Question

The resistance of each branch for the arrangement shown below is R = 5 $\Omega$ and vertices are connected with circular ideal conducting wire. The power dissipation in the circuit is

Answer 1

235.2 W

Answer 2

1. Identify the grid as a $4 \times 4$ array of squares, meaning $5 \times 5$ nodes.

2. The statement "vertices are connected with circular ideal conducting wire" implies all outer perimeter nodes are shorted, i.e., at the same potential.

3. Assume the battery's positive terminal is connected to the center node O=$(2,2)$ and the negative terminal to the shorted outer boundary (which includes A=$(4,4)$). This assumption simplifies the problem significantly and aligns with typical grid problems.

4. Calculate the equivalent resistance ($R_{eq}$) between the center node and the shorted outer boundary using nodal analysis and symmetry.

   • Let $V_O$ be the potential at the center node, and $0$ V at the boundary.

   • Due to symmetry, nodes at the same "distance" from the center have the same potential.

   • Nodes adjacent to O (e.g., $(1,2)$) are at potential $V_1$.

   • Nodes adjacent to $V_1$ nodes but not O or boundary (e.g., $(1,1)$) are at potential $V_2$.

   • Applying Kirchhoff's Current Law (KCL) at nodes $(1,2)$ and $(1,1)$ (or any representative nodes for $V_1$ and $V_2$):

     • At $V_1$ node (e.g., $(1,2)$): $4V_1 - V_O - 2V_2 = 0$
     • At $V_2$ node (e.g., $(1,1)$): $4V_2 - 2V_1 = 0 \implies V_1 = 2V_2$

   • Solving these equations gives $V_2 = V_O/6$ and $V_1 = V_O/3$.

   • The total current $I$ from O is $I = \frac{4(V_O - V_1)}{R} = \frac{4(V_O - V_O/3)}{R} = \frac{8V_O}{3R}$.

   • The equivalent resistance $R_{eq} = \frac{V_O}{I} = \frac{3R}{8}$.

5. Substitute $R = 5 \, \Omega$: $R_{eq} = \frac{3 \times 5}{8} = \frac{15}{8} \, \Omega$.

6. Calculate power dissipation: $P = \frac{V^2}{R_{eq}} = \frac{(21 \, \text{V})^2}{15/8 \, \Omega} = \frac{441 \times 8}{15} = \frac{1176}{5} = 235.2 \, \text{W}$.