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Battery and current

  1. Mar 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Please help!!!:confused:

    My problem is: (with figure attached, the figure explains the majority of the ptorblem)
    (a) What is the size and direction of current i1 in Fig. 27-40, where each resistance is 2.2 and V1 = 18 V?

    (b) What is the power of the 18 V battery, and is energy being supplied or absorbed by the battery?

    (c) What is the power of the 10 V battery, and is energy being supplied or absorbed by the battery?

    (d) What is the power of the 5.0 V battery, and is energy being supplied or absorbed by the battery?

    2. Relevant equations
    V=iR

    3. The attempt at a solution
    For part (a) I'm having trouble firguring out which resistances to use. Do I use all of the resistances and add them up according to whether they are parallel or perpendicular? Then, when I get the resistance that I need, do I just use the 10 V battery as my voltage to calculate the current? I tried adding all of the resistances and I got 8.64 ohms. Is this correct?

    For part (b) I need to know again what resistance to use to calculate the current, which I would need to calculate power. If I can get some help on (b), I can probably do (c) and (d) by myself.
     

    Attached Files:

  2. jcsd
  3. Mar 25, 2007 #2
    In general, the best place to start such problems is to compute your lump resistances together whenever possible using the rules for addition in series and parallel. Then redraw the diagram with these. In this case you should end up with three R's and three voltages. Then one can start to develop some meaningful equations which will answer all of the above questions. Out of curiosity, what course is this for?
     
  4. Mar 25, 2007 #3
    This is for Physics II. I understand how to lump the resistances together, i.e. parrallel and perpendicular. However, I am not sure how to calculate the amps. For example, for the left side, do I take the 5 volts and divide it by 0.88 ohms and then add that to 10 volts divided by 3.3 ohms (the resistance at the bottom left and bottom of the picture?
     
  5. Mar 27, 2007 #4
    Nevermind. I figured out the problem.
     
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