Brightness of light bulbs in a circuit

[Tokspor]

Two light bulbs, one 60 W bulb (higher resistance) and one 100 W bulb (lower resistance) are placed in series. A current is run through them. Which bulb glows brighter?

A. 60 W bulb
B. 100 W bulb
C. Same
D. One bulb would not light

The answer is A. I'm not sure how to arrive at the answer. I thought the answer was B because I thought that power would be a measure of the brightness of a bulb. Apparently not. How would you reason that a 60 W bulb would be brighter in series with a 100 W bulb?

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[ese]

A series circuit => current is the same for both bulbs. I'm guessing "glowing brighter" means which has a higher Power output in which case P=I^2 R since 60W has higher resistance it has a higher power output.

 

[stockraider]

ok, this is testing one critical concept, remember in series the current is the same, but the voltage drops through EACH resistor are not (this is vice versa for parallel circuits). so now using this critical concept, lets say we only had 5 amps running through this series circuit. since the resistance is higher in the 60 watt bulb, it will have a higher voltage drop through it, then the 100 watt bulb (which the question states has a lower resistance). now P=IV, and since current is the same and the 60 watt bulb has the higher voltage drop, the 60 watt bulb has more power running through it, meaning there are more joules per second going through that thing than the 100 watt bulb, hence the 60 watt is brighter. DO NOT get confused here with the wattage, i know from working for years as a laborer in elecrticity that the wattage on the bulb is only there to tell you what is the maximum watts that the bulb can handle, anything over that will blow out the filament....in this system, as long as P=IV is less than 60 watts neither bulb will blow out. the watts are there to simply confuse you, the thing you should pay attention to is the resistance.

 

[stockraider]

or yes, a faster way would just use p=I^R, the higher resistance gives the higher power. remember the watts of a bulb says nothing about how many watts it actually puts out when plugged into a circuit.

 

[ese]

To further clarify and check against the opposite case: if the situation was a parallel circuit then the 100W would indeed be brighter. Normal houses are wired in parallel thats why a 100W is brighter than a 60W. Since In parallel voltage is the same, but current is not (and it favors the path of least resistance i.e. the 100W bulb). Once again using I^2 R and knowing Ohmic materials (as usually assumed on the MCAT V=IR) so a linear relationship between resistance can be assumed. Since I is squared the difference is much greater than the differences in resistance resulting in the 100W having more power output.

 

[Tokspor]

To further clarify and check against the opposite case: if the situation was a parallel circuit then the 100W would indeed be brighter. Normal houses are wired in parallel thats why a 100W is brighter than a 60W. Since In parallel voltage is the same, but current is not (and it favors the path of least resistance i.e. the 100W bulb). Once again using I^2 R and knowing Ohmic materials (as usually assumed on the MCAT V=IR) so a linear relationship between resistance can be assumed. Since I is squared the difference is much greater than the differences in resistance resulting in the 100W having more power output.

Thanks for the responses. From the way you reasoned it here, was it vital that the question stem mentioned that the 100 W bulb has less resistance? If it did not mention this, would we not be able to determine which path the current is more likely to take now?

 

[ese]

Thanks for the responses. From the way you reasoned it here, was it vital that the question stem mentioned that the 100 W bulb has less resistance? If it did not mention this, would we not be able to determine which path the current is more likely to take now?

I'm answering this based on the original question (circuit in series):
I feel I would hopefully realize that from taking physics classes, but for the MCAT my gut feeling is that they would give you that information. If that relationship is not realized or given then you probably would not be able to get the right answer.

 

[iPodtosis]

I have encountered a question before that's related to this. If you have 3 identical light bulbs connected in a parallel circuit, now one of the bulb burned out, what will be the brightness of the 2 remained, increase, decrease, or remain the same?

Will reveal the answer tomo

 

[stockraider]

I have encountered a question before that's related to this. If you have 3 identical light bulbs connected in a parallel circuit, now one of the bulb burned out, what will be the brightness of the 2 remained, increase, decrease, or remain the same?

Will reveal the answer tomo

will remain the same

 

[Tokspor]

I have encountered a question before that's related to this. If you have 3 identical light bulbs connected in a parallel circuit, now one of the bulb burned out, what will be the brightness of the 2 remained, increase, decrease, or remain the same?

Will reveal the answer tomo

The voltage will remain the same in parallel resistors (bulbs) as stockraider pointed out in a previous reply. The two remaining resistors will have the same voltages and resistances as before, so they should have the same current. So they should give the power output as before, right?