When two capacitors are connected in parallel and then connected to a battery, the total stored energy is 6.7 times greater than when they are connected in series and then connected to the same battery. What is the ratio of the two capacitances?

4.48 and 0.22

Explanation:

Below are the stated formulas for capacitors in series and parallel.

Let both capacitors be C1 and C2 respectively.

For parallel:

Total C = C1 + C2

For series (after rearranging the formular):

Total C = C1C2 / (C1 + C2)

Now, formular to calculate total energy stored is given below.

Parallel:

E = 1/2 * CV^2

= 1/2 * (C1 + C2) * V^2

Series:

E = 1/2 * CV^2

= 1/2 * (C1C2 / (C1 + C2)) * V^2

In the question, it states that energy stored in parallel is 6.7 times the energy when stored in series.

E (parallel) = 6.7E (series)

1/2 * (C1 + C2) * V^2 = 6.7 * 1/2 * (C1C2 / (C1 + C2)) * V^2

(C1 + C2) = 6.7C1C2 / (C1 + C2) (cancelling out V^2 and 1/2 on both sides)

(C1 + C2) ^2 = 6.7C1C2

C1^2 + 2C1C2 + C2^2 = 6.7C1C2

C1^2 - 4.7C1C2 + C2^2 = 0

Now, we need to divide through by C2^2 to get an equation of the form ax^2 + bx + c = 0, which we can then solve using the quadratic equation.

C1^2 / C2^2 - 4.7C1 / C2 + 1 = 0

let x = C1 / C2 (x is the ratio of the capacitance of the capacitors)

x^2 - 4 x + 1 = 0

Using the quadratic equation, the solutions are:

x = C1 / C2 = 4.48 and x = C1 / C2 = 0.22 in two decimal places.