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

[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.

 

[kuruman]

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.

 

[OmCheeto]

...
Someone had told me that the outlets would have the same voltage drop,
...

This is correct.

 

[avaz567]

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.)

 

[kuruman]

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.

 

[avaz567]

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?

 

[avaz567]

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?

 

[kuruman]

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.

 

[kuruman]

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.

 

[avaz567]

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?

 

[avaz567]

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!

 

[kuruman]

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