# 100 W and 60W bulb plugged in separately....Which with more resistance?

• avaz567
In summary, the 100 W bulb is brighter, but it is unclear why the resistance is higher for the 60 W bulb. It is possible to solve this problem algebraically, but it is also possible to understand it conceptually by comparing powers.
avaz567

## Homework Statement

If you have a 100 W bulb and a 60 W bulb connected to separate electrical outlets, which will be brighter? Which has more resistance? Explain.

2. Necessary formulas
P=IV and (maybe?) I=V/R

3. The Attempt at a Solution

I know for sure the 100 W bulb is brighter. However, when it came to the resistance part, I’m confused. (When we solved this in class, it was that the 100 W had less resistance, but I don’t understand how we got to that part. Someone had told me that the outlets would have the same voltage drop, but I’m not sure about that.)
Thanks in advance to those who can help.

avaz567 said:
P=IV and (maybe?) I=V/R
Both are correct but there is another equation for the power that does not involve current and is more useful for this question.

On Edit: Perhaps it would be useful to know that the stamp 100 W - 120 V on a light bulb means that the bulb will consume 100 W of power when connected to a 120 V outlet. Outlets in the US are assumed to provide 120 V across their terminals regardless of their load.

avaz567 said:
...
Someone had told me that the outlets would have the same voltage drop,
...
This is correct.

kuruman said:
Both are correct but there is another equation for the power that does not involve current and is more useful for this question.
I think I’ve seen that other equation you’re talking about, yet our teacher did not give that to us (I believe we are using in AP Physics II.)

avaz567 said:
I think I’ve seen that other equation you’re talking about, yet our teacher did not give that to us (I believe we are using in AP Physics II.)
You don't need a teacher to give it to you. Start with P = IV and V = IR and use some simple algebra to eliminate the current I from the power equation.

CWatters
OmCheeto said:
This is correct.
So, if I were to plug numbers in given that the voltage drop was the same, I see that the 100 W has a lower amount of resistance, but why is this so? I thought that if a bulb has higher resistance, there would be a greater power value?

kuruman said:
You don't need a teacher to give it to you. Start with P = IV and V = IR and use some simple algebra to eliminate the current I from the power equation.
V^2/R, correct? Then, since I know voltage is the same, I would be able to determine resistance values from there.
However, is there a way to prove this not algebraically and more conceptually?

avaz567 said:
So, if I were to plug numbers in given that the voltage drop was the same, I see that the 100 W has a lower amount of resistance, but why is this so? I thought that if a bulb has higher resistance, there would be a greater power value?
Higher resistance dissipates more power when the two resistors are in series. In that case you can write P = I2R by eliminating V. Since resistors in series share the same current, the one in the series combination with higher resistance will dissipate more power. The opposite is true for resistors in parallel which share the same voltage.

avaz567 said:
V^2/R, correct? Then, since I know voltage is the same, I would be able to determine resistance values from there.
Correct. A rule of thumb is that, to compare powers, when two resistors are in parallel, you use P = V2/R because they share the same voltage; when they are in series you use P = I2R because they share the same current. That keeps you from going around in circles.

kuruman said:
Higher resistance dissipates more power when the two resistors are in series. In that case you can write P = I2R by eliminating V. Since resistors in series share the same current, the one in the series combination with higher resistance will dissipate more power. The opposite is true for resistors in parallel which share the same voltage.
So, would I treat my scenario of the bulbs being plugged in at different outlets like a parallel circuit?

kuruman said:
Correct. A rule of thumb is that, to compare powers, when two resistors are in parallel, you use P = V2/R because they share the same voltage; when they are in series you use P = I2R because they share the same current. That keeps you from going around in circles.
Ok, I think I got a better grasp on this concept now. Thanks for the help!

Great, but please wait until you get an answer and then post.

## 1. What is the difference between the 100 W and 60 W bulb?

The main difference between these two bulbs is the wattage. The 100 W bulb consumes more power and produces more light compared to the 60 W bulb.

## 2. Which bulb has more resistance?

The 100 W bulb will have more resistance compared to the 60 W bulb. This is because the higher wattage bulb will draw more current and generate more heat, leading to an increase in resistance.

## 3. How does resistance affect the brightness of a bulb?

Resistance directly affects the brightness of a bulb. The higher the resistance, the lower the current flow, and the dimmer the bulb will be. This is why the 60 W bulb will be dimmer compared to the 100 W bulb.

## 4. Is there a relationship between wattage and resistance?

Yes, there is a relationship between wattage and resistance. The higher the wattage, the lower the resistance. This is because as the wattage increases, the current flow increases, causing a decrease in resistance.

## 5. Which bulb will have a longer lifespan?

In general, the 60 W bulb will have a longer lifespan compared to the 100 W bulb. This is because the higher wattage bulb will generate more heat, leading to a shorter lifespan. However, the lifespan of a bulb also depends on other factors such as the quality of the bulb and usage.

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