In the following circuit, the battery has an emf of 2 V and an internal resistance of $$\frac{2}{3}$$ $$\Omega$$. The power consumption in the entire circuit is ______ W.

$$ R_{left} = \frac{2 \times 2}{2 + 2} = 1 \, \Omega $$
Similarly, the two right-side resistors are in parallel between the center and Node B:
$$ R_{right} = \frac{2 \times 2}{2 + 2} = 1 \, \Omega $$
The total resistance of this middle path is the series combination of these two halves:
$$ R_{mid} = 1 + 1 = 2 \, \Omega $$
Now, we have three parallel branches between Node A and Node B, each with a resistance of $$2 \, \Omega$$. The equivalent external resistance ($$R_{ext}$$) is:
$$ \frac{1}{R_{ext}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \, \Omega^{-1} $$
$$ R_{ext} = \frac{2}{3} \, \Omega $$
The total resistance of the entire circuit ($$R_{total}$$) includes the external resistance and the internal resistance ($$r$$) of the battery:
$$ R_{total} = R_{ext} + r $$
Given $$r = \frac{2}{3} \, \Omega$$:
$$ R_{total} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \, \Omega $$
The power consumption in the entire circuit is given by the formula:
$$ P = \frac{E^2}{R_{total}} $$
where $$E = 2 \text{ V}$$ is the electromotive force of the battery.
$$ P = \frac{2^2}{\frac{4}{3}} $$
$$ P = \frac{4}{\frac{4}{3}} $$
$$ P = 4 \times \frac{3}{4} $$
$$ P = 3 \text{ W} $$