Question

In an electrical circuit with resistors placed in parallel, thereciprocal of the total resistance is equal to the sum of thereciprocals of each resistance:

Answer 1

C

`R_2=16.16\Omega`

Step-by-step explanation:

Given the below equation;

`(1)/(R_c)=(1)/(R_1)+(1)/(R_2)`

We're also given;

`\begin{gathered} R_1=25\Omega \\ R_c=10\Omega \end{gathered}`

Let's substitute the given values into the equation, we'll have;

`(1)/(10)=(1)/(25)+(1)/(R_2)`

Let's subtract 1/25 from both sides of the equation;

`\begin{gathered} (1)/(10)-(1)/(25)=(1)/(R_2) \\ (5-2)/(50)=(1)/(R_2) \\ (3)/(50)=(1)/(R_2) \end{gathered}`

Let's cross multiply;

`\begin{gathered} 3R_2=50 \\ R_2=(50)/(3) \\ R_2=16.16\Omega \end{gathered}`