Resistance, power and brightness of bulb?

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Discussion Overview

The discussion revolves around the relationship between resistance, power, and the brightness of a bulb in practical electricity. Participants explore how changes in resistance affect power output and brightness, using relevant equations.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant notes that power (P) can be expressed as P = VI, leading to two forms: P = (V^2) / R and P = (I^2) * R, raising the question of how power changes with varying resistance.
  • Another participant questions whether voltage (V) remains constant when using bulbs with different resistances (R) in an outlet.
  • A third participant argues that while resistance is necessary for power generation, increasing resistance leads to a decrease in current (I) for a fixed voltage, suggesting that the relationship is not straightforward.
  • This participant also highlights that in the equation P = (I^2) * R, the current is dependent on resistance, which complicates the interpretation of how power changes with resistance.

Areas of Agreement / Disagreement

Participants express differing views on whether power increases or decreases with resistance changes, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There is uncertainty regarding the constancy of voltage and current in the context of varying resistances, as well as the implications of the equations used to describe power.

Who May Find This Useful

This discussion may be useful for individuals interested in electrical engineering, physics, or practical applications of electrical concepts, particularly those exploring the relationships between voltage, current, resistance, and power.

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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