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My textbooks states that the brightness of a light bulb that is connected in a circuit is determined by its power. That is, a light bulb connected in series would be brighter if it has a higher resistance because of the equation $P=I^{2}R$ , but a light bulb connected in parallel would be brighter if it has a lower resistance, as shown by the equation: $P=\frac{V^{2}}{R}$.

Here are my questions:

  • Why did we choose each power formula for each type of circuits?

  • If I was given the power of a light bulb that is connected in series, but I only have its voltage, is it okay to solve for $R$ in the second formula I mentioned above and then substitute in the first formula to find the power?

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  • $\begingroup$ What is the light bulb connected in series with? $\endgroup$ – Bob D Apr 30 at 20:52
  • $\begingroup$ We can assume that two lightbulbs are connected in series with a battery . $\endgroup$ – Positron12 Apr 30 at 20:54
  • $\begingroup$ Identical light bulbs? $\endgroup$ – Bob D Apr 30 at 20:56
  • $\begingroup$ We can assume that they are identical. $\endgroup$ – Positron12 Apr 30 at 20:57
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The power tells us how much energy per second is absorbed by the device. Presumably, a higher power means a higher temperature for the filament in the bulb. The filament is simply a resistive wire.

The power consumed by an electrical device operating in direct current mode (DC) is the product of the current through and voltage across the device: $$P_x=I_x V_x.$$

Because a filament bulb (as opposed to an LED) acts in accordance with Ohm's Law, $$V_x=I_x R$$ the power consumed by the bulb can be written one of two ways: $$P_x=I_x (I_x R) = I_x^2 R $$ $$\text {or } P_x = \frac{V_x}{R}V_x = \frac{V_x^2}{R}.$$

The form you use is whichever one is convenient. For parallel connections of two bulbs, they will have the same voltage, $V_x$, so the difference in power depends on the individual resistances. The bulb with the smaller $R$ will absorb more power and be brighter.

In the series connection of two bulbs, the current, $I_x$, will be the same in each, so the bulb with the larger resistance will absorb more power.

If the bulbs have identical resistances, they will be equally bright, BUT, the bulbs in parallel will always be brighter than the bulbs in series if they are connected to the identical supplies for both circuits.

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  • $\begingroup$ Lets say I was given the power of two lightbulbs, one has a power of 60W and the other 100W, how could I determine which one is brighter? If I used this formula: P=V^2\P I will conclude that a greater power means lower resistance, but using this formula P=I^2R I will get opposite results. $\endgroup$ – Positron12 May 1 at 8:29
  • $\begingroup$ The power ratings on a filament light bulb refer to the power that bulb would absorb if (in the USA) it was attached to a voltage source of 120 V RMS (time averaged voltage of 120 V). So $120^2/R=P$ at operating temperature, so you can calculate the resistance of the filament when it is hot. Then you could calculate the current after knowing the resistance. If you don't use a 120 V source, or you have more or less than 120 V, the power absorbed will be different. The bulb does NOT always absorb the power value printed on the bulb. $\endgroup$ – Bill N May 1 at 13:40
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There are three common power formulas that you can use:

1) $P=IV$

2) $P=I^2R$

3) $P=V^2/R$

You may use any of the three where two variables are known, and one variable is unknown.

Regarding why you chose the equations that you did, for the bulbs connected in series, all bulbs in series have the same current going through them whether or not they have equal electrical resistance. Thus, the bulb with the highest resistance has the highest value if $P=I^2R$, so that bulb is the brightest.

For the bulbs connected in parallel, all bulbs in parallel have the same voltage drop across them whether or not they have equal electrical resistance. Thus, the bulb with the lowest resistance would have the highest value of $P=V^2/R$, and that bulb would be brightest.

For a bulb where you know the power of the bulb and the voltage drop across the bulb, it is OK to solve for the power of the bulb with two equations, but it would be somewhat more straight-forward to solve for power with equation 3 above.

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