# Capacitor Over-voltage Exercise

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 import math u = 10.0**(-6.0) C1 = 7.0*u C2 = 3.0*u C3 = 3.0*u C4 = 2.0*u V1max = 1.0 V2max = 2.0 V3max = 1.0 V4max = 2.0 dE = 10.0**(-5.0) ############################## E = 0.0 flag = 0 while flag == 0: Q12 = E/(1.0/C1 + 1.0/C2) Q34 = E/(1.0/C3 + 1.0/C4) V1 = Q12/C1 V2 = Q12/C2 V3 = Q34/C3 V4 = Q34/C4 if V1 > V1max: flag = 1 if V2 > V2max: flag = 1 if V3 > V3max: flag = 1 if V4 > V4max: flag = 1 E = E + dE ############################## print E #>>> #2.50001000001 

Note by Steven Chase
1 year ago

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@Steven Chase by the way what is the salary of a person in NASA?

- 11 months, 2 weeks ago

@Steven Chase seems that you are online and free. I have some doubts can you help me ? Please

- 11 months, 2 weeks ago

@Steven Chase have a look on 37th problem. Please

- 1 year ago

@Steven Chase Try my new problem.. You have also love affair with this types problem, therefore I posted.

- 1 year ago

@Steven Chase and the 3rd capacitor fails, not the first one

- 1 year ago

@Steven Chase This is the ideal explanation, I will give if anyone will ask me this problem

- 1 year ago

@Steven Chase voltage will get divided in both capacitor in each branch? So it will be less than 2.6.isn't it?

- 1 year ago

Yes, it is divided across both, but not evenly

- 1 year ago

@Steven Chase it will be like $(2,0.6) , (1.9, 0.7), (1,1.6)$ so what?

- 1 year ago

So sweep the source voltage upward until one of the capacitors reaches its limiting voltage

- 1 year ago

@Steven Chase so, can you show by sweeping?

- 1 year ago

It's in the note

- 1 year ago

@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.?

- 1 year ago

Once you have calculated the charge, you can calculate the capacitor voltages

- 1 year ago

@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???

- 1 year ago

No single capacitor will get the full battery voltage

- 1 year ago

@Steven Chase how can you say so confidently?

- 1 year ago

It is explained in the note. Both series capacitors in each branch have the same charge. It is therefore impossible for one to have voltage without the other one also having some. So neither one can have the full source voltage across it

- 1 year ago

@Steven Chase But how? Confusion is killing me.

- 1 year ago

@Lil Doug This was a good one

- 1 year ago