Power in resistive elements: which equation directly computes power dissipation when both the voltage drop (E) across the resistor and the current (I) through it are known?

Difficulty: Easy

Correct Answer: P = IE

Explanation:


Introduction / Context:
Power relationships in circuits are foundational. Engineers often select formulas based on which quantities are measured. When both current and voltage are known across a component, there is a direct expression to compute power dissipation without needing resistance.


Given Data / Assumptions:

  • Element is resistive for purposes of average power.
  • Voltage drop E (volts) and current I (amperes) are known.
  • We seek instantaneous or average power in the simple V–I form.


Concept / Approach:
The fundamental power relation is P = V * I. For a resistor, this gives the real power converted to heat. Alternative forms P = I^2 * R and P = V^2 / R come from substituting Ohm’s law when only one of V or I is known along with R.


Step-by-Step Solution:

1) Start from the definition of electric power: P = V * I.2) Identify the knowns: both E (same as V) and I are given.3) Conclude the direct formula: P = I * E.


Verification / Alternative check:
Cross-check by Ohm’s law: If R is known, P = I^2 * R = V^2 / R = V * I; all are consistent when the correct substitutions are made.


Why Other Options Are Wrong:

P = E^2 / R and P = I^2 * R: Correct in other contexts, but not needed when both V and I are already known.I = P / E: Rearranged form solving for I, not P.None of the above: Incorrect because P = I * E is valid and preferred.


Common Pitfalls:
Using a form requiring R when R is not known or mixing average power with reactive power in AC circuits.


Final Answer:
P = IE.

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