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    The first thing to bear in mind, is that Y and X also have a resistance.The Voltage is split between the resistor R1 and the parallel system of R2, Y and X.

    Within the parallel system, the voltage is common between the two branches, but split across R2 and Y. So, the voltage across R2 plus the voltage across Y is equal to the voltage across X. And this voltage (across X) plus the voltage across R1 is equal to the voltage of the battery (24V)
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    perhaps you should have included the earlier section... there doesn't seem to be enough information.
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    Just updated my first post. Didn't realise that the annotations on the diagram were his questions initially >_>
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    (Original post by Tynos)
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    Could I draw your attention to the guidelines for posting here. In particular

    When you do post, post the whole question instead of just a few bits from it; it may not be solvable or may be much harder without some of the details - help us to help you!
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    (Original post by Joinedup)
    perhaps you should have included the earlier section... there doesn't seem to be enough information.

    My bad X = 3A

    Y = 0.4A
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    (Original post by Tynos)
    My bad X = 3A

    Y = 0.4A
    Yea, so you can use those to work out the resistances of X and Y.

    The voltage is not split evenly, because everything seems to have a different resistance.

    Hang on, and I'll work out the resistances of everything and go through it for you .
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    (Original post by Stonebridge)
    Could I draw your attention to the guidelines for posting here. In particular
    There where no further information provided after this, just questions like calculate the pd across R1
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    (Original post by Qwertish)
    Yea, so you can use those to work out the resistances of X and Y.

    The voltage is not split evenly, because everything seems to have a different resistance.

    Hang on, and I'll work out the resistances of everything and go through it for you .

    But the Voltage seems to have been split evenly, the make scheme says 12V meaning each resistor has 12V and equal?
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    First you need to work out the total current running through the circuit, I_{Tot}. The current through R1 and the sums of the current through each of the parallel branches are equal to I_{Tot}, so:

    

I_{Tot} = I_X + I_Y

I_{Tot} = \SI{3.4}{A}

    If you know what reading the voltmeter in parallel with R1 has, we can work out the resistance of R1 .
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    (Original post by Tynos)
    But the Voltage seems to have been split evenly, the make scheme says 12V meaning each resistor has 12V and equal?
    Do you have the actual questions? The image you posted just seems to have the preamble and your annotations.
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    (Original post by Qwertish)
    Do you have the actual questions? The image you posted just seems to have the preamble and your annotations.

    Yes they are:

    6 (b) (i) Calculate the pd across R1.

    6 (b) (ii) Calculate the current in R1.

    6 (b) (iii) Calculate the resistance of R1.

    6 (b) (iv) Calculate the pd across R2.

    6 (b) (v) Calculate the resistance of R2.
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    (Original post by Tynos)
    There where no further information provided after this, just questions like calculate the pd across R1
    You have not provided the full question. This is 6b. What about 6a?

    For example, what information is given earlier about the bulbs?

    I repeat. Please provide the whole question. It saves us all a lot of time trying to guess.
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    (Original post by Tynos)
    Yes they are:

    6 (b) (i) Calculate the pd across R1.

    6 (b) (ii) Calculate the current in R1.

    6 (b) (iii) Calculate the resistance of R1.

    6 (b) (iv) Calculate the pd across R2.

    6 (b) (v) Calculate the resistance of R2.
    Thanks. And were those currents ALL the information you were provided? Or was there anything else?
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    (Original post by Qwertish)
    Thanks. And were those currents ALL the information you were provided? Or was there anything else?
    yeah there was this.

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    (Original post by Tynos)
    yeah there was this.

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    Ahhhhhhhh right. Now I can do the question, lol.

    So, 6b says that all lamps are running at their proper voltage. This means that X is dropping 12V. That leaves 24-12 = 12V for R1.

    As I said before, the current in R1 is 3.4A.

    Now, use ohms law to work out the resistance of R1:
    R = V/I
    R1 = 12/3.4
    R1 = 3.529ohms

    The PD across R2 (call it V2) plus the PD across Y is equal to the PD across X, so:
    V2 + 4.5 = 12
    V2 = 7.5V

    The current through Y is 0.4A, and the current through R2 is the same as this. So now we can work out R2:
    R2 = 7.2/0.4
    R2 = 0.53ohms
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    (Original post by Qwertish)
    Ahhhhhhhh right. Now I can do the question, lol.

    So, 6b says that all lamps are running at their proper voltage. This means that X is dropping 12V. That leaves 24-12 = 12V for R1.

    As I said before, the current in R1 is 3.4A.

    Now, use ohms law to work out the resistance of R1:
    R = V/I
    R1 = 12/3.4
    R1 = 3.529ohms

    The PD across R2 (call it V2) plus the PD across Y is equal to the PD across X, so:
    V2 + 4.5 = 12
    V2 = 7.5V

    The current through Y is 0.4A, and the current through R2 is the same as this. So now we can work out R2:
    R2 = 7.2/0.4
    R2 = 0.53ohms

    Can i just ask a question please.

    all lamps are running at their proper voltage

    can you explain to me how
    This means that X is dropping 12V. That leaves 24-12 = 12V for R1.

    i dont understand this part.

    So if someone didnt understand this part are they able to complete the quuestion?
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    (Original post by Tynos)
    Can i just ask a question please.

    all lamps are running at their proper voltage

    can you explain to me how
    This means that X is dropping 12V. That leaves 24-12 = 12V for R1.

    i dont understand this part.

    So if someone didnt understand this part are they able to complete the quuestion?
    In 6b, it says "the lamps are operating at their correct working voltage". At the start of the question, it says "X is rated at 12V". That means the working voltage for X is 12V. This means X 'uses' 12V of the battery's 24V leaving 12V for R1.

    And no, you can't do 6b if you don't understand this. Have you done Kirchhoff's Laws about parallel and series circuits?
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    (Original post by Qwertish)
    In 6b, it says "the lamps are operating at their correct working voltage". At the start of the question, it says "X is rated at 12V". That means the working voltage for X is 12V. This means X 'uses' 12V of the battery's 24V leaving 12V for R1.

    And no, you can't do 6b if you don't understand this. Have you done Kirchhoff's Laws about parallel and series circuits?
    Ohh thank you, and yes i have.
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    (Original post by Tynos)
    Ohh thank you, and yes i have.
    Okay, just checking haha, you should be fine then
 
 
 
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