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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?

Respuesta :

lcmendozaf

Answer:

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, [tex]Ct = (C1^{-1} + C2^{-1} + C3^{-1})^{-1}[/tex] = 1.09μF

To calculate the energies:

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

[tex]E1 / E2 = \frac{1/2*Q^2/C1}{1/2*Q^2/C2} = C2 / C1 = 2[/tex]

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

[tex]E1 / E2 = \frac{1/2*C1*V^2}{1/2*C2*V^2} = C1 / C2 = 0.5[/tex]

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