Here is a problem I received recently
The first basic principle is the relationship between the capacitor charge and voltage:
$Q = C V \\ V = \frac{Q}{C}$
The second basic principle is that since capacitors 1 and 2 are in series, they have the same charge $Q_{12}$. Capacitors 3 and 4 also have the same charge $Q_{34}$.
These two principles yield the following:
$E = \frac{Q_{12}}{C_1} + \frac{Q_{12}}{C_2} \\ E = \frac{Q_{34}}{C_3} + \frac{Q_{34}}{C_4}$
The procedure is therefore:
1) Sweep $E$ over a range
2) For each value of $E$, solve for $Q_{12}$ and $Q_{34}$
3) Then for each $E$, solve for the capacitor voltages after the charges are known
4) Stop sweeping $E$ when any of the capacitor voltages exceeds its limit
As it turns out, $E = 2.5$ is the source voltage at which the first capacitor fails.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 

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Top Newest@Lil Doug This was a good one
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@Steven Chase wait, if I applied a battery of $E=2.6$ it doesn't mean that the potential across that 1st capacitor will be 2.6 , so how??
There is one more capacitor in series with first one???
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No single capacitor will get the full battery voltage
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@Steven Chase how can you say so confidently?
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@Steven Chase But how? Confusion is killing me.
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@Steven Chase voltage will get divided in both capacitor in each branch? So it will be less than 2.6.isn't it?
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Yes, it is divided across both, but not evenly
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@Steven Chase it will be like $(2,0.6) , (1.9, 0.7), (1,1.6)$ so what?
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@Steven Chase so, can you show by sweeping?
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@Steven Chase why it getting so much hard.
Now I will show my work, let the charge in that $7 uF$ be x
Then, $E=\frac{x}{7}+\frac{x}{3}$
Which gives $x=5.46$ Can you resolve it, where I am lacking.?
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@Steven Chase This is the ideal explanation, I will give if anyone will ask me this problem
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@Steven Chase and the 3rd capacitor fails, not the first one
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@Steven Chase Try my new problem.. You have also love affair with this types problem, therefore I posted.
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@Steven Chase have a look on 37th problem. Please
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@Steven Chase seems that you are online and free. I have some doubts can you help me ? Please
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@Steven Chase by the way what is the salary of a person in NASA?
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