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.
$R _{eq}=\frac{4}{3} \Omega$
$\therefore P=\frac{V^{2}}{R _{eq}}=\frac{4}{4 / 3}=3 W$
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