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Chapter 6: Problem 84

Solve formula for the specified variable. \(I=\frac{E}{R+r}\) for \(r\)

Short Answer

Expert verified
\( r = \frac{E - IR}{I} \).

Step by step solution

01

Identify the equation and the variable to solve for

The equation given is \( I = \frac{E}{R + r} \) and we need to solve for the variable \( r \).
02

Isolate the fraction

Multiply both sides of the equation by \( R + r \) to get rid of the fraction: \( I(R + r) = E \).
03

Distribute \( I \) on the left side

Distributing \( I \) on the left side gives: \( IR + Ir = E \).
04

Isolate the variable \( r \)

Subtract \( IR \) from both sides to isolate \( Ir \): \( Ir = E - IR \).
05

Solve for \( r \)

Divide both sides by \( I \) to solve for \( r \): \( r = \frac{E - IR}{I} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

isolation of variables
When we're solving for a specific variable, we need to isolate it.
This means we want the variable we're solving for to be by itself on one side of the equation.
For example, in the given equation, we need to solve for the variable \( r \).
We need to isolate \( r \) from the rest of the equation to figure out its value.
distributive property
The distributive property is a helpful algebra principle.
It states that \( a(b + c) = ab + ac \).
This allows us to multiply a single term by each term inside a parenthesis. In our equation, we use this property after multiplying both sides by \( R + r \). So, we get:
\( I(R + r) = E \).
Applying the distributive property here gives us:
\( IR + Ir = E \).
Now, the equation is simpler and has no fractions.
fraction elimination
When we start with an equation involving a fraction, like \( I = \frac{E}{R + r} \), we often want to remove the fraction to simplify our work.
To do this, we need to perform an operation that will eliminate it. Here, multiplying both sides by \( R + r \) helps:
\( I(R + r) = E \).
With this multiplication, we have no more fraction, making it easier to isolate the variable.
solving for variables
Solving for a variable means finding its value.
Here, to solve for \( r \), we use various algebraic tools including isolating the variable, applying the distributive property, and eliminating fractions.
Our final steps to solve for \( r \) are subtracting \( IR \) from both sides and dividing by \( I \):
\( Ir = E - IR \).
Then, divide by \( I \) to get:
\( r = \frac{E - IR}{I} \).
This process uses logical steps to systematically find the desired variable's value.

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Most popular questions from this chapter

Match each division problem in Column I with the correct quotient in Column II. (a) \(\frac{5 x^{3}}{10 x^{4}} \div \frac{10 x^{7}}{4 x}\) (b) \(\frac{10 x^{4}}{5 x^{3}} \div \frac{10 x^{7}}{4 x}\) (c) \(\frac{5 x^{3}}{10 x^{4}} \div \frac{4 x}{10 x^{7}}\) (d) \(\frac{10 x^{4}}{5 x^{3}} \div \frac{4 x}{10 x^{7}}\) A. \(\frac{5 x^{5}}{4}\) B. \(5 x^{7}\) C. \(\frac{4}{5 x^{5}}\) D. \( \frac{1}{5 x^{7}}\)

Multiply or divide. Write each answer in lowest terms. $$ \frac{2 m^{2}-5 m-12}{m^{2}-10 m+24} \cdot \frac{m^{2}-9 m+18}{4 m^{2}-9} $$

Multiply. Write each answer in lowest terms. $$ \frac{12 x^{4}}{18 x^{3}} \cdot \frac{-8 x^{5}}{4 x^{2}} $$

Multiply. Write each answer in lowest terms. $$ \frac{4(y-2)}{x} \cdot \frac{3 x}{6(y-2)^{2}} $$

Find the greatest common factor of each group of terms. $$ 24 m, 18 m^{2}, 6 $$
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