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Electronics - Battery voltage & load

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data
    A 12V battery has an internal resistance of 6Ω.

    a) If this battery is connected to a 10Ω load, calculate the power dissipated by the load.

    b) Draw a graph showing how the terminal voltage of the battery and the current it can supply vary with load.

    c) Derive a value for the maximum power this battery can deliver. What load resistance should be applied to the battery for this to occur?

    2. Relevant equations
    V=I(R+r)
    P=IW
    V=IR
    I=v/(R+r)

    3. The attempt at a solution
    I have completed part a) My answer is 14.4W

    I am stuck on part 2, I have made a table in excel which I will use to plot a graph. I am slightly confused. I am probably wrong here but I have established that the equation V=I(R+r) is used for working out the terminal voltage and current supplied across a load.

    If this is the case then only one variable can be worked out at one time (with a varying load), either terminal voltage or the current supplied.

    Therefor do I draw 2 seperate graphs: 1 of a 'load' vs 'terminal voltage' with the current constant. Another of 'load' vs 'current' with the terminal voltage constant?

    Or am I not understading it properly?

    My intention is to derive the maximum power of part C after I have suitable graphs for part B.

    Thanks :smile:
     
  2. jcsd
  3. Apr 11, 2012 #2

    SammyS

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    Hello jannos. Welcome to PF !

    Your answer to part (a) is incorrect. 14.4 Watts is the power dissipated by the whole circuit, including the internal resistance of the battery. The battery's internal resistance is not considered to be part of the load.

    For part (b), the terminal voltage is 12 Volts minus the voltage drop across the internal resistance of 6 Ω .
    VTerminal = 12 - Ir = 12 - 6I.​
     
  4. Apr 11, 2012 #3

    Curious3141

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    Could you show working? I'm getting a very different answer.


    V = I(R+r) relates the EMF of the battery V to the current I, the load resistance R and the internal resistance r.

    The terminal voltage is the actual "useful" voltage that can be drawn from the ends of the battery. It's given by the EMF minus the voltage drop across the battery internal resistance. In this case, the terminal voltage VT is given by:

    VT = IR --- eqn(1)

    Now I varies with the load. But it can be calculated very easily because the EMF remains constant. So,

    I = V/(R+r)

    This is the expression you need to plot the graph of I against R. Just use the given numerical values for V and r and a graph plotting software.

    If you put that back into equation 1, you should be able to get the expression for the terminal voltage and plot the corresponding graph.

    I suggest instead that you get an expression for the power dissipated across the load with the only variable being the load R. Then differentiate the expression with respect to R and set the derivative equal to zero.

    This exercise is presumably to introduce you to the concept of 'impedance matching'.
     
  5. Apr 11, 2012 #4

    nsaspook

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    Because this is a resistive DC circuit I think the concept of "IR loss" is what you mean as there is only real power in this circuit with no (imaginary) power stored in fields to worry about. While impedance matching is possible with resistive networks, it's usually only seen in low-level audio padding due to high losses.
    http://en.wikipedia.org/wiki/L_pad#Impedance_matching
     
  6. Apr 11, 2012 #5
    This exercise should show that there is a definite value of load to get maximum power transfer and that is one feature of 'impedance' matching.
     
  7. Apr 12, 2012 #6
    Thanks for that everyone.

    A 12V battery has an internal resistance of 6Ω.
    a) If this battery is connected to a 10Ω load, calculate the power dissipated by the load.

    b) Draw a graph showing how the terminal voltage of the battery and the current it can supply vary with load.

    c) Derive a value for the maximum power this battery can deliver. What load resistance should be applied to the battery for this to occur?


    I did question A) again and got 5.625W by doing the following:
    I=V/(R+r) = 0.75A
    P=I2R, 0.752x10 = 5.625W

    For part B I plotted a graph by using calculated values from a table within excel. Initially I didn't expect the graph to be exponential as shown.

    As for part C from my table I get 6W at a load of 6Ω for the maximum power output of the battery.

    Does this seem right?
    Thanks
     

    Attached Files:

  8. Apr 12, 2012 #7

    Curious3141

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    Yes, looks right. :smile:

    You've also got the right answer for part A this time.

    The graph is not exponential, BTW, it's a hyperbola.

    So what can you say about the load where maximum power delivery occurs?
     
  9. Apr 12, 2012 #8
    That is reassuring!

    Looking at the graph without sensibile consideration the obvious answer is that the load must equal the terminal voltage to maximize power transmission. (Which I know is wrong)

    For Pmax: RLoad = VT

    The other obvious answer is that the Internal resistance of the battery must equal the Load resistance which can be seen in this case:

    Rinternal = RLoad (which is the correct answer)

    Which follows the Maximum Power Transfer Theorem. Therefor maximum power transmission will occur at this point but not maximum efficiency.

    That makes sense now right?
     
  10. Apr 12, 2012 #9

    Curious3141

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    Yes, the second part (equal load and source resistance for max power transfer) is right.
     
  11. Apr 12, 2012 #10

    nsaspook

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    I guess I'm old school where 'impedance' means AC and DC means 'resistance'.
    http://technologyinterface.nmsu.edu/Spring08/30_Cartwright/pdf.cgi [Broken]
     
    Last edited by a moderator: May 5, 2017
  12. Apr 12, 2012 #11
    Thanks for your help everyone!

    I thought 'Resistance' was the resistance to flow of electrons through a conductor whether it is AC or DC.

    Impedance being the sum of (the circuits 'Resistance' through the circuit conducting materials & components) and (the resistance of electron flow caused by the collapsing sinusoidal magnetic field being out of phase?) In the case of DC the impedance is equal to the resistance because there is no oscillating magnetic field?

    I am probably out of my depth here to be honest!
     
    Last edited by a moderator: May 5, 2017
  13. Apr 13, 2012 #12

    nsaspook

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    My quibble with "Impedance matching" and impedance being used where resistance is the correct term comes from the origin of the words and common usage in electronics.

    Resistance: opposition. (to stop or reduce)
    Impedance: to hinder or delay. (to slow down or store)

    "Impedance matching" normally involves the movement of real power from a source via a transmission medium (not only cables but transformers, a PCB trace, etc ...) to a load where the medium uses as little real power as possible. Power is stored in fields in the transmission medium but not reduced or stopped by pure reactance. At RF the load might be "free space" and the transmission medium coax and an antenna. "Impedance matching" the transmission medium in a DC circuit is simple, make it as close to zero resistance from load to source as possible.
     
    Last edited: Apr 13, 2012
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