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 

Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Sort by:
Top Newest@Lil Doug This was a good one
Log in to reply
@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???
Log in to reply
No single capacitor will get the full battery voltage
Log in to reply
@Steven Chase how can you say so confidently?
Log in to reply
Log in to reply
@Steven Chase But how? Confusion is killing me.
Log in to reply
@Steven Chase voltage will get divided in both capacitor in each branch? So it will be less than 2.6.isn't it?
Log in to reply
Yes, it is divided across both, but not evenly
Log in to reply
@Steven Chase it will be like $(2,0.6) , (1.9, 0.7), (1,1.6)$ so what?
Log in to reply
Log in to reply
@Steven Chase so, can you show by sweeping?
Log in to reply
Log in to reply
@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.?
Log in to reply
Log in to reply
@Steven Chase This is the ideal explanation, I will give if anyone will ask me this problem
Log in to reply
@Steven Chase and the 3rd capacitor fails, not the first one
Log in to reply
@Steven Chase Try my new problem.. You have also love affair with this types problem, therefore I posted.
Log in to reply
@Steven Chase have a look on 37th problem. Please
Log in to reply