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    According to me the answer should be B, because when the slider will b at Y, the Voltmeter should read half the value of EMF! But the mark scheme gives A as the answer - why is that so?! Explanation, please.
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    When you increase the resistance current goes down but total voltage stays the same.
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    Strange as it may seem (and it did to me on first look), answer A is correct.
    I've redrawn the circuit here so that at the top we have the case with the contact at X.
    Here the pd across the voltmeter is without doubt 4V.
    With the contact at Y I have redrawn the circuit.
    On the right at the bottom you need to see that it is the same circuit.
    The clue is the statement that "the voltmenter has a very high resistance".
    This means that the current through P and the meter is the same (they are in series) and is small because the voltmeter has a high resistance.
    As they are in series and the resistance of the meter is very high, nearly all of the 4 volts is across the meter and almost zero across P.
    So the voltmeter will still read 4 volts, or at least, so close to 4 that you would not see the difference. The meter would typically have a resistance of mega ohms.
    So the split of voltage between P and the meter would be of the order of 3.99999 across V and 0.00001 across P
    So graph A is right.
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    (Original post by Zishi)
    According to me the answer should be B, because when the slider will b at Y, the Voltmeter should read half the value of EMF! But the mark scheme gives A as the answer - why is that so?! Explanation, please.
    The voltmeter shows potential difference across its terminals. If the voltmeter has infinitely high resistance, i.e. no current goes through it, then when slider is at Y no current passes through P, so there is no potential drop across P, so the voltmeter shows the electromotive force all the time.

    However, that's an ideal situation, in fact there would be a small decrease of the measured voltage. No voltmeter has infinitely high resistance which means that there would be some very small current going through the voltmeter, and through P from slider to the voltmeter. That means that there would be some potential drop across P and the voltmeter would show a smaller voltage. Also, internal resistance of the source (here neglected) would contribute to the decrease in the voltage measured by the voltmeter.

    But in this case, when we assume every element of the circuit to be ideal, there is no potential drop between the slider and voltmeter, so the measured value is always the same.

    I hope I haven't complicated it too much, I've really tried my best
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    (Original post by Stonebridge)
    Strange as it may seem (and it did to me on first look), answer A is correct.
    I've redrawn the circuit here so that at the top we have the case with the contact at X.
    Here the pd across the voltmeter is without doubt 4V.
    With the contact at Y I have redrawn the circuit.
    On the right at the bottom you need to see that it is the same circuit.
    The clue is the statement that "the voltmenter has a very high resistance".
    This means that the current through P and the meter is the same (they are in series) and is small because the voltmeter has a high resistance.
    As they are in series and the resistance of the meter is very high, nearly all of the 4 volts is across the meter and almost zero across P.
    So the voltmeter will still read 4 volts, or at least, so close to 4 that you would not see the difference. The meter would typically have a resistance of mega ohms.
    So the split of voltage between P and the meter would be of the order of 3.99999 across V and 0.00001 across P
    So graph A is right.

    (Original post by jaroc)
    The voltmeter shows potential difference across its terminals. If the voltmeter has infinitely high resistance, i.e. no current goes through it, then when slider is at Y no current passes through P, so there is no potential drop across P, so the voltmeter shows the electromotive force all the time.

    However, that's an ideal situation, in fact there would be a small decrease of the measured voltage. No voltmeter has infinitely high resistance which means that there would be some very small current going through the voltmeter, and through P from slider to the voltmeter. That means that there would be some potential drop across P and the voltmeter would show a smaller voltage. Also, internal resistance of the source (here neglected) would contribute to the decrease in the voltage measured by the voltmeter.

    But in this case, when we assume every element of the circuit to be ideal, there is no potential drop between the slider and voltmeter, so the measured value is always the same.

    I hope I haven't complicated it too much, I've really tried my best
    Thanks a lot to both of you! I got it.
 
 
 
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