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Confused with the relationship between Volts, Current, and Power

  1. May 31, 2017 #1
    1. The problem statement, all variables and given/known data

    For an engineering assessment (first year with no electrical background), we are using a solar panel to power an element for a distillation system. we have calculated our heat energy output requirement that we need from our heating element as being 800 J/s. For simplicity we are assuming 100% heat transfer into the system.

    My understanding is that if we have a 12V 140W panel, we have 11.67A, similarly 5.83A for a 24V 140W. Is it wrong to combine joule's law and Ohms law to end up with P=I2R, and then apply that to calculate the heat energy released from the element to then suit it to our initial heat energy requirement? P's units being in Joules and the resistance can vary based on the current from the solar panels.

    In looking online for potential elements, many have high Watt ratings, what would happen if you attempted to use a 400W DC element in that 140W solar panel system. Currently i understand it as the voltages need to match, i.e a 12V panel needs to supply a 12V element. Is that train of thought wrong?

    Any help would be appreciated,
    Thanks in Advance!

    2. Relevant equations

    P=IV
    P=I2R
     
    Last edited: May 31, 2017
  2. jcsd
  3. May 31, 2017 #2

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    Yes you sure can use Ohm's Law to substitute a quantity for combinations of the other two. Here is a link to a Formula Wheel, which should help you. This is just one link I found. http://www.sengpielaudio.com/FormulaWheel-ElectricalEngineering.htm
    Print it out and slide it in your binder or put it on a bulletin board, for reference, until you have it down and don't need to refer to it as often.
     
  4. May 31, 2017 #3
    Do you want to connect a heating resistor to a solar panel? The power rated for the panel is the power it can deliver. The power rated for a heating resistor is the power it can disipate with out being damaged. So the 400w resistor will deliver 140w of heat because that is the power from the source of energy. It is ok to use Ohm and Joule's laws you will need the R value. Is important to check current I needed to feed the resistor for example a 0,5ohm conected to 12V source will consume 24Amp that is 12x24 =288 watts more than the source can deliver.
     
  5. May 31, 2017 #4
    Yeah essentially running a heating resistor straight from a panel. OK i think i have a better understanding of it now, just because an element has a rating of 400W, it doesn't mean that a 100W panel can't power it, more that it will only operate at what is available. Would that mean the resistor would only work at 140/400 of it's heat capacity? The R value is yet to be determined as i was thinking i could pick it to suit the energy needs once i understood what i could work with in regards to the potential resistors based on their wattage rating and voltage.

    Was i right in the thought that given the panel is 12V, the resistor/element also needs to operate at 12 volts? like you can't run a 24V resistor/element off of a 12V panel, or vice versa? or can you run a 12V off of a 24V source?

    thanks for that response btw, already helpful!
     
    Last edited: May 31, 2017
  6. Jun 1, 2017 #5
    Yes to the first question. Your resistor will work with less power than it can stand. But the same happens in an electrical oven whe you regulate temperature to less than the maximun. Maybe you will have to check also if with that % of power the temperature will reach the right value.
    To the second. Generally the power source deliver a constant voltage within a range of currents. And a resistor can stand a range of voltage (there is a maximun for ex. 200V) always keeping that VI=I2R<=400W. Maybe some resistor has a specific voltage and current to match in that case you will have to match voltage also.

    You will have to work with both the source and the resistor until you get the right pair.
     
  7. Jun 1, 2017 #6
    perfect, thanks a lot for your help. That cleared it up for me, really appreciated :)
     
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