A student measures the electric current, I (in amperes, \text{A}), passing through a resis… — ACT

A student measures the electric current, $I$ (in amperes, $\text{A}$ ), passing through a resistor in a closed circuit with a constant voltage. The current is inversely proportional to the resistance, $R$ (in ohms, $\Omega$ ). When the resistance is $4.0\text{ }\Omega$ , the current is $3.0\text{ A}$ . What is the current, in amperes, when the resistance is changed to $6.0\text{ }\Omega$ ?

Answer: 2 A

Answer

The current is

2.0\text{ A}

when the resistance is changed to

6.0\text{ }\Omega

Since current and resistance are inversely proportional, their product remains constant:

I_1 R_1 = I_2 R_2

. Substituting the values gives

(3.0)(4.0) = I_2 (6.0)

, which simplifies to

12.0 = 6.0 I_2

. Solving for

I_2

yields

2.0\text{ A}

Step-by-Step Solution

State the inverse proportionality relationship between current and resistance.

I \times R = k

, where

k

is a constant.

Since the voltage is constant, current and resistance share an inverse relationship.

Calculate the constant value (

k

) using the initial measurements.

k = 3.0\text{ A} \times 4.0\text{ }\Omega = 12.0

Multiplying the known corresponding current and resistance values yields the proportionality constant.

Use the constant to calculate the new current at the new resistance.

I = \frac{12.0}{6.0} = 2.0\text{ A}

Dividing the constant by the new resistance value of

6.0\text{ }\Omega

gives the new current.

Key Concept

In an inverse proportionality relationship, the product of the two variables remains constant (

y \times x = k

). If one variable increases, the other must decrease proportionally.

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