# The current in a circuit varies inversely as the resistance. The current is 21 Amperes, when the...

## Question:

The current in a circuit varies inversely as the resistance. The current is 21 Amperes, when the resistance is 5 ohms. Find the current for existence of 20 Ohms.

## Proportions and Variation:

When we have a numerical ratio, it indicates the relationship between two numbers {eq}a {/eq} and {eq}b {/eq}. On the other hand, a proportion indicates the similarity between two ratios. When two variables are interdependent, changes in the value of one variable will have a proportional effect on the other. Variations occur when there is an increase or decrease of a variable {eq}a {/eq} with respect to another {eq}b {/eq}, for a ratio or constant {eq}k {/eq}. In the circumstance that we have an inverse variation, it occurs that when one variable decreases the other increases, which we can also write as {eq}a = \dfrac{k}{b}{\text{,}}\,{\text{ or }}\,k = a \cdot b {/eq}.

Given Data:

• Current {eq}1{/eq}, {eq}(I_1) = 21\ \rm A {/eq}
• Resistance {eq}1{/eq}, {eq}(R_1) = 5\ \Omega {/eq}
• Resistance {eq}2{/eq}, {eq}(R_2) = 20\ \Omega {/eq}

Required:

• Current {eq}2{/eq}, {eq}(I_2) = ? {/eq}

In this specific case, we have two values, {eq}I{/eq} for the current and {eq}R{/eq} for the resistance given in the problem. Since {eq}I{/eq} and {eq}R{/eq} vary inversely, the equation for the ratio is:

\begin{align} I &= \frac{V}{R} \\[0.3cm] V &= IR \end{align}

where {eq}V{/eq} is the constant voltage. For this reason, it must be satisfied that:

\begin{align} {R_2} \cdot {I_2} &= {R_1} \cdot {I_1}\\[0.3cm] {I_2} &= \frac{{{R_1} \cdot {I_1}}}{{{R_2}}} & \left[ \text{Solving for } I_2 \right] \\[0.3cm] {I_2} &= \frac{{5\,\Omega \cdot 21\,A}}{{20\,\Omega }} & \left[ \text{Substituting the given values} \right] \\[0.3cm] {I_2} &= 5.25\ \rm A \end{align}

Therefore, when the resistance increase from {eq}5\ \Omega {/eq} to {eq}20\ \Omega {/eq}, the current decreases from {eq}21\ \rm A {/eq} to {eq}5.25\ \rm A {/eq}. 