Match each circuit modification on the left with its corresponding effect on power dissipa… — Armed Services Vocational Aptitude Battery (ASVAB)

Match each circuit modification on the left with its corresponding effect on power dissipation relative to the baseline power ( $P_0$ ) on the right.

Doubling the applied DC voltage across a constant load resistance ( $V \to 2V$ , $R$ constant)Power dissipation increases to $4P_0$ (quadrupled)
Halving the current flowing through a fixed-value resistor ( $I \to \frac{1}{2}I$ , $R$ constant)Power dissipation decreases to $\frac{1}{4}P_0$ (quartered)
Doubling the load resistance connected across a constant-voltage source ( $R \to 2R$ , $V$ constant)Power dissipation decreases to $\frac{1}{2}P_0$ (halved)
Tripling the current through a load while simultaneously reducing its resistance to one-third ( $I \to 3I$ , $R \to \frac{1}{3}R$ )Power dissipation increases to $3P_0$ (tripled)

Doubling voltage with constant resistance matches

4P_0

; Halving current with constant resistance matches

\frac{1}{4}P_0

; Doubling resistance with constant voltage matches

\frac{1}{2}P_0

; Tripling current with one-third resistance matches

3P_0

Each modification is correctly determined by applying Watt's Law formulas (

P = \frac{V^2}{R}

and

P = I^2 R

). Doubling voltage quadruples power (

2^2 = 4

), halving current reduces power to one-fourth (

(\frac{1}{2})^2 = \frac{1}{4}

), doubling resistance at constant voltage reduces power by half (

\frac{1}{2}

), and tripling current while reducing resistance to one-third increases power threefold (

3^2 \times \frac{1}{3} = 3

Select the appropriate form of Watt's Law for each scenario depending on which parameters are held constant and which are modified.

Use

P = \frac{V^2}{R}

when voltage and resistance are given, and

P = I^2 R

when current and resistance are given.

Using the formula directly containing the relevant independent variables simplifies scaling calculations.

Calculate the scaling factor for doubling voltage with constant resistance.

P' = \frac{(2V)^2}{R} = 4\frac{V^2}{R} = 4P_0

Power is directly proportional to the square of the voltage across a fixed load.

Calculate the scaling factor for halving current with constant resistance.

P' = \left(\frac{1}{2}I\right)^2 R = \frac{1}{4}I^2 R = \frac{1}{4}P_0

Power is directly proportional to the square of current through a fixed load.

Calculate the scaling factor for doubling resistance at constant voltage.

P' = \frac{V^2}{2R} = \frac{1}{2}\frac{V^2}{R} = \frac{1}{2}P_0

At constant voltage, power is inversely proportional to resistance.

Calculate the scaling factor for tripling current while reducing resistance to one-third.

P' = (3I)^2 \cdot \left(\frac{1}{3}R\right) = 9I^2 \cdot \frac{1}{3}R = 3I^2 R = 3P_0

The squared current factor of 9 multiplied by the fractional resistance factor of 1/3 results in an overall factor of 3.

Watt's Law Power Scaling Relationships:

P = VI = I^2 R = \frac{V^2}{R}

Estimated Time:1m 30s

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