Ohm’s law power check: With 1 ampere flowing through a 1 ohm resistor, what is the electrical power dissipated in the resistor?

Introduction / Context:Relating current, resistance, voltage, and power is foundational in circuit analysis. When a known current flows through a known resistance, we can compute both voltage drop and power dissipation. This question reinforces the correct unit and formula for power in resistive elements.

Given Data / Assumptions:

• Current I = 1 A through a pure resistor.
• Resistance R = 1 Ω (ohm).
• Steady-state DC conditions; no reactive components.

Concept / Approach:

Ohm’s law: V = I * R. Power relations for resistors: P = V * I = I^2 * R = V^2 / R. Using any of these forms yields the same result if variables are consistent. Units: watt (W) is joule per second, the SI unit of power (rate of energy transfer).

Step-by-Step Solution:

Verification / Alternative check:

Dimensional analysis confirms that ampere^2 * ohm simplifies to watt. Measuring with a wattmeter or by monitoring temperature rise (calorimetry) also validates dissipation in practical setups.

Why Other Options Are Wrong:

• 1 horsepower and 1 Btu: units of power and energy respectively but not equal to this electrical case; their magnitudes differ widely.
• 1 joule: a unit of energy, not power (energy per unit time).
• 1 volt: a voltage value (which is present) but the question asks specifically for power.

Common Pitfalls:

• Confusing power (watts) with energy (joules).
• Assuming “1 A through 1 Ω” implies only a voltage answer; power requires P relations.

Final Answer:

1 watt