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Three capacitors C1, C2, and C3, whose values are 10 µF, 5 µF, and 2 µF respectively, have breakdown voltages of 10 V, 5 V, and 2 V respectively. For the interconnection shown, the maximum safe voltage in Volts that can be applied across the combination and the corresponding total charge in µC stored in the effective capacitance across the terminals are respectively?

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By simple calculation we can find out that the max voltage that can be applied is 2.8 volts.
Now, if we take the equivalent capacitance C = (5 in series 2) || 10 = 80/7 micro farads.
Charge stored across this is Q = CV = 2.8 * 80/7 = 32 micro coulombs.

But if you calculate charges stored for the individual capacitances and add them up, it comes as 36 micro coulombs.
2 volts across 2 μF => q= 2*2 = 4 μC
0.8 volts across 5 μF => q= 0.8 * 5 = 4 μC
2.8 volts across 10 μF => q= 2.8 * 10 = 28 μC

so total q= 36 μC.

What am I doing wrong here?

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  • \$\begingroup\$ The 2nd sentence in your first paragraph is incomplete. \$\endgroup\$
    – Andy aka
    Commented Oct 8, 2023 at 10:30
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    \$\begingroup\$ When C2 and C3 have different breakdown voltages and capacities, chances are that they have different leakage characteristics. That makes it impossible to predict the charge distribution under DC conditions and consequently the "safe voltage" for their combination in series. This may be a textbook question, but it is not safely applicable to the real world. \$\endgroup\$
    – user107063
    Commented Oct 8, 2023 at 10:31

2 Answers 2

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Consider two 2uF 1V capacitors in series.

Series capacitance is 1uF at 2V, with the voltage splitting equally across the two caps.

Using your analysis technique, the max charge on the total capacitor is \$2*1=2uC\$. However the charge on each cap is \$1*2=2uC\$, so the "total" is 4uC.

Why the difference?

It is because the charges on the "inner" plates (those at the series connection point) are balanced "induced" charges, of equal but opposite polarity.

However, being at the internal plates, they never flowed "in" from the external circuit, and similarly if the total capacitor was discharged, they wouldn't flow "out" into the external circuit, they would just cancel internally

In you example the amount of induced charge in the series cap is the same as the charge on the smaller cap (4uC), and if you subtract this from the "total" of 32uC, you then get back to the total "external" charge of 28uC which you correctly calculated by the other technique.

This is similar to how the induced/aligned dipoles in a capacitor dielectric (assuming it isn't a vacuum capacitor), reduce the field strength between the plates of a capacitor. The induced charge is present (and can be calculated), but doesn't flow in the external circuit.

Having said that, there are circumstances where the "inner" charge does come in from the external circuit, for example in a capacitive charge pump. In this circuit, the capacitors are charged in parallel, then discharged in series.

The discharge voltage is twice the charge voltage (given the series connection), but only half the charge which flowed in during charging, flows out round the external circuit during discharge, The other half cancels out internally on the inner plates. This has to be the case, else there would not be conservation of energy.

Both the dielectric and charge pump effects can be nicely demonstrated in physics lessons with "disectable capacitors" made up of physically separated elements whose physical and electrical connections can be reconfigured.

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What am I doing wrong here?

There is a mistake in your calculations. You assumed the voltage drop across C2 is 0.8V, which is correct, but then used 2V for the charge calculation, which is incorrect.

There is a misunderstanding of the charges on capacitors when connected in series.

When capacitors are connected in series, the charge (Q) on each capacitor is the same. This is different than resistors in series, where currents are the same but voltages are different.

Given: (C1 = 10µF) (C2 = 5µF) (C3 = 2µF) and maximum voltages of 10V, 5V, and 2V respectively.

The capacitors (C2) and (C3) are in series and have a combined voltage of (V2 + V3 = 5V + 2V = 7V). Since (C1) has a maximum voltage of 10V and (C2 + C3) together have a maximum voltage of 7V, it is clear that the weakest link in this configuration is (C3) with a maximum of 2V.

Therefore, the maximum safe voltage across the entire combination is 2.8V.

For capacitors in series: (\frac{1}{C_{equivalent}} = \frac{1}{C2} + \frac{1}{C3}) (\frac{1}{C_{equivalent}} = \frac{1}{5} + \frac{1}{2}) (C_{equivalent} = \frac{10}{7} µF)

Now, (C1) is in parallel with (C_{equivalent}), so the total capacitance is: (C_{total} = C1 + C_{equivalent}) (C_{total} = 10 + \frac{10}{7}) (C_{total} = \frac{80}{7} µF)

The total charge on this effective capacitance is: [Q = C_{total} × V = \frac{80}{7} × 2.8 = 32 µC]

Now, let's verify the charges for individual capacitors:

  1. (Q3 = C3 × V = 2µF × 2.8V = 5.6 µC)
  2. (Q2 = C2 × V2 = 5µF × 0.8V = 4 µC) (because for capacitors in series, the voltage drop is not equal, (V = V2 + V3 = 2.8V) and (V2) is 0.8V due to the 7V maximum limit)
  3. (Q1 = C1 × V = 10µF × 2.8V = 28 µC)

Adding up: (Q_{total} = Q1 + Q2 + Q3 = 28 + 4 + 5.6 = 37.6 µC)

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