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=CVV=QC 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 Q12Q_{12} . Capacitors 3 and 4 also have the same charge Q34 Q_{34} .

These two principles yield the following:

E=Q12C1+Q12C2E=Q34C3+Q34C4 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 E over a range
2) For each value of E E , solve for Q12Q_{12} and Q34Q_{34}
3) Then for each E E , solve for the capacitor voltages after the charges are known
4) Stop sweeping E E when any of the capacitor voltages exceeds its limit

As it turns out, E=2.5 E = 2.5 is the source voltage at which the first capacitor fails.

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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
2 weeks, 2 days ago

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1 vote

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Comments

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@Lil Doug This was a good one

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase wait, if I applied a battery of E=2.6E=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???

Lil Doug - 2 weeks, 2 days ago

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No single capacitor will get the full battery voltage

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase how can you say so confidently?

Lil Doug - 2 weeks, 2 days ago

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@Lil Doug 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

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase @Steven Chase But how? Confusion is killing me.

Lil Doug - 2 weeks, 2 days ago

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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?

Lil Doug - 2 weeks, 2 days ago

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Yes, it is divided across both, but not evenly

Steven Chase - 2 weeks, 2 days ago

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

Lil Doug - 2 weeks, 2 days ago

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

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase @Steven Chase so, can you show by sweeping?

Lil Doug - 2 weeks, 2 days ago

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@Lil Doug It's in the note

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase @Steven Chase why it getting so much hard.
Now I will show my work, let the charge in that 7uF7 uF be x
Then, E=x7+x3E=\frac{x}{7}+\frac{x}{3}
Which gives x=5.46x=5.46 Can you resolve it, where I am lacking.?

Lil Doug - 2 weeks, 2 days ago

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@Lil Doug Once you have calculated the charge, you can calculate the capacitor voltages

Steven Chase - 2 weeks, 2 days ago

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

Lil Doug - 2 weeks, 2 days ago

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

Lil Doug - 2 weeks, 2 days ago

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

Lil Doug - 2 weeks, 1 day ago

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

Lil Doug - 2 weeks, 1 day ago

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