Respuesta :
Answer:
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.
