1. Limited time only! Sign up for a free 30min personal 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!

Effeciency and Current Relationship

  1. Aug 21, 2016 #1
    1. The problem statement, all variables and given/known data

    I'm doing an investigation for my physics class, where we were told to investigate the relationship between current flowing through and the efficiency of an immersion heater. My hypothesis was that as current increased, the efficiency would decrease due to the relationship P=I^2R, where as current is doubled, the power loss would quadruple.

    The way I conducted the experiment was to use a power pack connected to an immersion heater inside a calorimeter and measured the rise in temperature of the water over 2 minutes. I then raised the voltage, which increased the current and recorded the new change in temperature.

    I then compared the heat absorbed by the calorimeter and the water inside it using the equation q=mcT versus the actual power emitted from the power pack. What I found was that as the current rose, the efficiency actually increased, seen in the data below.

    Average Current (A)

    0.616

    1.206

    1.834

    2.445

    3.046


    Average Energy Supplied (J)
    (P=W/t)

    55.

    238

    653

    1349

    2356

    Average Energy Required (J)
    (q=mcT)

    140

    564

    1284

    2346

    3655

    Average Efficiency (%)
    (Efficiency = (Energy output/ energy input) x 100)

    39

    42

    50

    57

    64

    Does anybody know why the efficiency would actually increase when the current was raised?
    (sorry for the data being spread out, I couldn't find out how to insert a table)

    2. Relevant equations

    P=VI
    P=I^R
    P=W/t
    Q=mcT
    (Efficiency = (Energy output/ energy input) x 100)

    3. The attempt at a solution

    The only reason I can think of for the efficiency increasing would be having something to do with how we did not maintain the same voltage throughout the experiemnt, otherwise I'm reeally not sure.

    Thanks in advance
     
  2. jcsd
  3. Aug 21, 2016 #2

    Tom.G

    User Avatar
    Science Advisor

    Here are some questions to consider. They may lead you to an answer.

    Was the resistance of the heater constant regardless of temperature and current?
    Was the calorimeter temperature still rising at the end of the 2 minute measurement period?
    If so, was it rising at the same rate regardless of the input power?
    Is there another formula than the one you used for computing the power (energy) delivered to the heater?
     
  4. Aug 21, 2016 #3
    Yes, the resistance was kept the same at 3.28 Ohms.
    We stopped recording data after the 2 minute period, so any heating after that would not affect the results.

    The other formula for power would be P=V^2/R, so does that mean that the efficiency was increased through the increase in voltage of the circuit?
     
  5. Aug 22, 2016 #4

    Tom.G

    User Avatar
    Science Advisor

    I asked this because practically all materials change resistance with temperature, some a little, some a lot. So is this a measurement or an assumption?

    There is one more formula for power that does not require constant resistance. It could be useful here.

    Did the calorimeter temperature stabilize between measurements at different power levels?
     
  6. Aug 22, 2016 #5
    Sorry, yes it was calculated, it did change slightly but across the whole experiment it stayed at between 3.27 and 3.28 Ohms


    I'm not sure if I know that one, are you reffering to P=VI?


    Ahh yes ok, we didn't leave much time between increasing the voltages so the remaining heat in the heater could have influenced the recorded heating of the water.
     
  7. Aug 22, 2016 #6

    Tom.G

    User Avatar
    Science Advisor

    Yes, that's the one.
    Apparently you calculated the resistance from the voltage/current then used the resistance to calculate power. That works, it just wasn't clear to me from the description.

    And this one covers the other likelyhood.
    Good Show!

    I would be curious to know your results after you modify the experimental procedure. If convenient, please post any new results you get/procedures you use.
     
  8. Aug 22, 2016 #7

    billy_joule

    User Avatar
    Science Advisor

    What power loss?
    You are transforming electrical energy into thermal energy - There really is nowhere to lose energy. Even if the element glows and generates some light, that energy will become heat within the system. Googling 'Electrical heater efficiency' will likely be enlightening...

    I assume you didn't measure the resistance of a live circuit (don't, it won't give a correct reading and can destroy your meter), so we still don't know the resistance when the element is hot (and it'll change with varying voltage as the current & temp. varies), which is why you really need to measure the current to get decent results. Doing that, along with waiting for the water temp. to stabilise will likely give much better results.
     
  9. Aug 22, 2016 #8
    I see, so would that mean that increasing the current shouldn't really affect the efficiency of the heater, as pretty much all of the energy should be going into the water anyway? The only problem that may arise from that was that the heater I was using was only a substitute immersion heater made from wrapping 30cm of steel wire around a piece of plastic, which I would assume wouldn't be very efficient.

    Yeah sorry I wasn't very clear, the data for resistance that I got was found through using V=IR, as I knew the voltage and current going through the circuit, so I could find the resistance of the circuit.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted