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If R is the total resistance for a parallel circuit with two resistors of resistances of r1 and r2 then 1/r=1/r1+1/r2 find the resistance r1 if the total resistance is 20 ohms and r2 is 75 ohms round your answer to the nearest ohms if neccesary


a.16 ohms

b.1405 ohms

c.27 ohms

d.102 ohms

Verified Answer :

jtvatsim

A little Algebra, and we're home free!

We are given R = 20 and r2 = 75. We need to find r1.

The formula is 1/R = 1/r1 + 1/r2. Plug-in:
1/20 = 1/r1 + 1/75
1/20 - 1/75 = 1/r1
0.0366 = 1/r1
r1 = 1/0.03666
r1 = 27.272

Our answer is r1 = 27 ohms.

sebasgadu

Answer:

The correct answer is c. \(27\Omega\)

Step-by-step explanation:

- The first step is to get the \(r_1\) variable for alone on one side.

\(\frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}\)

  • Pass the term \(\frac{1}{r_2}\) to subtract to the left side.

\(\frac{1}{r}-\frac{1}{r_2}=\frac{1}{r_1}\)

  • Pass the term \(\frac{1}{r}-\frac{1}{r_2}\) to divide to the right side and pass the term \(r_1\) to multiply to the left side.

\(r_1=\frac{1}{\frac{1}{r}-\frac{1}{r_2}}\)

- The second step is to replace the values of \(r\) and \(r_2\) into the previous expression.

\(r_1=\frac{1}{\frac{1}{20}-\frac{1}{75}}\)

- The third step is to perform the operations to simplify the expression.

\(r_1=\frac{1}{0.05-0.01333}\)

\(r_1=\frac{1}{0.03666}\)

\(r_1=27.2727\Omega\)

As it is necessary to round the answer, the value of \(r_1\) is \(27\Omega\).

Thus, the correct answer is c. \(27\Omega\)

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