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Terminal voltage

  1. Jun 1, 2006 #1
    "A battery labeled as 12.0-V is measured to supply 1.90 A to a 6.00-W resistor (see figure). (a) What is the terminal voltage of the battery? (b) What is its internal resistance?"
    I'm not sure how to solve this problem...

    I originally did R = V/I = 6.32 Ohms, and E = 12V, so V = E - IR = 0... but that is obviously wrong
    Last edited: Jun 1, 2006
  2. jcsd
  3. Jun 1, 2006 #2


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    Note that the resistor value is given in watts, which is a measure of power.

  4. Jun 1, 2006 #3
    Your method would work had the cell had negligible internal resistance .
    Note that the battery is only marked 12V. This is not the actual voltage drop across the load resistance .
    You can start off by finding the value of the resistor . It is given that current through the resistor is 1.90 A .
    What is the relation connecting power and current ?

    Then you must find what the value of a resistance connected in series should be , so that from the 12 V source, a current of 1.90 A is drawn. This is the internal resistance .
    Now find the voltage drop across load resistor, this will be terminal voltage.
    Can you follow ?

    Edit : 6W may be the power consumed by the resistor when 1.9 A flows theough it .
  5. Jun 1, 2006 #4
    P = I2R
    R = 6/1.92 = 1.66 ohms
    now what?
    not really...:confused:
  6. Jun 1, 2006 #5
    [tex]I = 1.90A, P = 6.00W, V = 12V[/tex]

    [tex]R_{load} = \frac{P}{I^2} = 1.66\Omega[/tex]

    [tex]V_{emf} = 12V = I \cdot R_{total} = I \cdot R_{load} + I \times R_{internal}[/tex]

    [tex]R_{internal} = 4.65\Omega[/tex]

    but can't I just skip all that and do

    [tex]V_{terminal} = I \cdot R_{load} = 3.16 V[/tex]

    The answer is supposed to be 11.4 V...what am I doing wrong?:frown:
    Last edited: Jun 2, 2006
  7. Jun 2, 2006 #6
    What you have done is completely correct . In your original question, it was asked to calculate the internal resistance as well as the terminal potential difference .And of course you should skip and use
    [tex]V_{terminal} = I \cdot R_{load}[/tex].
    However, from the answer that you have provided , it seems that in the question, it is given that the value of the resistance is indeed 6 ohms .
    So you need not consider power at all and I am sure you can do the rest.
    It is much simpler as compared to the question that we have been discussing .

  8. Jun 2, 2006 #7
    so i guess it was a typo...
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