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Basic Elecricity

  1. Oct 28, 2006 #1
    Can someone please check my answers here, there are loads of formulaes been give that I haven't used so I'm not sure if they are right or not, thanks.
     

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  3. Oct 29, 2006 #2

    Doc Al

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    The first problem (resistors in series) is done correctly, but the second (resistors in parallel) is not. Recalculate the total resistance in that second circuit. Hint: When resistors are in parallel, the total resistance will be less than that of any single resistor.
     
  4. Oct 29, 2006 #3

    andrevdh

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    The total resistance of resistors in parallel can be calculated with

    [tex]\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} [/tex]

    this can be changed to

    [tex]R_p=\frac{R_1R_2}{R_1 + R_2}[/tex]
     
  5. Oct 29, 2006 #4
    When I do that I get 6x2 = 12ohm 12ohm/8 ohms = 1.5 ohms, is that right?
     
  6. Oct 29, 2006 #5

    andrevdh

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    Yes. That seems to be it.
     
  7. Oct 29, 2006 #6
    Am I right in thinking that if you have a 12v battery and short it the current would be very high?
     
    Last edited: Oct 29, 2006
  8. Oct 29, 2006 #7

    andrevdh

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    Not neccessarily. The amp-hours rating is a complicated affair. It is the amount of current it can supply (chosen for "normal" [designed] operating conditions) for a certain set period (I think it is normally for 20 hours, but I might be wrong) before the battery is completely drained and dies out. So you might get more out of it for a shorter period but only to a certain limit (everyone has his limitations!).
     
  9. Oct 29, 2006 #8
    Ok I think I'm not sure If I've got this right. I worked out I1 and I2 using fractions.
     

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  10. Oct 30, 2006 #9
    Anyone?.........
     
  11. Oct 30, 2006 #10

    Doc Al

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    Rethink your answer for C (the total current).
     
  12. Oct 30, 2006 #11

    andrevdh

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    The current through the 2 ohm resistor is the current through all of the circuit since

    [tex]I = I_1 + I_2[/tex]

    So [tex]I[/tex] will be be determined by the total resistance of the circuit and the voltage of the power supply.
     
  13. Oct 31, 2006 #12
    Ok we went through that question today and I got A-F right but G was 0.9A.
     
  14. Oct 31, 2006 #13
    12/5 does not equal 1.2, maybe you sent us the wrong version of the problem, or the teacher did it wrong
     
  15. Oct 31, 2006 #14
    Ah sorry I put 12v instead of 6, doh.
     
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