A current of 3 A flows through the $2\Omega$ resistor as shown in the circuit. The power dissipated in the $5\Omega$ resistor is

Voltage across $2\Omega$ is the same as voltage across arm containing $1\Omega$ and $5\Omega$ resistances. Voltage across $2\Omega$ resistance,

$V=2\times 3=6V$

So, voltage across lowest arm,

V₁ = 6 V

Current across $5\mathrm{\Omega },\mathrm{I}=\frac{6}{1+5}=1\mathrm{A}$

Thus, power across $5\Omega$

P = I² R = (1)² x 5 = 5 W