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Electric Circuit Problem - Help Asap Final In 3 Days

  1. Jun 23, 2005 #1
    Hello! i'm just preparing to write my physics 12 final and can't seem to solve a problem form a past exam that i was doing

    Here is the question:

    a circuit using a new battery which has an emf of 6.00 V and an internal resistnace of 1.00 ohms is shown on the left( In the diagram there also is a current of 0.375 A and a resistor place in series labeled R)
    THe battery is then replaced with a used one that has the same emf of 6.00 V bu ta different internal resistance.
    If resistor R now dissipates 1.75 W, waht is the internal resistance of the used battery??????????

    (diagram the same but with 1.75 W at the R (resistor in series placed in circuit)

    THe answer is 2.57 ohms

    i tryed using Vterminal =6.00- (.375)(1) in regards to the orginal battery
    and v terninal is then equal to 5.625
    You then need to figure out I in the next situiation i'm assuming using Power. P=IV... i dunno i got lost and tryed to work backwards but no dice

    any help woudl be awesome
    email me at liajohnson55@hotmail.com
  2. jcsd
  3. Jun 23, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Use this information to find the value of R.
    Start by finding the current.
  4. Jun 24, 2005 #3
    1)Since the resistor R is in series with internal resistance of the battery , add them up in series , apply Ohm's Law to get the value of internal resistance.

    2)Power generated by a resistor is given by [itex]i^2R[/itex]

    Last edited: Jun 24, 2005
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