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Capacitors of different capacity in series

Discussion in 'General Electronics Discussion' started by pidja105, May 15, 2016.

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  1. pidja105

    pidja105

    106
    1
    Oct 16, 2015
    Hello,
    If I have one capacitor of 2uF 2000V and the other of 470nF 1000V, and wire them in series what I will get?
     
  2. duke37

    duke37

    5,364
    771
    Jan 9, 2011
    Add the reactances for series connection.
    So 1/C = 1/2 + 1/0.47 to give C in μF.

    Do not put more than 1kV on the pair since the voltages may not be shared as you might expect due to leakages.
     
  3. Bluejets

    Bluejets

    4,761
    999
    Oct 5, 2014
    Final substitution being .......C = 1/((1/2) +(1/.047)) uF

    moderator edit for clarification​
     
    Last edited by a moderator: May 16, 2016
  4. hevans1944

    hevans1944 Hop - AC8NS

    4,606
    2,151
    Jun 21, 2012
    Substitution of 0.47 μF = 470 nF yields a series value of 0.3806 μF = 380.6 nF.
     
  5. Ratch

    Ratch

    1,093
    334
    Mar 10, 2013
    Well, let's see. Caps work by storing and releasing energy. It takes energy to imbalance the charge on the cap plates. Since the two caps are in series, each cap can only be imbalanced by the same amount of charge.

    V = Q/C . Assume the charge imbalance on each cap is 1 coulomb. Then the voltages on each cap is (1 coulomb)/(0.47 uf) and (1 coulomb)/(2 uf) = 2.128E6+500000 volts = 2.628E6 volts total. The effective capacitance is then (1 coulomb)/(2.628E6 volts) = 0.381 uf.

    Ratch
     
  6. duke37

    duke37

    5,364
    771
    Jan 9, 2011
    If you have a 1/X button on your calculator, this comes in handy to calculate capacitors in series or resistors in parallel.

    In reverse Polish, which I prefer:-
    2 1/x 0.47 1/x + 1/x Shows 0.381
     
  7. hevans1944

    hevans1944 Hop - AC8NS

    4,606
    2,151
    Jun 21, 2012
    So, do you mean you have charged the 2 μF capacitor to a potential of 2000 V between its two terminals, and you have also separately charged the 470 nF (0.47 μF) capacitor to 1000 V between its two terminals, and then you have connected the two capacitors in series? Let us assume this is so.

    If we assume "perfect" capacitors without leakage, the result will depend on how you make the series connection, series aiding or series opposing. In either case, there will be different charges, Q = CV, on each capacitor, but the equivalent series capacitance is still 0.381 μF. If series-aiding, you have 3000 V across the two capacitors and if series opposing you have 1000 V across the two capacitors. In both cases C = 0.381 μF. Using Q = CV yields two different stored charges, so what is going on here? In particular, what happens to the stored charge if you try to discharge the capacitors connected series-aiding versus series-opposing?

    If you calculate the energy in each capacitor as 1/2 CV^2 and add the energies of each capacitor, you get 4.000 J for the 2 μF cap charged to 2000 V, 0.2350 J for the 0.47 μF cap charged to 1000 V, and 4.235 J for the energy stored in both caps. When connected in series, the capacitance of 0.381 μF has a stored energy of 1.7145 J if connected series-aiding for 3000 V terminal potential and only 0.1905 J if connected series-opposing for 1000 V terminal potential. So where did that (4.235 J - 1.7145 J) = 2.5205 J or (4.235 J - 0.1905 J) = 4.0445 J of energy disappear to? Energy cannot just disappear into thin air, or can it?

    Hint: if the capacitors were equal and both charged to the same voltage and then connected in series-opposing, there would be zero energy available from the series-connected pair, although both capacitors are storing energy in the electrical field between their plates. Sometimes you just can't win.
     
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