Answer:
C
[tex]R_2=16.16\Omega[/tex]
Explanation:
Given the below equation;
[tex]\frac{1}{R_c}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
We're also given;
[tex]\begin{gathered} R_1=25\Omega \\ R_c=10\Omega \end{gathered}[/tex]
Let's substitute the given values into the equation, we'll have;
[tex]\frac{1}{10}=\frac{1}{25}+\frac{1}{R_2}[/tex]
Let's subtract 1/25 from both sides of the equation;
[tex]\begin{gathered} \frac{1}{10}-\frac{1}{25}=\frac{1}{R_2} \\ \frac{5-2}{50}=\frac{1}{R_2} \\ \frac{3}{50}=\frac{1}{R_2} \end{gathered}[/tex]
Let's cross multiply;
[tex]\begin{gathered} 3R_2=50 \\ R_2=\frac{50}{3} \\ R_2=16.16\Omega \end{gathered}[/tex]
