Ohm's Law & Power: Bicycle LED Headlight & 4 AA Batteries

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Discussion Overview

The discussion revolves around the application of Ohm's Law and power calculations related to a bicycle LED headlight powered by four AA rechargeable batteries. Participants explore concepts such as resistance, power consumption, and battery life without seeking definitive answers.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant states the LED headlight requires 5 V and 250 mA, and uses Ohm's Law to propose calculating the equivalent resistance.
  • Another participant notes the relationship between power and voltage, questioning how to incorporate time into the power calculation.
  • A participant mentions that 1 watt equals one joule per second and provides a conversion for current to electrons per second.
  • One participant calculates that each battery, rated at 2300 mA-Hour, would last for 10 hours if supplying 230 mA, and proposes a formula for determining battery life based on current draw.

Areas of Agreement / Disagreement

Participants express uncertainty about specific calculations and methods, indicating that multiple approaches and interpretations exist without a consensus on the best way to proceed.

Contextual Notes

Participants have not fully resolved how to apply time in power calculations or the implications of battery specifications on operational duration.

Who May Find This Useful

Readers interested in electrical engineering, physics, or practical applications of Ohm's Law and power calculations may find this discussion relevant.

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A bicycle LED headlight requires 5 V, 250 mA to operate. Using Ohm’s law, determine the
equivalent resistance of the light. What is the power consumption rate in W? How many
electrons pass through per second? It is powered by four AA rechargeable battery in series. Each battery provides 1.25 V and has a specification of 2300 mA-Hour. How many second would the battery last?

The first part i used R = v/I

The second part I know P = VI but am not sure what to use for time because it does't give one.

The third part i don't know about. I am not looking for the answer but help on how to get there thank you
 
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can somebody give me an idea
 
Isn't 1 watt = one joule per second?

Maybe better: 1 ampere = 6.242*1018 electrons/second
 
Last edited:
Each battery provides 1.25 V and has a specification of 2300 mA-Hour. How many second would the battery last?

This means each battery would last for an hour if it was supplying 2300 mA. It would last for 10 hours if it was supplying 230 mA.

So product of (current * time) = 2300 mAH
or
time = 2300 mAH / current

So, how long would it last if it was supplying 250 mA?
 

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