Total Energy Stored in Seven Capacitors

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If you knew the voltage at c and at d, then you would know the voltage everywhere, and then find the energy in each capacitor, and so you could add it up to get all the energy.

I'd also try a simple circuit with two capacitors and recall a technique that works there and see if it simplifies your problem at all. Then try three. If you can't find an existing problem, make up some reasonable voltages (or build it an measure it!) and then place charges in a way that conserves charge (for the parts not directly connected to the battery), then based on the charge and voltage compute the capacitance, then hide the voltages from yourself and try to find them from just knowing the capacitance and the circuit wiring (and the battery).

Edit to respond to the edited question

Congratulations on finding the effective capacitance. All the effective capacitance tells you is how the total charge relates to the total voltage. So use that (and the 100V) to find the total charge on each side that is connected to each side of the battery. Now you're going to have to go backwards. Your last step had two (effective) capacitors in parallel, so each had 100V across it and you know the effective capacitance of each, so you know the charge on each.

Knowing how much charge is on the actual (not effective) $16\mu F$ capacitor tells you the voltage drop from b to c. Similarly, using the appropriate effective capacitance find out how much charge is on the actual (not effective) $6\mu F$ capacitor to tell you the voltage drop from a to d.

This is enough to tell you the voltage drop across every actually existing capacitor, so you can now compute the energy of all seven actual existing capacitors.

It should work out fine. Just remember that when you add capacitors you only get something with the right voltage and charge relationship, it doesn't have the same energy as the actual real capacitors.

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Updated on June 09, 2022

Comments

  • CerebralCortexan
    CerebralCortexan over 1 year

    The capacitive network shown in the figure is assembled with initially uncharged capacitors. A potential difference, Vab = +100V, is applied across the network. The switch S in the network is kept open. Assume that all the capacitances shown are accurate to two significant figures. What is the total energy stored in the seven capacitors?

    I'm really struggling on this one. The answer is 72 mJ but I'd like to know why exactly.

    My thought process has been something like this: I know that to find the total energy stored I'm going to need to use .5*C*V^2. In this case, my voltage is 100, so all I need now is the capacitance. The 9 μF capacitor and the 15 μF one are in parallel so I'll just add those to get 24 μF. The 24 and the 16 are in series, and treating them as such gets me a combined 9.6 μF. Moving on to the far right side, I see the 4 μF, 8 μF, and 12 μF all in parallel, and I add them to get 24 μF. Then the 24 μF and the 6 μF are in series, and adding them with the series equation gets me 4.8 μF on that side. So I have 4.8 μF and 9.6 μF on either side. They're in parallel, and adding them together gets me 14.4 μF which, when plugged into my .5*C*100^2 equation does not yield my desired 72 mJ answer. This whole time I chose to ignore the switch because I assumed that is being open meant that it wasn't functioning and really wasn't changing the problem. Clearly I have gone astray somewhere but don't know where.

    Capacitor Diagram

    • Alfred Centauri
      Alfred Centauri over 8 years
      Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better
    • Timaeus
      Timaeus over 8 years
      What kinds of things did you try? Did you try to identify any equivalent or effective circuits? Did you try a simpler circuit to see what parts gave you problems? For instance if the switch is an issue, you can tell us what you know about switches. If it is finding the energy in one capacitor you can ask about that. There are too many things you might struggle with and might not, tell us what you tried, tell us how it went and ask about a physics concept.
    • dmckee --- ex-moderator kitten
      dmckee --- ex-moderator kitten over 8 years
      Treat it as two separate circuits (one for each position of the switch) and work it as a straight equivalent capacitance problem.
    • Timaeus
      Timaeus over 8 years
      Since being edited the question now shows effort, and there is a clear question about a physics concept (whether it is OK to ignore the switch or if switches create complications other than easily making different circuits).
  • CerebralCortexan
    CerebralCortexan over 8 years
    Using this technique got the job done. Thank you so much. You should ultimately get 14.4μF for the whole thing which, when plugged into the equation .5CV^2 get's you your answer. .5(14.4μF)(100V)^2 ultimately equals 72 mJ. Thanks again.