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Internal voltage and resistance of a power supply

  1. Jan 3, 2007 #1
    1. The problem statement, all variables and given/known data
    Ok, you'll have to bare with me on this problem, this topic is not covered as part of my second level education. I think that ye do this in the US though.

    You are given a circuit containing a non-ideal power supply, an NTC resistor, an ammeter, a voltmeter, and a switch. The circuit is at room temp. The ammeter and voltmeter are both ideal.

    After closing the switch, the temp of the NTC changes, and the voltage across and the current through the NTC are measured at regular intervals until the temp of the NTC remains constant.

    You are given a graph of V (in Volts) versus I (in Amps). -> (V, A)

    Data points: (0.16, 6.9) (0.25, 6.84) (0.33, 6.9)

    Q. Determine the internal voltage and the internal resistance of the power supply.


    2. Relevant equations
    Well, since we don't cover this topic, I haven't got the equations straight from a textbook. But, I've come up with the following ideas:

    EMF = V + v
    and EMF = I(R + r)
    and EMF = V + v = I(R + r)
    and v = Ir


    3. The attempt at a solution
    Well, I'm not completely sure where to go from here. Ah, something just came into my head, using one set of those values, does the internal voltage = internal resistance?

    To be honest, I'm stumped. Could someone give me an idea about what to do, the concepts involved, etc.?

    Cheers!
     
  2. jcsd
  3. Jan 3, 2007 #2

    Hootenanny

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    Welcome to the Forums,
    Can you determine the load resistance using those points...?
     
    Last edited: Jan 3, 2007
  4. Jan 3, 2007 #3
    Yeah, you can using V = IR. That's part of what I did.
     
  5. Jan 3, 2007 #4

    berkeman

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    The NTC is just giving you several sample points of what the voltage and current are with different load resistors. Draw the circuit showing the internal voltage source and the Rs source resistance, and the resistor outside. Vs and Rs are the unknowns. Now just write the equations for Vs and Rs, and use the fact that you have multiple datapoints to solve for the unknowns.
     
  6. Jan 3, 2007 #5
    Ok, I probably should have posted my workings:

    E = V + v = I(R + r)

    6.79 + v = 0.33(20.576 + r)
    6.79 + v = 6.79 + 0.33r
    A: v - 0.33r = 0

    6.9 + v = 0.16(43.125 + r)
    6.9 + v = 6.9 + 0.16r
    B: v - 0.16r = 0

    Or those equations can be achieved from v = Ir

    Therefore, v = 0 and r = 0.................?

    That was one of the solutions I came up with. I doubt it's right.
     
  7. Jan 3, 2007 #6

    berkeman

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    In your original post, you said that the numbers were (V,I), but in your latest post, you are using them as (I,V) (which makes more sense, given the numbers). Can you please clarify the data?
     
  8. Jan 3, 2007 #7
    It was meant to be (I, V). Just made a mistake when typing it out at first.
     
  9. Jan 3, 2007 #8

    berkeman

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    So,

    "Data points: (0.16, 6.9) (0.25, 6.84) (0.33, 6.9)"

    160mA @ 6.9V => Rntc = 43.125 Ohms
    250mA @ 6.84V => Rntc = 27.36 Ohms
    330mA @ 6.9V => Rntc = 20.91 Ohms

    So at least the NTC's resistance is dropping now as it heats up. That's a good thing :)

    Now you have the currents and voltages for three different load resistances. How should you set up some simultaneous equations to solve for Vs and Rs?
     
  10. Jan 3, 2007 #9
    I had the resistances worked out. (The figures I had supported that the NTC's resistance dropped as it heated up. I know that's the basis on which thermistors work.) That's how I set up the simultaneous equations in my post above!

    As above, I put the figures into the equation
    V + v = I(R + r)

    I only set up two of the equations because with my calculations v = 0 and r = 0 (using lower case letter for internal v and internal r)

    Is this right or wrong?
     
  11. Jan 3, 2007 #10

    Hootenanny

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    Set up the following system of equations, putting in each of the appropriate figures for V, I & R;

    [tex]\left\{\begin{array}{cccc}
    6.9+v & = & 160\times10^{-3}(43.125+r) & (1)\\
    6.84+v & = & 250\times10^{-3}(27.36+r) & (2)\\
    6.9+v & = & 330\times10^{-3}(43.125+r) & (3)
    \end{array}\right\}[/tex]

    Do you follow? I know the numbers look a little messy, but I'm sure you can solve the above system for v and r.
     
  12. Jan 3, 2007 #11
    I noticed one other slip up when I posted the question, very sorry.

    Data points: (0.16, 6.9) (0.25, 6.84) (0.33, 6.79)

    Therefore my equations above should be right. I was thrown because v = 0 = r because it is described as a non-ideal power supply in the question.
     
  13. Jan 3, 2007 #12
    The change in temp per second of an NTC will be constant when the power generated in the NTC (temp) is greater than the rate at which the NTC loses heat to its environment and when the NTC is at approx. room temp?

    Edit: If you are given a graph of Power (j-axis) versus Temp (i-axis), and it takes 1.4 J of energy to raise NTC temp by 1 degree celsius, how do you work out the change in temp per second at a certain temp from the graph?
     
    Last edited: Jan 3, 2007
  14. Jan 3, 2007 #13

    berkeman

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    Sorry, I'm not tracking at all what you are saying now. Is this some other part of the problem? Have you done what Hootenanny suggested, solved for Vs and Rs, and are now moving on to other parts of the same homework problem?
     
  15. Jan 4, 2007 #14
    I solved for v and r back here. I still don't know if that's right because I didn't expect v = 0 = r because it is described as a non-ideal power supply in the question. Is it right or wrong? Edit: I just worked it out using Thevenin's Theorem and v = 0 and r = 0.


    "The change in temp per second of an NTC will be constant when the power generated in the NTC (temp) is greater than the rate at which the NTC loses heat to its environment and when the NTC is at approx. room temp?"
    That is just an observation I made, and I was wondering if that was right. It's not part of the question.


    "Edit: If you are given a graph of Power (j-axis) versus Temp (i-axis), and it takes 1.4 J of energy to raise NTC temp by 1 degree celsius, how do you work out the change in temp per second at a certain temp from the graph?"
    I wasn't sure which method I should use for this.
     
    Last edited: Jan 4, 2007
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