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Physics help - electricity

  1. Oct 24, 2005 #1
    OK I have been Taking AS Level Physics and my teacher isnt very good.
    I havent done well in this test and I need to do the corrections over the half term holiday.
    http://img443.imageshack.us/img443/7987/scannedat2410200518561nu.jpg [Broken]
    I'll be frank OMGWTFBBQ!! How on earth do you do this?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 24, 2005 #2


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    Homework Helper

    I will never understand what BBQ has to do with things like that. Beef, chicken, or pork?

    Anyway. It may seem silly, but I would strongly suggest you get into the habit of sketching the circuit in cases like this, even simple ones. Remember, a battery with an internal resistance can be treated as an ideal battery with a resistor in series. That resistance will be added to the total resistance of the actual load circuit. Because of that, you're going to end up in your first problem with two equations in two unknowns - the unknown voltage of the ideal battery and the unknown internal resistance of the battery.

    Set it up using Ohm's Law in both cases. You should be able to determine the effective resistance of the two resistors in parallel - add that to R for the battery to get the total resistance of the circuit and Ohm's Law it. It's even easier to get the effective resistance of the two resistors in series - add it (again) to the internal resistance of the battery, use Ohm's Law, and you'll have your two equations. After that, it's just algebra.

    For the second problem, I presume your teacher's comments (possibly added to mine) can tell you what you're missing? With the ammeter, again, model it as an ideal ammeter (resistance = 0) in series with a resistance corresponding to the internal resistance of the ammeter. In this case, the problem will be easier because you'll only have one equation, but you'll have to remember to take into account the resistors in parallel, the internal resistances of the cels in the battery, and the internal resistance of the ammeter.

    Does that do it?
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