Parallel Christmas tree light string: Each bulb has a 1.8 kΩ filament and is connected across a 110 V source in parallel. Approximately what current flows through each individual bulb?

Difficulty: Easy

Correct Answer: 61 mA

Explanation:


Introduction / Context:
In household lighting and basic electronics, bulbs connected in parallel each experience the full source voltage. This problem checks your application of Ohm's law to determine branch current when resistance and applied voltage are known.


Given Data / Assumptions:

  • Supply voltage, V = 110 V (ac RMS).
  • Each bulb's filament resistance, R = 1.8 kΩ = 1,800 Ω.
  • Bulbs are in parallel, so each bulb has 110 V across it independently of the others.
  • Filament resistance is treated as constant for the calculation (temperature effects ignored).


Concept / Approach:
Use Ohm's law for a single branch: I = V / R. In a parallel network, branch currents are computed individually because the voltage across each branch is the same as the source voltage.


Step-by-Step Solution:

Convert resistance: R = 1.8 kΩ = 1,800 Ω.Apply Ohm's law: I = V / R = 110 / 1,800 A.Calculate: 110 / 1,800 ≈ 0.0611 A = 61.1 mA.


Verification / Alternative check:
Rough estimation: 100 V across 2 kΩ would give 50 mA; with 110 V and 1.8 kΩ, expect slightly higher than 50 mA, matching ~61 mA.


Why Other Options Are Wrong:

  • 18 mA: Too low; would require a much larger resistance.
  • 110 mA and 610 mA: Too high; these imply unrealistically low filament resistances for 110 V.


Common Pitfalls:

  • Accidentally dividing by 1.8 instead of 1,800 (forgetting the kilo prefix).
  • Thinking series behavior applies; in parallel, each branch gets full voltage.


Final Answer:
61 mA

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