Remember that if two resistors are conected in series, the effective resistance of both of them is given by:
[tex]R=R_1+R_2[/tex]
On the other hand, when two resistors are connected in parallel, the effective resistance of both of them is given by:
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
Notice that the resistors R₂ and R₃ are connected in parallel. Then, the effective resistance of that array is:
[tex]\begin{gathered} \frac{1}{R_{2,3}}=\frac{1}{R_2}+\frac{1}{R_3} \\ \\ =\frac{1}{6.00\Omega}+\frac{1}{13.00\Omega} \\ \\ =\frac{13.00\Omega+6.00\Omega}{(6.00\Omega)(13.00\Omega)} \\ \\ =\frac{19.00}{78.00\Omega} \\ \\ \\ \Rightarrow R_{2,3}=\frac{78.00\Omega}{19}\approx4.1\Omega \end{gathered}[/tex]
On the other hand, the array of resistors 2 and 3 is connected in series to the resistor 1. Then, the effective resistance is given by:
[tex]\begin{gathered} R_{1,2,3}=R_1+R_{2,3} \\ \\ =1.00\Omega+4.1\Omega \\ \\ \therefore R_{1,2,3}=5.1\Omega\approx5\Omega \end{gathered}[/tex]
Therefore, the effective resistance of the entire circuit to the nearest whole number is 5Ω.
