Design a combination of Capacitors

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Both the user's and textbook approaches to designing capacitor combinations are valid, but the choice between them should consider the failure modes of the components. The discussion highlights that a series array of capacitors in parallel is less affected by an open failure, while a parallel array in series is less impacted by a closed failure. The prevalence of specific failure types, such as ceramic capacitors typically failing closed, can influence the preferred design. Additionally, cascading failures can lead to short circuits regardless of the initial configuration. Ultimately, the best approach depends on the specific application and failure considerations.
Shreya
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Homework Statement
It is required to construct a 10mF capacitor which can be connected to a 200V Battery. Capacitors of Capacitance 10mF are available but they can withstand only 50V. Design a combination which can yield the desired result.
Relevant Equations
C=Q/V
My attempt was this:
IMG_20211016_153820.jpg

The textbook answer was:
Screenshot_20211016-153914_SelfStudys.png

Is my approach wrong?
 
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Shreya said:
Is my approach wrong?
Both approaches are valid.

If one wanted to choose between those two approaches (or other possible approaches), then it would be interesting to consider the effect of a failure (fail open or fail closed) of one of the components.
 
jbriggs444 said:
interesting to consider the effect of a failure (fail open or fail closed) of one of the components.
On failure of components, the textbook answer would be preferred,right?
 
Shreya said:
On failure of components, the textbook answer would be preferred,right?
Not necessarily. It depends on what failure type is most prevalent.

A series array of capacitors in parallel will not change its capacitance much if one element fails open.
A parallel array of capacitors in series will not change its capacitance much if one element fails closed.

But then one should also be concerned with the possibility of cascading failure. I do not build such things for a living, so I cannot offer an opinion on which approach is "best" in real life.

Edit: a quick trip to Google says that ceramic capacitors tend to fail closed (short circuit) on overload.

Cascading failure yields a short circuit either way.
 
Last edited:
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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