A student conducts three trials to investigate the mathematical relationships between volt… — ACT

A student conducts three trials to investigate the mathematical relationships between voltage ( $V$ ), current ( $I$ ), resistance ( $R$ ), and electric power ( $P$ ) in a DC circuit. The data collected from these trials are shown in the tables below:

**Trial 1 (Constant Resistance of

10\ \Omega

)**

Voltage ( $V$ , $\text{V}$ )	Current ( $I$ , $\text{A}$ )
$2.0$	$0.20$
$4.0$	$0.40$
$6.0$	$0.60$

**Trial 2 (Constant Voltage of

12\ \text{V}

)**

Resistance ( $R$ , $\Omega$ )	Current ( $I$ , $\text{A}$ )
$2.0$	$6.0$
$4.0$	$3.0$
$6.0$	$2.0$

**Trial 3 (Constant Resistance of

2.0\ \Omega

)**

Current ( $I$ , $\text{A}$ )	Power ( $P$ , $\text{W}$ )
$1.0$	$2.0$
$2.0$	$8.0$
$3.0$	$18.0$

Based on the tables, match each trial to the mathematical relationship that best describes the variables in that trial.

Trial 1: Current ( $I$ ) as a function of Voltage ( $V$ )Direct linear proportionality ( $y \propto x$ )
Trial 2: Current ( $I$ ) as a function of Resistance ( $R$ )Inverse proportionality ( $y \propto \frac{1}{x}$ )
Trial 3: Power ( $P$ ) as a function of Current ( $I$ )Direct quadratic proportionality ( $y \propto x^2$ )

Answer

Trial 1 matches direct linear proportionality; Trial 2 matches inverse proportionality; Trial 3 matches direct quadratic proportionality.

The correct matches align each trial with its proportional trend: Trial 1 displays a constant ratio between current and voltage, indicating direct linear proportionality. Trial 2 shows a constant product between current and resistance, indicating inverse proportionality. Trial 3 shows that power increases with the square of the current, indicating direct quadratic proportionality.

Step-by-Step Solution

Analyze Trial 1 to determine the relationship between Voltage (

V

) and Current (

I

As voltage increases, current increases at a constant rate. Specifically, doubling voltage from

2.0\ \text{V}

4.0\ \text{V}

doubles the current from

0.20\ \text{A}

0.40\ \text{A}

. The constant ratio

\frac{I}{V} = 0.10\ \text{A/V}

confirms direct linear proportionality (

I \propto V

This determines the constant of proportionality and the nature of the relationship when both variables change in the same direction at a constant ratio.

Analyze Trial 2 to determine the relationship between Resistance (

R

) and Current (

I

As resistance increases, current decreases. Doubling the resistance from

2.0\ \Omega

4.0\ \Omega

halves the current from

6.0\ \text{A}

3.0\ \text{A}

. The product

I \times R = 12.0

remains constant, confirming inverse proportionality (

I \propto \frac{1}{R}

This identifies whether the variables have a constant product, which is the defining characteristic of an inverse relationship.

Analyze Trial 3 to determine the relationship between Current (

I

) and Power (

P

As current increases, power increases non-linearly. When current doubles from

1.0\ \text{A}

2.0\ \text{A}

, power increases by a factor of

4

(

2.0\ \text{W}

8.0\ \text{W}

). When current triples from

1.0\ \text{A}

3.0\ \text{A}

, power increases by a factor of

9

(

2.0\ \text{W}

18.0\ \text{W}

). This is a quadratic relationship, representing direct quadratic proportionality (

P \propto I^2

This distinguishes a linear increase from an exponential or power-based increase by calculating the factor changes.

Key Concept

Identifying direct linear, inverse, and quadratic proportional relationships from experimental tables by analyzing how proportional changes in the independent variable affect the dependent variable.

Estimated Time:1m 30s

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