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:

concave lenses are used here. they spread the light out before it reaches the convex lens in the eye, therefore letting the image focus directly on the retina. hyperopia is known to you probably as farsightedness.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)