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    How do I work out the voltage for ii)?
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    (Original post by scientific222)
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    How do I work out the voltage for ii)?
    It's a potential divider, I believe.
    So: \ V_{out} = \frac{R_1}{R_1 + R_2} \times V_{in}
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    Vout is Vin times R1/R1 + R2
    So pd = 5* 1.8/2.8 = 3.2 V
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    (Original post by scientific222)
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    How do I work out the voltage for ii)?

    \ V_{out} = \frac{R_1}{R_1 + R_2} \times V_{in}

    R1 = resistor your trying to find the voltage for

    R2 = other resistor in series

    Vin = total voltage in circuit

    Vout = voltage across R1 resistor
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    (Original post by joostan)
    It's a potential divider, I believe.
    So: \ V_{out} = \frac{R_1}{R_1 + R_2} \times V_{in}
    The answer using the above comes out correct for my question. Odd thing however is that these are questions that come before the topic of "potential divider" in the textbook (which I still haven't learnt), so working through the book in order I perhaps shouldn't be applying that equation. Is that the only possible way there is for working the voltage out in this situation?
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    (Original post by scientific222)
    The answer using the above comes out correct for my question. Odd thing however is that these are questions that come before the topic of "potential divider" in the textbook (which I still haven't learnt), so working through the book in order I perhaps shouldn't be applying that equation. Is that the only possible way there is for working the voltage out in this situation?
    Its just a matter of ratios. The equation simply defines them more neatly.
    It seems a bit odd though.
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    well if you work out differently, you will eventually kind of prove that it is a potential divider
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    (Original post by scientific222)
    The answer using the above comes out correct for my question. Odd thing however is that these are questions that come before the topic of "potential divider" in the textbook (which I still haven't learnt), so working through the book in order I perhaps shouldn't be applying that equation. Is that the only possible way there is for working the voltage out in this situation?
    Is this the OCR heinemann Physics A book?
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    (Original post by ZakRob)
    Is this the OCR heinemann Physics A book?
    Yes it is
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    (Original post by scientific222)
    Yes it is
    Thought so :L it tends to ask questions on topics which come later on in the book. Useful when you've covered everything but not so much if you're just moving through it topic by topic.
 
 
 
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