Using the relation P = V^2 / R for a resistive load at fixed voltage, what happens to power when the resistance decreases?

Introduction / Context:Power relationships are central to sizing components and predicting thermal behavior. For purely resistive loads driven by a constant-voltage source, power depends on both the resistance and the current drawn.

Given Data / Assumptions:

• Resistive load, no reactance.
• Applied voltage is fixed.
• Formula of interest: P = V^2 / R.

Concept / Approach:The formula P = V^2 / R shows an inverse relationship between power and resistance at constant voltage. Decrease R → denominator decreases → P increases. This also aligns with P = I^2 * R and I = V / R; as R drops, I rises, and the net effect in P = V^2 / R is increased power dissipation.

Step-by-Step Solution:Start from P = V^2 / R (V fixed).Let R decrease to R_new < R_old.Then P_new = V^2 / R_new > V^2 / R_old = P_old.Therefore, power increases as resistance decreases.

Verification / Alternative check:Pick numbers: V = 10 V. If R = 10 Ω, P = 10^2 / 10 = 10 W. If R halves to 5 Ω, P = 10^2 / 5 = 20 W. Power doubled as resistance halved, confirming the inverse proportionality.

Why Other Options Are Wrong:

• A decrease in power: Opposite of the formulaic result.
• An increase in ohms: Resistance is a parameter, not an outcome here.
• A decrease in current: As R drops, I = V / R increases, not decreases.

Common Pitfalls:

• Mixing formulas and forgetting which variable is held constant (voltage vs current).
• Confusing instantaneous changes with steady-state relations.

Final Answer:an increase in power