In the given figure, the emf of the cell is $2.2$ V and if internal resistance is $0.6\ \Omega$. Calculate the power dissipated in the whole circuit:
Answer: (D) $2.2$ W
The left side of the rectangle is a plain wire joined to $A$ and the right side a plain wire joined to $B$, so every resistor path runs from $A$ to $B$. The two diagonals cross without touching.
Four parallel paths between $A$ and $B$: $4\ \Omega$ (top), $8\ \Omega$ (bottom), $4 + 8 = 12\ \Omega$ and $2 + 4 = 6\ \Omega$ (diagonals).
$$\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{6} = \frac{15}{24} \Rightarrow R = 1.6\ \Omega$$
$I = \dfrac{2.2}{1.6 + 0.6} = 1$ A. Power dissipated in the whole circuit (including the cell) $= \varepsilon I = 2.2$ W.
Solution by Sreeraj P, M.Sc Physics