Given:
• R1 = 24.8Ω
,• R2 = 24.8Ω
,• R3 = 12.7Ω
From the diagram, let's find R4 and R5.
We can see that the 3 resistors R1, R2, and R3 are connected in parallel.
Where:
R1 + R2 = R4
To solve for R4, we have:
[tex]\frac{1}{R_4}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]Thus, we have:
[tex]\begin{gathered} \frac{1}{R_4}=\frac{1}{24.8}+\frac{1}{24.8} \\ \\ \frac{1}{R_4}=\frac{1+1}{24.8} \\ \\ \frac{1}{R_4}=\frac{2}{24.8}=\frac{1}{12.4} \\ \\ R_4=12.4\text{ \Omega} \end{gathered}[/tex]Now, to solve for R5 since R3 and R4 are now in series, we have:
[tex]\begin{gathered} R_5=R_3+R_4 \\ \\ R_5=12.7+12.4 \\ \\ R_5=25.1Ω \end{gathered}[/tex]Therefore, we have:
R4 = 12.4 Ω
R5 = 25.1 Ω
ANSWER:
• R4 = 12.4 ,Ω
,• R5 = 25.1 ,Ω