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    Doing a multiple choice question and I can't work out what I'm doing wrong.

    When fully charged the 2.0 mF capacitor used as a backup for a memory unit has apotential difference of 5.0 V across it. The capacitor is required to supply a constantcurrent of 1.0 μA and can be used until the potential difference across it falls by 10%.For how long can the capacitor be used before it must be recharged?

    A 10 s
    B 100 s
    C 200 s
    D 1000 s

    The answer is D.

    I did C=Q/V and Q = It.

    Therefore (2x10^-3)*(0.9*5)/(1x10^-6) which gives 9000 seconds. Anyone got any ideas?
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    You have calculated how long it would take for the potential difference to fall TO 10% instead of how long it would take to fall BY 10%.
 
 
 
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