Capacitors connected in parallel to store a charge of 1.49 C?

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SUMMARY

To store a charge of 1.49 C with a potential of 105 V, a total of 9589 capacitors rated at 1.48 µF must be connected in parallel. The calculation requires using the formula for capacitance in parallel, where the total capacitance is the sum of individual capacitances. Rounding is necessary to ensure a whole number of capacitors, as fractional capacitors cannot be purchased.

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McAfee
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Homework Statement


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How many 1.48 µF capacitors must be connected in parallel to store a charge of 1.49 C with a potential of 105 V across the capacitors?


Homework Equations


Refer to the attempt at a solution.

The Attempt at a Solution


10AcM.jpg

Also, for the answer exact number, no tolerance. I tried just rounding the number but that didn't work. I not sure. Does it mean I should round the answer up to 9589?
 
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Hi McAfee! :smile:

(i haven't checked your figures)
McAfee said:
How many 1.48 µF capacitors must be connected in parallel to store a charge of 1.49 C with a potential of 105 V across the capacitors?

Also, for the answer exact number, no tolerance. I tried just rounding the number but that didn't work. I not sure. Does it mean I should round the answer up to 9589?

yes, of course

you can't buy 0.16 of a capacitor, can you? :biggrin:
 
tiny-tim said:
Hi McAfee! :smile:

(i haven't checked your figures)


yes, of course

you can't buy 0.16 of a capacitor, can you? :biggrin:


You are right. Thanks for the help.
 

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