Question

Suppose, you are given three capacitors C1 = 2.0μF, C2 = 4.0μF, and C3 = 6.0μF.

Answer 1

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, $Ct = (C1^{-1} + C2^{-1} + C3^{-1})^{-1}$ = 1.09μF

To calculate the energies:

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

$E1 / E2 = \frac{1/2*Q^2/C1}{1/2*Q^2/C2} = C2 / C1 = 2$

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

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