calculating a resistance of a bulb

Discussion in 'Homework Help' started by electronicsnewb, Nov 8, 2013.

  1. electronicsnewb

    Thread Starter New Member

    Dec 8, 2012
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    Hi,

    I'm stuck on a problem where its asking to calculate the resistance of a 300 Watt bulb connected across a 120 V power line, using two different methods to arrive at the same answer.

    I understand that P=V^2/R and solving for R will give me R=V^2/P so the resistance here is (120V)^2/300W=48 ohms

    But it is asking to arrive at this answer using two different methods, and with the given info, I'm not sure how else I can calculate the resistance? I appreciate any help in advance. Thanks!
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    What is the current in the bulb?

    Does that give you any ideas?

    Note that it is debatable whether these two methods are truly different.
     
  3. wayneh

    Expert

    Sep 9, 2010
    12,101
    3,033
    You could make it a lot more complicated by recognizing the effect of temperature. I mean, the question could ask what the room temperature resistance is. But that would require a lot more information than what is given.

    I think identifying all the usual parameters, as wbahn has hinted, will likely satisfy the teacher.
     
  4. electronicsnewb

    Thread Starter New Member

    Dec 8, 2012
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    Yes. After solving for R=V^2/P, the resistance here is (120V)^2/300W=48 ohms.

    I can then find the current using I=V/R=120V/48=2.5 A.

    With current known, I can solve for R using V/I=120V/2.5A=48Ω.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    You've just made things circular. You solved for R and then used that value of R to solve for R.

    The idea is to find another way to find the UNKNOWN value of R.

    You know that IF you have the voltage and the current, then you can find R.

    The question is, given just the power and the voltage, can you find the current?
     
  6. electronicstech07

    Member

    Nov 16, 2007
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    I see. So solving for R in the equation P=V^2/R will give me R=V^2/P.

    So R=(120)^2/300=48Ω
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    But using P=V^2/R is the FIRST method. It is asking you to use a SECOND method.

    And don't drop your units. You did it just right the first time

    R=(120V)^2/300W=48Ω

    But now see if you can get the same answer WITHOUT using that equation.

    So let me ask again. Given just the power and the voltage, can you find the CURRENT?
     
  8. bwilliams60

    Active Member

    Nov 18, 2012
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    Can you not just use I = P/E ? What am I missing?
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    That's what I've been trying to get the OP to come up with. Thanks for doing his thinking for him so that he doesn't have to.
     
  10. electronicsnewb

    Thread Starter New Member

    Dec 8, 2012
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    Sorry, I guess I wasn't reading and thinking straight when I replied. Since it involves the relationship of power voltage, current, and resistance, using the formula I=P/V with P and V known to find the current, and then using the formula R=P/I^2 to find the resistance will be the right steps to take.

    I=300W/120V=2.5A

    R=300W/2.5^2A=48Ω
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    Or yet a "third" way.

    I = 300W/120V = 2.5A

    Now use Ohm's Law

    R = V/I = 120V/2.5A = 48Ω

    But, at the end of the day, these are ALL the same approach.

    What have two relationships:

    P = V*I
    V = I*R

    All each of these three methods are doing is applying these two relationships in slightly different order. The first method (the most direct), uses the second relation to replace I in the first in order to use it with only V. The second method uses the second relation to find I and then uses the second relation to replace V in the first in order to use it with only I. The third (the most fundamental), use the first to find I and the second to find R.
     
  12. bwilliams60

    Active Member

    Nov 18, 2012
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    WBahn, I admire the fact that you are trying to help him but sometimes all people are not geniuses and need a little help. It is nice that you are very knowledgeable and willing to help, but the rest of us are not at the genius level yet.
     
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