Solving Ampere-Hour Problem for Camcorder Battery

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SUMMARY

The original ampere-hour (AH) rating of the camcorder battery is 0.625 AH. The current battery, which operates at 1A for 30 minutes, has an effective capacity of 0.5 AH, reflecting a 20% reduction in charge capacity over two years. The user initially miscalculated the original capacity by not properly setting up the proportion for the percentage decrease. Correctly applying the proportion yields the original capacity of 0.625 AH.

PREREQUISITES
  • Understanding of ampere-hour (AH) ratings
  • Basic knowledge of electrical current and charge (Coulombs)
  • Proportional reasoning in mathematical calculations
  • Familiarity with battery performance metrics
NEXT STEPS
  • Research battery chemistry and degradation effects on capacity
  • Learn about calculating ampere-hours for different battery types
  • Explore methods for measuring battery performance over time
  • Study the principles of electrical current and its relationship to time and charge
USEFUL FOR

Electrical engineering students, battery technology enthusiasts, and anyone involved in the design or maintenance of camcorder batteries will benefit from this discussion.

Lildon
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Homework Statement


A 5 volt camcorder now holds 20% fewer electrons than it did two years ago. Today, with a fully charged battery, a user can get 30 minutes of operation before the battery is dead. What was the original AH rating of the battery? The camcorder consumes 1A when operating.

2. The attempt at a solution
I'm in my second class of EE and my professor didn't really teach us anything about problems like this. It says the answer is 0.625AH. I started out thinking since it uses 1A for 30 minutes, the current battery AH rating would be 0.5 so the original battery should be 20% more which is 0.6AH. I also did 1A = 1C/s so 30 minutes would be 1800C x 0.20 = 360C + 1800C = 2160C/3600C = 0.6 AH. Am i doing something wrong?
 
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1A for 30 mins is 0.5Ahr
If this is what you get with only 80% of the original charge what would 100% equate to?
 
Haha wow. so i was just doing the percentage wrong. I set up proportions for a bunch of my other problems but didn't for this one. 0.5Ah/80 = x/100, x = 0.625Ah. Thanks.
 

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