The equivalent resistance of the parallel combination of the 4Ω and 6Ω resistors is calculated as follows:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}
\]
\[
R_{\text{eq}} = \frac{12}{5} = 2.4 \, \Omega
\]
This equivalent resistance is then in series with the 2Ω resistor, yielding a total circuit resistance:
\[
R_{\text{total}} = R_{\text{eq}} + 2 = 2.4 + 2 = 4.4 \, \Omega
\]
Ohm's law is applied to determine the total current in the circuit:
\[
I = \frac{V}{R_{\text{total}}} = \frac{12}{4.4} = 2.73 \, \text{A}
\]
The total power dissipated is then calculated using the formula \( P = I^2 R \):
\[
P = (2.73)^2 \times 4.4 = 7.46 \times 4.4 = 32.7 \, \text{W}
\]
The total power dissipated by the circuit is 32.7 W.