The total power dissipated (in $watt$ ) in the circuit shown is
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as $6\, \Omega \| 3\, \Omega \Rightarrow \mathrm{R}_{\mathrm{P}}=\frac{3 \times 6}{3+6}=2 \,\Omega$
$P=\frac{V^{2}}{R}=\frac{18 \times 18}{2+4}=3 \times 18=54 \mathrm{\,W}$
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