Brightness of Light Bulbs Ranking Task

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SUMMARY

The discussion focuses on ranking the brightness of five identical light bulbs (A through E) connected in a circuit with an ideal battery. The conclusion is that Bulb C is the brightest, followed by Bulbs A and B, which are equivalent in brightness, and finally Bulbs D and E, which are also equivalent. The reasoning is based on the principles of electrical resistance and power, specifically using the equations V=IR and P=IV. Bulb C has a higher voltage due to its configuration, leading to greater power output despite all bulbs having the same resistance.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Familiarity with electrical power calculations (P=IV)
  • Knowledge of series and parallel circuits
  • Basic concepts of voltage and current in electrical circuits
NEXT STEPS
  • Study the effects of series vs. parallel configurations on circuit behavior
  • Learn about equivalent resistance in series and parallel circuits
  • Explore the relationship between voltage, current, and power in electrical systems
  • Investigate practical applications of circuit analysis in real-world scenarios
USEFUL FOR

Students studying electrical engineering, educators teaching circuit theory, and anyone interested in understanding the principles of electrical circuits and light bulb brightness ranking.

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Homework Statement


Consider a circuit containing five identical light bulbs and an ideal battery. Assume that the resistance of each light bulb remains constant. Rank the bulbs (A through E) based on their brightness.
Rank from brightest to dimmest. To rank items as equivalent, overlap them.
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Homework Equations



V=IR, junction law and loop law

The Attempt at a Solution



i use that fact that more resistance means that less current flow by looking at the pic the current in A&B are the same, but the current in C is greater than in D&E where current in D&E are same b/c of series. there for conclude that C>(A=B)>(D=E)

can any 1 tell me if this is right please!
 
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its right
 
If anyone didn't mind me asking, I was wondering WHY Bulb C has less resistance compared to Bulb A or B?
 
It doesn't have less resistance, all resistances are the same. It has more voltage because the equivalent resistance of RcRdRe is greater than that of RaRb. If you link brightness to power (P=I*V) and Rab has the same current as Rcde, then because the voltage is higher Rc has more power.
 

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