5. (25 points) Six resistors with resistances $R_{1}=1$ Ohm, $R_{2}=2$ Ohm, $R_{3}=3$ Ohm, $R_{4}=4$ Ohm, $R_{5}=5$ Ohm, and $R_{6}=6$ Ohm are connected in series and form a loop. A source of constant voltage is connected to the resulting circuit such that the resistance between its terminals is maximized. The voltage of the source is $U=12 \mathrm{~V}$. Find the power $P_{3}$ dissipated on the resistor $R_{3}$.
5. When connecting a source to any two contacts, the external circuit will represent two parallel branches. Then the total resistance of the circuit is

$R_{\text {total }}=\frac{R_{I} R_{I I}}{R_{I}+R_{I I}}$

For any method of connection, the sum of the resistances in the branches is the same (21 Ohms). The total resistance will be the greatest when the product of the resistances in the branches is maximized. (See the diagram). The power dissipated in the third resistor is

$P_{3}=I_{3}^{2} R_{3}=\frac{U^{2} R_{3}}{\left(R_{1}+R_{2}+R_{3}+R_{4}\right)^{2}}=4.32$ W.

(25 points)