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Capacitors' safe working voltage Watch

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    I'm having problem with part (ii) of this question:


    As the potential difference across two things in parallel is same, so the total safe working should be 12 V. 6V for the left hand side capacitor and 6V for the capacitors in parallel. But mark scheme says:
    "p.d. across parallel combination = ½ ×p.d. across single capacitor
    maximum is 9V"

    Why is it dividing the potential difference by two for parallel capacitors? Like I said, isn't the potential difference same when two capacitors are in parallel?
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    The two in parallel have a total of 60uF and the other is 30uF
    If you connect across this combination the ratio of the pd across the parallel to the pd across the single will be 30 / 60 (=1/2) because V= Q/C (V is inversely prop to capacitance)
    So it must be 3V across the parallel pair and 6V across the single. This gives the maximum 6V across the single capacitor.
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    Hmm, things are rarely straightforward as they seem! Anyway, many thanks.
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    what if all the three have different values of both capacitance and breakdown voltage value.???????/
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    (Original post by nidhi dahiya)
    what if all the three have different values of both capacitance and breakdown voltage value.???????/
    better to start a new thread than add to an old one that's marked 'answered'

    but work out the voltage ratios based on the capacitance as shown by stonebridge

    use the ratios to solve with algebra or by trial & error (if the overall voltage isn't limited by one of the parallel capacitors it must be the series capacitor)
 
 
 
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