Resistance, power and brightness of bulb?

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SUMMARY

The discussion centers on the relationship between resistance (R), power (P), and brightness in electric bulbs. The equations P = V^2 / R and P = I^2 * R illustrate conflicting outcomes regarding power changes with varying resistance. When resistance increases, power decreases according to the first equation, while the second suggests power increases with resistance. The conversation concludes that voltage (V) remains constant in a typical outlet, while current (I) inversely varies with resistance.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Familiarity with electrical power equations (P = VI)
  • Basic knowledge of electrical circuits and components
  • Concept of thermal energy release in resistive materials
NEXT STEPS
  • Explore the implications of Ohm's Law in practical circuits
  • Research the effects of varying resistance on circuit performance
  • Learn about the thermal dynamics of resistive heating in electrical components
  • Investigate the role of voltage regulation in electrical systems
USEFUL FOR

Electrical engineering students, hobbyists working with circuits, and anyone interested in understanding the principles of electricity and power management in bulbs and other devices.

pizzaboyx
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Hey,

I have a question regarding practical electricity. Since we know that power (rate of energy released) determines brightness of bulb, then:

Since P = VI and V = RI, it implies
1. P =(V^2) / R
2. P = (I^2) * R

How can we tell if power increases or decreases when resistance (R) of the bulb increases? Using equation 1 it seems P decreases as R increases, but using equation 2 it seems P increases as R increases.

Thanks!
 
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Do you think V remains constant if you put various bulbs with different R's in an outlet?

Do you think VIremains constant if you put various bulbs with different R's in an outlet?
 
Surely you need resistance (R) for there to be power in the form of thermal energy release, but also as it increases the current flow must decrease given a fixed potential difference (V) being applied. You can see this clearly with EQ 1. In EQ 2, it may not be as obvious. (I) is not constant like (V) is in EQ 1, but rather dependent on (R) -- I = V/R.

Considering that the (I) term is squared and the (R) term is not in EQ 2 and (I) is inversely proportional to (R); as (R) decreases, (I) squared "takes over". This is obvious in EQ 1. That is P -> ∞ as R -> 0.
 
Hey guys,

Thanks so much for the replies! Helped me a lot :)

Cheers :)
 

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