Household bulbs rated at power P and operating at 120 V: which equation most directly gives the current rating I?

Difficulty: Easy

Correct Answer: I = P / E

Explanation:


Introduction / Context:
Labeling on household bulbs typically specifies power (watts) at a nominal line voltage (e.g., 120 V). To determine current draw, the simplest relation uses the definition of electric power for a load: P = V * I. This gives a direct way to compute the current rating from nameplate values.


Given Data / Assumptions:

  • Nominal operating voltage E = 120 V (rms).
  • Power rating P given in watts.
  • We seek the current I under nominal conditions.


Concept / Approach:

  • Use P = V * I for steady-state AC with resistive loads (or apparent power S = V * I for general loads; bulbs are predominantly resistive when hot).
  • Rearrange to I = P / E.
  • Other power identities (I^2 R and V^2 / R) are equivalent but require R, which is not given.


Step-by-Step Solution:

Start with P = V * I.Solve for I: I = P / V.Substitute E = 120 V to get I numerically if P is known (e.g., 60 W → 0.5 A).


Verification / Alternative check:

If the filament resistance R were known, you could also use P = V^2 / R and then I = V / R, yielding the same I.


Why Other Options Are Wrong:

  • P = I^2 R and P = E^2 / R: Correct identities but not the most direct here because R is unknown.
  • P = I E: True but not explicitly solved for I; the option 'I = P / E' directly gives the current.
  • None of the above: Incorrect because 'I = P / E' is correct and most direct.


Common Pitfalls:

  • Confusing peak and rms values; nameplate values use rms.
  • Ignoring that filament resistance changes with temperature; nameplate current corresponds to hot operating condition.


Final Answer:

I = P / E

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