Which lamp is the brightest?

There are four nodes in the circuit, which we label by numbers from 1 to 4. Node 1 and 4 are connected to the battery so that there is a fixed voltage $U$ between them . If we set node 4 to ground, the potentials at the nodes result $\left( \begin{array}{c} \phi_1 \\ \phi_2 \\ \phi_3 \\ \phi_4 \end{array} \right) = U \left( \begin{array}{c} 1 \\ \xi \\ \eta \\ 0 \end{array} \right)$ where $\xi,\eta \in [0,1]$ are unknowns.

All bulbs have the same ohmic resistance $R$ . The resistance $R_ {ij} = n_ {ij} R$ between two nodes $i \not = j$ is therefore proportional to the number $n_ {ij}$ of light bulbs. The current flowing between the points $i$ and $j$ results according to Ohm's law $I_{ij} = \frac{\phi_{i} - \phi_{j}}{R_{ij}}$

For the nodes 2 and 3 the Kirchhoff's current law applies: The sum of all currents emanating from this point must be zero $\begin{aligned} 0 = \sum_{j \not= 2} I_{2j} &= -I_{12} + I_{23} + I_{24} \\ &= - \frac{\phi_1 - \phi_2}{R_{12}} + \frac{\phi_2 - \phi_3}{R_{23}} + \frac{\phi_2 - \phi_4}{R_{24}} \\ &= \left(- \frac{1 - \xi}{2} + (\xi - \eta) + \frac{\xi}{4} \right) \frac{U}{R} \\ 0 = \sum_{j \not= 3} I_{3j} &= - I_{13} - I_{23} + I_{34} \\ &= - \frac{\phi_1 - \phi_3}{R_{13}} - \frac{\phi_2 - \phi_3}{R_{23}} + \frac{\phi_3 - \phi_4}{R_{34}} \\ &= \left(- \frac{1 - \eta}{2} - (\xi - \eta) + \frac{\eta}{3} \right)\frac{U}{R} \end{aligned}$ This results in a linear system of equations for the unknowns $\xi, \eta$ . In matrix form the system of equations can be represented and solved as follows $\begin{aligned} & & \left(\begin{array}{cc} \tfrac{7}{4} & - 1 \\ -1 & \tfrac{11}{6} \end{array} \right) \cdot \left( \begin{array}{c} \xi \\ \eta \end{array} \right) &= \left( \begin{array}{c} \tfrac{1}{2} \\ \tfrac{1}{2} \end{array} \right) \\ \Rightarrow & & \left( \begin{array}{c} \xi \\ \eta \end{array} \right) &= \frac{1}{\tfrac{7}{4} \cdot \tfrac{11}{6} - 1} \left(\begin{array}{cc} \tfrac{11}{6} & 1 \\ 1 & \tfrac{7}{4} \end{array} \right) \cdot \left( \begin{array}{c} \tfrac{1}{2} \\ \tfrac{1}{2} \end{array} \right) \\ & & &= \frac{1}{53} \left( \begin{array}{c} 34 \\ 33 \end{array} \right) \approx \left( \begin{array}{c} 0.6415 \\ 0.6226 \end{array} \right) \end{aligned}$ With the help of the potentials $\phi_i$ all currents $I_ {ij}$ can now be calculated according to Ohm's law $\begin{aligned} I_{12} = \frac{1 - \xi}{2} \frac{U}{R} &\approx 0.1792 \cdot \frac{U}{R} \\ I_{13} = \frac{1 - \eta}{2} \frac{U}{R} &\approx 0.1887 \cdot \frac{U}{R} \\ I_{14} = \frac{1}{5} \frac{U}{R} &= 0.2 \cdot \frac{U}{R} \\ I_{23} = (\xi - \eta) \frac{U}{R} &\approx 0.0189 \cdot \frac{U}{R} \\ I_{24} = \frac{\xi}{4} \frac{U}{R} &\approx 0.1603 \cdot \frac{U}{R} \\ I_{34} = \frac{\eta}{3} \frac{U}{R} &\approx 0.2075 \cdot \frac{U}{R} \end{aligned}$ Thus, the current flow between 3 and 4 is the highest, so that the lamps shine the brightest here. (However, the differences are very small.)