx 6)/3+6 = 4 Omega`
Thus resistance R' is in parallel with resistance `4 Omega`, so total resistance of circuit is
`R = (R' xx 4)/(R + 4) + 1 = (4 xx 4)/(4 + 4) + 1 = 2 +1 = 3 Omega`
Current in circuit, `I = epsilon/r = 4.5/3 = 1.5A`
As arms AD and BC have same resistance so current in arm AD
` =I/2 = 1.5/2 = 0.75 A`
Current in arm FE is, `I = 0.75/(3+6) xx 6 = 0.50 A`
Power dissipated in `3 Omega`,
`P = I_(1)^(2) R = (0.50)^(2) xx 3 = 0.75 W`
