In the following circuit, the battery has an emf of $2\ \text{V}$ and an internal resistance of $\dfrac23\ \Omega$. The power consumption in the entire circuit is ______ W.
Numerical value type. Enter your answer.
Answer: 3
The left side and the right side of the rectangle are plain wires, so the network is connected between two nodes $L$ and $R$.
- Top edge: $2\ \Omega$; bottom edge: $2\ \Omega$.
- Each diagonal: two $2\ \Omega$ resistors, $4\ \Omega$. By symmetry the crossing point is at the mid potential, so it makes no difference whether the diagonals are joined there.
$$\frac1{R_{LR}} = \frac12 + \frac12 + \frac14 + \frac14 = \frac32 \Rightarrow R_{LR} = \frac23\ \Omega$$
Total resistance $= \dfrac23 + \dfrac23 = \dfrac43\ \Omega$, $I = \dfrac{2}{4/3} = 1.5\ \text{A}$.
Power in the entire circuit (including the internal resistance): $P = EI = 2\times1.5 = 3\ \text{W}$.
Solution by Sreeraj P, M.Sc Physics