Solve Battery Problem: Calculate Resistance & Power Increase

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SUMMARY

The discussion focuses on calculating the resistance and power increase of a flashlight bulb powered by D-cell batteries. The bulb operates at 450mA with two 1.5V batteries, resulting in a power dissipation of 1.35W. When using four D-cell batteries in series, the power increases to 5.4W, demonstrating that doubling the voltage quadruples the power due to the relationship P=IV. The confusion arose from incorrectly maintaining the original current instead of recalculating it based on the new voltage.

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  • Understanding of Ohm's Law (V=IR)
  • Familiarity with power calculations (P=IV, P=I²R)
  • Basic knowledge of series circuits and voltage addition
  • Concept of current constancy in resistive loads
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  • Study the relationship between voltage, current, and power in electrical circuits
  • Learn about series and parallel circuit configurations
  • Explore the implications of Ohm's Law in practical applications
  • Investigate the effects of battery configurations on power output
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Homework Statement


An Ordinary flashlight uses two D-cell 1.5V batteries connected in series. The bulb draws 450mA when turned on. A)Calculate the resistance of the bulb and the power dissipated. B) By what factor would the power increase if 4 D-cell batteries in series were used with the same bulb? (neglect heating effects)



2. Homework Equations
P=IV=I^2R=V^2/R V=IR


3. The Attempt at a Solution
I understand part A, and I got 1.4W for the power. For part B though, I keep getting that it's 2 times greater by doing P=IV where (.45A)(4 x 1.5) =2.7. but I know the answer is supposed to be 4 times. I'm pretty sure the current stays constant and that you add voltages, so I'm confused why it would be 4 and not 2.
 
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You should post these questions either in introductory physics or in the engineering homework help subforum.

I get P = 1.35 W for part a), and 5.4 W for part b)

Doubling the voltage doubles the current, which quadruples the power.

EDIT: Your mistake is that you used the original current in your calculation for part b), rather than using the new current once having calculated it.
 
Last edited:
Thank you. I did post this in the introductory forum about 3 hours ago but no one was able to respond so I thought maybe I needed to put it here, figuring someone would be able to answer it quickly--which fortunately was the case. Sorry to take your time.
 

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