In the following circuit, the battery has an emf of 2 V and an internal resistance of 2 3 Ω . The power…

As we can see, two resistances of $2 \Omega$ in the middle portion(in both left and right side) are in parallel, so their combined resistance will be $\frac{2\times 2}{2+2}+\frac{{\displaystyle 2\times 2}}{{\displaystyle 2+2}}=2 \Omega$. Now this resistance along with the topmost & bottommost resistance are in parallel again, so total resistance of the circuit will be,

$R_{\mathrm{eq}}=\frac{2}{3}+\frac{2}{3}=\frac{4}{3} \Omega$

$\therefore P=\frac{V^2}{R_{\mathrm{eq}}}=\frac{4}{{\displaystyle \frac{4}{3}}}=3 \mathrm{W}$