# The resistance in an electric circuit varies inversely with the current. If the resistance is 12...

## Question:

The resistance in an electric circuit varies inversely with the current. If the resistance is 12 ohms when the current is 4 amps, what is the resistance when the current is 8 amps?

## Proportions and Variation:

We can say that rational numbers are related to ratios, proportions, and variations since they can be expressed as a fraction. A ratio shows us the relationship between two numbers or variables {eq}x {/eq} and {eq}y {/eq}. While a proportion shows us the equality of two ratios. When two variables are interdependent, changes in the value of one variable will have a predictable effect on the other. Variations occur when there is the increase or decrease of a variable {eq}x {/eq} with respect to another {eq}y {/eq}, for a ratio or constant {eq}k {/eq}. In the case that we have an inverse variation, it occurs that when one variable decreases the other increases, which we can express as {eq}x = \frac{k}{y}{\text{,}}\,{\text{ or }}\,k = x \cdot y {/eq}.

## Answer and Explanation:

{eq}\eqalign{ & {\text{In this particular case we have two values }}I\,{\text{(current) and }}R\,{\text{(resistance) that have }} \cr & {\text{a variation in inversely proportional form}}{\text{. So we have:}} \cr & \,\,\,\,{I_1} = 4\,A \cr & \,\,\,\,{R_1} = 12\,\Omega \,\,\,\,\,\,\left[ {resistance{\text{ }}when{\text{ }}the{\text{ }}current{\text{ }}is{\text{ }}{I_1} = 4\,A} \right] \cr & \,\,\,\,{I_2} = 8\,A \cr & \,\,\,\,{R_2}:\,\,\,\left[ {resistance{\text{ }}when{\text{ }}the{\text{ }}current{\text{ }}is{\text{ }}{I_2} = 8\,A} \right] \cr & {\text{Since }}I{\text{ and }}R{\text{ vary inversely}}{\text{, the equation for the ratio is }}I = \frac{V}{R}{\text{, or }}V = R \cdot I{\text{, where }} \cr & V{\text{ is the constant voltage}}{\text{. For this reason}}{\text{, it must be satisfied that:}} \cr & \,\,\,\,{R_2} \cdot {I_2} = {R_1} \cdot {I_1} \cr & {\text{Now}}{\text{, solving for }}\,{R_2}{\text{:}} \cr & \,\,\,\,{R_2} = \frac{{{R_1} \cdot {I_1}}}{{{I_2}}} \cr & {\text{So}}{\text{, substituting the given values:}} \cr & \,\,\,\,{R_2} = \frac{{12\,\Omega \cdot 4\,A}}{{8\,A}} = \left. {\underline {\, {6\,\Omega } \,}}\! \right| \cr & {\text{Therefore}}{\text{, when the current increases from }}4\,A{\text{ to }}8\,A{\text{, the resistance decreases }} \cr & {\text{from }}12\,\Omega {\text{ to }}6\,\Omega {\text{ ohm}}{\text{.}} \cr} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from Geometry: High School

Chapter 7 / Lesson 1