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    (Original post by megara)
    http://www.bbc.co.uk/scotland/learni...uiz/resistors/

    Questions 3, 4 & 5. I cant do this for some reason? *useless*
    3) How can you 'simplify' two resistors in parallel?
    4) What can you do to the top two resistors, so that you're left with two resistors in parallel, from which you can easily find the overall resistance?
    5) Same parallel stuff applies here
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    (Original post by CallumFR)
    3) How can you 'simplify' two resistors in parallel?
    4) What can you do to the top two resistors, so that you're left with two resistors in parallel, from which you can easily find the overall resistance?
    5) Same parallel stuff applies here
    So I put them over each other as fractions?
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    (Original post by megara)
    So I put them over each other as fractions?
    If you have two resistors in parallel, R1 and R2, their effective resistance is:

    \displaystyle \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}
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    (Original post by CallumFR)
    If you have two resistors in parallel, R1 and R2, they're effective resistance is:

    \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}
    Ja, I get that..but do i do that before i add the one in series?
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    (Original post by megara)
    Ja, I get that..but do i do that before i add the one in series?
    It depends on how the resistors are arranged. In Q3, you need to add the two resistors in parallel BEFORE you add on the series one, whereas in Q4 you need to add the two series resistors, and then carry out the parallel calculation
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    (Original post by megara)
    Thank you so much<3!
    This seems weird, but our teacher hinted there would be a lot of unit comparison, like 1 N = 1 J s^-1. Could you give me all of these things? n__n
    For this all you need to do is write down all of the units of the things that you are using e.g F=MA, if M=Kg and A=ms-2, then the Newton (N) can also be written as Kgms-2.

    Hope that helped a little.
 
 
 

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