1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Resistors (lightbulbs) Power & Brightness

  1. Mar 30, 2016 #1
    1. The problem statement, all variables and given/known data
    A 60 W lightbulb and a 100 W lightbulb are placed one after another ( in series, the 60 W battery closer to the positive terminal of the battery) in a circuit. The battery's emf is large enough that both bulbs are glowing. Which is the true statement?
    A. The 60 W bulb is brighter
    B. Both bulbs are equally bright
    C. The 100 W bulb is brighter.
    D. There's not enough information to tell which is brighter.

    2. Relevant equations
    Pr = Pbat = I*Epsilon
    I = Pr/ delta Vr
    Pr= I * delta Vr = (I^2)*R = ( ( delta Vr)^2)/R


    3. The attempt at a solution

    I know the answer is A because the book says so, I just don't understand why. I would have gone with D.
    The current is the same in both bulbs ( conservation of current). And R = Pr/(I^2). So the 60 W bulb should have a smaller resistance than the 100 W ( the internet says otherwise ). Stuck. Please help.
     
  2. jcsd
  3. Mar 30, 2016 #2
    Sorry - I misread that. The 60W bulb has a higher resistance that the 100W - so it draws less current.

    BTW: I think it's worth mentioning that incandescent lamps present a "lamp load". But for the purpose of this exercise, they can be considered resistors.
     
    Last edited: Mar 30, 2016
  4. Mar 30, 2016 #3

    cnh1995

    User Avatar
    Homework Helper

    The bulbs are rated for the same voltage.
    Since R=V2/P, resistance of bulb is inversely proportional to its power rating. Hence, 60W bulb has higher resistance than 100W bulb.
     
  5. Mar 30, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    Start by assuming that the lightbulbs are simple resistors (not really true as there are certain heat-related effects that change their resistance values as they warm up to their proper operating temperature, but we can ignore this for now). Define a fixed battery voltage that is their normal operating source, say 12 V. Use one or more of your relevant formulas to determine a resistance for each bulb when it's operating at its designed power.

    Then connect the bulbs in series with the same battery and determine power consumed by each bulb. What do you find?
     
  6. Mar 30, 2016 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    @.Scott
    If you are saying that the 60 W bulb has smaller resistance, then you are incorrect .
     
  7. Mar 30, 2016 #6
    Your right. It's a higher resistance - and draws less current.
     
  8. Mar 31, 2016 #7

    I don't know of I understood you well, but here's what I tried:

    - The resistors are in series, so the same current passes through them
    - Let the voltage of the battery be 12 V
    - Let the 60 W bulb be bulb 1 & the 100 W one be bulb 2
    --> The voltage of the battery is the same as that of bulb 1; hence Pr_1=I*12V --> I = 60W / 12V = 5A.
    --> Voltage of bulb 2 = Pr_2/I = 100W/5A = 25V

    Then, to find the resistance of each bulb, Ohm's law:
    -> (delta Vr_1 )/ I = R_1 =12V/5A=2.4 ohms
    -> (delta Vr_2)/ I = 25V/5A = 5 onhd....I know this is wrong, but I can't think of the system another way...
     
  9. Mar 31, 2016 #8

    cnh1995

    User Avatar
    Homework Helper

    Resistances of the bulbs are determined by the power rating(P) and voltage rating(V) as R=V2/P. Since they are rated for the same voltage V whose value is not mentioned, you can only conclude that R60>R100. Since they are in series, same current will flow through both of them. Power consumed in each will be I2R, I being same for both. Which bulb will glow brighter?
     
  10. Mar 31, 2016 #9

    cnh1995

    User Avatar
    Homework Helper

    How so? The bulbs are in series, so there must be voltage division.
     
  11. Mar 31, 2016 #10

    gneill

    User Avatar

    Staff: Mentor

    To find the resistance of the individual bulbs, do not place them in series yet. Consider them individually, attached to the battery alone. (Clearly the voltage you found for the 100W bulb above is incorrect, as it exceeds the battery voltage!).
    Your resistance value for the 60W bulb is good because you assumed 12 V across it. Do the same for the 100W bulb.

    Next place the two resistances in series with the battery. Find the power dissipated by each resistor.
     
  12. Mar 31, 2016 #11
    Ok, I'll try again:

    1. Define a fixed voltage and determine the resistance for each bulb.
    - Delta Vr = 12 V for each bulb
    - R = ((delta Vr)^2)/ Pr
    --> R_1= ((12V )^2)/60W = 2.4 ohms
    --> R_2= ((12V)^2)/100W = 1.44 ohms

    Now I see that bulb 1 has a higher resistance. However, I don't understand that the two bulbs operate at the same voltage. Shouldn't the battery's voltage (12V) be divided between the two bulbs?

    2. Connect bulbs in series
    --> Req = 2.4 ohms + 1.44 ohms = 3.84 ohms
    3. Find the current through each bulb ( resistors in series have the same current )
    --> I = Epsilon/Req = 12V / 3.84 ohms = 3.125 A.

    4. Power consumed in each bulb is (I^2)*R
    --> Pr_1 = ((3.125A)^2)*2.4 ohms = 23.44 j/s
    --> Pr_2= ((3.125A)^2)*1.44 ohms =14.06 j/s

    I see that more power is consumed by bulb 1. ....so it glows brighter....however, I don't understand the difference in meaning between the power I was given ( 60W and 100W ) and the power I found. Can someone please explain?
     
  13. Mar 31, 2016 #12

    cnh1995

    User Avatar
    Homework Helper

    The power you were given is for when the bulbs operate at the same voltage i.e. when they are conncted in parallel(in your house, all the bulbs operate in parallel). The power you calculated is for when the bulbs are in series. Since same current flows through both the bulbs,the bulb having higher resistance glows brighter.
     
  14. Mar 31, 2016 #13

    gneill

    User Avatar

    Staff: Mentor

    Remember that I said:
    This is to establish the individual resistance values for the bulbs. Then after doing this you can connect them in series, knowing what those resistor values are.
    The first setup was to obtain the individual bulb resistance values. The second (series) setup corresponds to the problem statement's arrangement.
     
  15. Mar 31, 2016 #14
    Ok, I think I get it now! Thank you all very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Resistors (lightbulbs) Power & Brightness
  1. Lightbulb Brightness (Replies: 3)

Loading...