Suppose, you are given three capacitors c1 = 2.0μf, c2 = 4.0μf, and c3 = 6.0μf.
1. what is the minimum and the maximum capacitance you can obtain using all three capacitors in a circuit?
2. if you apply a voltage vab = 12.0v to the terminals of the series and parallel capacitor circuits of part [a] above to charge the capacitors, what is the ratio of the energy that can be stored in the 2.0μf and the 4.0μf capacitors when they are parts of
[a] the minimum capacitance?
[b] the maximum capacitance configuration?

1.a) Cmax = 12μF

 b) Cmin = 1.09μF

2.a) E1/E2 = 2

  b) E1/E2 = 0.5

Explanation:

For the maximum Capacitance, we have to connect them in parallel. In this configuration, Ct = C1 + C2 + C3 = 12μF

For the minumum Capacitance, we have to connect them in parallel. In this configuration, = 1.09μF

To calculate the energies:

For the minimum capacitance configuration, the charge is the same on all of the capacitors, so:

For the maximum capacitance configuration, the voltage is the same on all of the capacitors, so:

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explanation:

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.