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Homework Help: Looking for some help

  1. Mar 6, 2005 #1
    Ok this question is kinda weird b/c I don't think there's enough info, even though there probably is:

    One heater uses 340W of power when connected by itself to a battery. Another heater uses 240 W of power when connected by itself to the same battery. How much total power do the heaters use when they are both connected in series across the battery.

    I figure it's just P1+P2=Ptotal. So 340+240=580W? :uhh:
  2. jcsd
  3. Mar 6, 2005 #2
    Personally I would agree that you add them together.

    The Bob (2004 ©)
  4. Mar 6, 2005 #3
    Yeah, but apparently I got that wrong...
  5. Mar 6, 2005 #4
    In that case there might need to be more information.

    The Bob (2004 ©)
  6. Mar 6, 2005 #5
    This was one of our practice problems from the text, exactly, and the answer was something like 140W, but I don't know how they possibly got that.
  7. Mar 6, 2005 #6
    Sorry, I don't know where to take this. I can't seem to think of anything that would help. No equation or anything. I thought of [tex]P = \frac{V}{I}[/tex] but I can't see how that will help.

    The Bob (2004 ©)
  8. Mar 6, 2005 #7
    Anyone else want to help? Please. o:)
  9. Mar 6, 2005 #8


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    Well for a start, P=IV, not V/I....
  10. Mar 6, 2005 #9
    Do I even need that forumla, I mean I have the power already for both.
  11. Mar 6, 2005 #10
  12. Mar 7, 2005 #11
    There's got to be a way to solve this, without saying too little info :mad:
  13. Mar 7, 2005 #12
    Connected in series, so it's the same current I. Use formula to get total V, i.e.

    (1/V) = (1/V1) + (1/V2).

    Then use P=IV.

    I think that's the way to do it. Not sure though......
  14. Mar 7, 2005 #13
    Wouldn't really work because there's no way I can get V, I only have power. I can go:



    Ptotal =580W...Still not getting anywhere, feels like I'm going in circles here :(
  15. Mar 7, 2005 #14


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    Is the answer 140.7 W ?

    Treat the heaters as simple resistances [tex]R_1[/tex] and [tex]R_2[/tex]. Using [tex]P_i = \frac{V^2}{R_i}[/tex] find an expression for each of the resistances in terms of the power dissipated by each and the voltage.

    When the circuit has the heaters in series, the resistances are additive. The voltage remains the same. You should be able to derive an expression for the new power [tex]P' = \frac{V^2}{R_1 + R_2}[/tex] purely in terms of the [tex]P_1[/tex] and [tex]P_2[/tex].
  16. Mar 8, 2005 #15
    Wait wait... I still don't get where that equation is supposed to go. For example,
    P=V^2/R1+R2 is just saying P=V^2/R1 + V^2/R2, and Since P=V^2/R, we already have the power for both so it just goes back to Ptotal=P1+P2, I'm going in circles here. ahhhh!
  17. Mar 8, 2005 #16


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    Curious has the correct approach. Treat each heater like a simple resistor. Then for battery Voltage "V" and Heater Resistance "R", we have:
    {Power Dissipated by Resistance R with Voltage V} = E2/R :::: Eq #1

    Applying Eq #1 to each heater separately:
    Heater #1 ---> E2/R1 = 340 ⇒ R1 = E2/340
    Heater #2 ---> E2/R2 = 240 ⇒ R2 = E2/240

    When the heaters are connected in series, the total (equivalent) circuit resistance equals the sum of each individual resistance (that is, R = R1 + R2), so that Eq #1 applied to the new 2-heater circuit becomes:
    {Power Dissipated by R1 + R2 In Series with Voltage V} = E2/{R1 + R2} =
    = E2/{E2/340 + E2/240} =
    = 1/{1/340 + 1/240} =
    = (140.7 watts)

    Last edited: Mar 8, 2005
  18. Mar 8, 2005 #17
    Huck damn, an equation within an equation...sigh.

    Thanks, I knew my one liner for work wasn't correct lol.
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