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    Hi Here's the question...

    'A battery has e.m.f. E and internal resistance, r. How does the terminal p.d. change when the current drawn from the battery increases?'

    The answer is:

    'The terminal p.d. decreses because there's a greater voltage drop across the internal resistance.'

    ^^I really have no idea what this means, can anyone explain please?
    What exactly is the voltage drop, and why would increasing the current increase the voltage drop on the internal resistance but not the external resistance?

    Thank you!
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    (Original post by tea&biscuits<3)
    Hi Here's the question...

    'A battery has e.m.f. E and internal resistance, r. How does the terminal p.d. change when the current drawn from the battery increases?'

    The answer is:

    'The terminal p.d. decreses because there's a greater voltage drop across the internal resistance.'

    ^^I really have no idea what this means, can anyone explain please?
    What exactly is the voltage drop, and why would increasing the current increase the voltage drop on the internal resistance but not the external resistance?

    Thank you!
    For a fixed e.m.f, E=I(R+r) as r is a fixed quantity, if I increases, with a constant e.m.f, if Ir increases then IR must decrease.
    As V=IR if IR decreases then V decreases. Where V is voltage out.
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    (Original post by joostan)
    For a fixed e.m.f, E=I(R+r) as r is a fixed quantity, if I increases, with a constant e.m.f, if Ir increases then IR must decrease.
    As V=IR if IR decreases then V decreases. Where V is voltage out.
    That's alot clearer now, cheers
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    (Original post by tea&biscuits<3)
    That's alot clearer now, cheers
    ...and I would give you +ve rating but it says I can't, so thank you anyway
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    (Original post by tea&biscuits<3)
    ...and I would give you +ve rating but it says I can't, so thank you anyway
    Np
 
 
 
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