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    7) Does resistance DECREASE as voltage increases??????????
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    6 marker - what did everyone put
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    (Original post by micycle)
    This is how I answered the heater question:

    R = 12

    V = 230

    Resistor:

    R = 0.015 (x2)

    New current = 230/(12+2*0.015) = 19.11

    V lost over cable (2 wires) = 19.11*0.015*2 = 0.574V

    New V = 230 - 0.574 = 229.4V
    It's 6.3 x resistance because they were 3.15m each.
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    (Original post by Yavish13)
    7) Does resistance DECREASE as voltage increases??????????
    I put decrease
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    (Original post by micycle)
    This is how I answered the heater question:

    R = 12

    V = 230

    Resistor:

    R = 0.015 (x2)

    New current = 230/(12+2*0.015) = 19.11

    V lost over cable (2 wires) = 19.11*0.015*2 = 0.574V

    New V = 230 - 0.574 = 229.4V
    Seems right. I STUPIDLY took away the resistance of the wires from 12 as a pose to adding them thinking 12 was the resistance of the whole circuit...And the length was 3.15m. You had to take that into consideration.
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    (Original post by Charlie543)
    Yeah but 229.5v to the same sig figs as question is 230, so mustve been series to get an acceptable different answer of 228v
    I used a parallel arrangement and still got 229 volts? Found the lost volts in the wire and subtracted it from 230 - same method as someone above who used a series arrangement
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    Okay i'll try and explain that heater question.

    Firstly you consider the wires to have NO resistance therefore the power of the heater is simply V^2/R

    this is 230^2/12 = 4408 = 4400

    After this it tells you that in reality the cables will have an internal resistance

    You can consider the two wires as resistors (each with 3.15 * 0.0150 ohms of resistance = 0.04725) and place these in series with the 12 ohm resistor (heating element)

    the terminal voltage was 230 volts meaning that the voltage would be shared (as a potential divider) between the three different resistors.

    this means the voltage across the heating element would be (12/12+3.15*2*0.0150)*230. In other words, the resistance of the heating element over the total resistance of the circuit multiplied by the voltage. This gave 228.2029021 volts.

    The energy wasted in the two wires would be as follows: 230-228.2029021 = 1.797...= voltage shared between two cables. One cables voltage =1.797/2 = 0.8985...V^2/R again = 0.8985^2/0.04725 * 2 (there are two cables) = 34.1 W
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    (Original post by Einstein1997)
    It's 6.3 x resistance because they were 3.15m each.
    what did you put for the 6 marker
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    (Original post by Rid123)
    For 5.
    People whacking in a variable resistor in there to alter current?
    Also to get readings before the spike in graph would you turn diode to reverse bias
    And would resistance increase for increasing pd for last in 5?
    Yes, variable resistor or source was needed. And correct on the reverse bias. However resistance would decrease as voltage increased
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    (Original post by TheLifelessRobot)
    Why would the wires be in parallel? The circuit would be a cell and a resistor (filament) with each side of the resistor connected to the cell therefore it is in series.
    Surely they are in parallel because you can't model it as a cell as there is no negative terminal, they are both connected to the positive terminal... I don't think it can make a full circuit because the voltage coming in would be equal to the voltage going out so no current current would flow... idk
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    (Original post by TajwarC)
    June 15 - Q7

    There's a cable that's made from other cables, each with their own resistance. When calculating the resistance of the entire cable, you had to assume they were in parallel
    That was the example i was trying to find!!!
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    (Original post by Kmeister77)
    Surely they are in parallel because you can't model it as a cell as there is no negative terminal, they are both connected to the positive terminal... I don't think it can make a full circuit because the voltage coming in would be equal to the voltage going out so no current current would flow... idk
    That's what I thought.
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    (Original post by TajwarC)
    I used a parallel arrangement and still got 229 volts? Found the lost volts in the wire and subtracted it from 230 - same method as someone above who used a series arrangement
    Surely though anyway theyd had to have been in series othwerwise you'd only connevt to the heater by two wires in one end but no wire to connect to other ebd, so circuit wouldn't be complete
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    (Original post by micycle)
    This is how I answered the heater question:

    R = 12

    V = 230

    Resistor:

    R = 0.015 (x2)

    New current = 230/(12+2*0.015) = 19.11

    V lost over cable (2 wires) = 19.11*0.015*2 = 0.574V

    New V = 230 - 0.574 = 229.4V
    I did what you did the first time on that question, but the question says that the resistance of the wires were 0.0150 ohms per meter (length of wires were 3.15m). Therefore I did 0.0150*3.15*2 to get the total resistance of the wires and added that to 12 to get total overall resistance.
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    (Original post by Einstein1997)
    It's 6.3 x resistance because they were 3.15m each.
    So was the R of 0.015 for one meter of wire? I don't remember reading that!
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    Personally I found that exam extremely easy and thrilling to read. I agree with most of the answers given in this post, however it must be said that some of the answers given here do not quite stand up to my par.
    I was aiming to achieve 80UMS, but after sitting it I expect myself to get- at least- 98UMS. Question 6+7 were rather quite easy, and a lot of the people on this post seem to completely blow the questions out of proportion.

    I'm praying to the Gods for you to all do well, however we all know that if you failed the exam then you yourself have no one to blame other than yourself- and no amount of attribution bias that is present will remove this fact.

    At the end of the day I will be achieving an A*, whereas most of you commoners will, maybe, scrap a C overall.
    See you all in the next post I grace you with my presence.
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    (Original post by micycle)
    This is how I answered the heater question:

    R = 12

    V = 230

    Resistor:

    R = 0.015 (x2)

    New current = 230/(12+2*0.015) = 19.11

    V lost over cable (2 wires) = 19.11*0.015*2 = 0.574V

    New V = 230 - 0.574 = 229.4V
    I did it by calculating current (230/12)

    Calculating lost volts - (3.15*0.0150*2)^-1 * 230

    Then found the difference which was 229.5 or something
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    (Original post by HARRY108)
    I put decrease

    Thanks, Same
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    (Original post by TajwarC)
    June 15 - Q7

    There's a cable that's made from other cables, each with their own resistance. When calculating the resistance of the entire cable, you had to assume they were in parallel
    Think about it this way. It's a power source. 2 cables on either side of it. The other end of the cables connect to the heating system. How is it possible to make a parallel circuit from that set up?

    (Original post by micycle)
    This is how I answered the heater question:

    R = 12

    V = 230

    Resistor:

    R = 0.015 (x2)

    New current = 230/(12+2*0.015) = 19.11

    V lost over cable (2 wires) = 19.11*0.015*2 = 0.574V

    New V = 230 - 0.574 = 229.4V
    It's 0.015 per meter. And there's 3.15 meters in each wire. I'm assuming factoring that in gets you 228V so your so your method is slightly incorrect. That being said, idk if 228V is right either
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    (Original post by Yavish13)
    Thanks, Same
    Constant up to ~0.7V then rapidly decreases.


    (Original post by micycle)
    So was the R of 0.015 for one meter of wire? I don't remember reading that!
    Yeah it said per metre.
 
 
 
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