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    Hi all,

    Could anyone explain why the answer is 6 please?

    Thank you!
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    (Original post by londoncricket)
    Hi all,

    Could anyone explain why the answer is 6 please?

    Thank you!
    ok so, first step:
    the reading when switch is closed is 3 ohms for the resistance and 24v
    2nd : as the switch is closed the resistance changes ( product over sum it is 2.25 ohms)
    3rd: 3 / 2.25 is 0.75
    4th:
    0.75 x 24 = 18
    5th: 24 - 18 = 6
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    Nevermind i messed up
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    (Original post by EggFriedRai)
    ok so, first step:
    the reading when switch is closed is 3 ohms for the resistance and 24v
    2nd : as the switch is closed the resistance changes ( product over sum it is 2.25 ohms)
    3rd: 3 / 2.25 is 0.75
    4th:
    0.75 x 24 = 18
    5th: 24 - 18 = 6
    Was the load connexted in parallel? I couldnt tell
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    (Original post by EggFriedRai)
    ok so, first step:
    the reading when switch is closed is 3 ohms for the resistance and 24v
    2nd : as the switch is closed the resistance changes ( product over sum it is 2.25 ohms)
    3rd: 3 / 2.25 is 0.75
    4th:
    0.75 x 24 = 18
    5th: 24 - 18 = 6
    Thank you for your reply

    What does "product over sum" mean?
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    (Original post by metrize)
    Was the load connexted in parallel? I couldnt tell
    you would assume it was set up reguarly
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    (Original post by EggFriedRai)
    you would assume it was set up reguarly
    I thought total resistance would be 12ohms
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    (Original post by londoncricket)
    Thank you for your reply

    What does "product over sum" mean?
    when you have two resistances in parallel, you do the product ( times them together ) nd then divide it by the sum ( add them up ) for the total resistance
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    (Original post by metrize)
    I thought total resistance would be 12ohms
    Follow a logical thought pattern
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    (Original post by EggFriedRai)
    Follow a logical thought pattern
    I couldnt tell from the question that it was in parallel so i assumed it was in series so i added the resistance
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    (Original post by metrize)
    I couldnt tell from the question that it was in parallel so i assumed it was in series so i added the resistance
    easily done mate dont worry
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    (Original post by EggFriedRai)
    when you have two resistances in parallel, you do the product ( times them together ) nd then divide it by the sum ( add them up ) for the total resistance
    Okay so this is the same as doing (1/A+1/B)^-1 if "A" and "B" were to be two resistances in parallel.

    And so why would you calculate 0.75?
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    (Original post by londoncricket)
    Okay so this is the same as doing (1/A+1/B)^-1 if "A" and "B" were to be two resistances in parallel.

    And so why would you calculate 0.75?
    to get the ratio to divide by
 
 
 
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