Solving for Capacitors in Parallel and Series Configurations

Electric circuits often involve capacitors connected in various configurations, either in parallel or in series. The equivalent capacitance in each configuration can be used to determine the individual capacitor values. In this article, we look at an example where two capacitors, when connected in parallel, have an equivalent capacitance of 10 microfarads (μF), and when connected in series, the equivalent capacitance is 2 farads (F).

Graphic Method to Solve the Capacitors

To solve the values of the two capacitors, we can use the known values of their equivalent capacitance in parallel and series. Let's denote the two capacitors as C1 and C2.

In parallel, the equivalent capacitance (Ceq) is given by the sum of the individual capacitances:

C1 C2 10 μF

In series, the equivalent capacitance is given by the reciprocal sum of the individual capacitances:

1/Ceq 1/C1 1/C2

1/2 F 1/C1 1/C2

Algebraic Solution

Since the series equivalent capacitance is 2 F, we can convert it to μF for easier calculations:

1/2 F 1/2,000,000 μF 2,000,000/1,000,000 2 μF

Given:

C1 C2 10 μF

1/C1 1/C2 1/2,000,000 μF

Let C1 x μF and C2 y μF.

From C1 C2 10 μF:

x y 10 … (1)

From 1/C1 1/C2 1/2,000,000 μF:

1/x 1/y 1/2,000,000 … (2)

From equation (1):

y 10 - x … (3)

Substitute equation (3) into equation (2):

1/x 1/10 - x 1/2,000,000

10 - x x / x(10 - x) 1/2,000,000

10 1 - x2 / x(10 - x) 1/2,000,000

2,000,000(10 1 - x2) 10 - x

20,000,000 20,000,00 - 2,000,002 10 - x

2,000,002 - 20,000,001x - 19,999,990 0

Using the quadratic formula:

x [-b ± √(b^2 - 4ac)] / (2a)

a 2,000,000, b -20,000,001, c -19,999,990

x [20,000,001 ± √(20,000,001^2 - 4 * 2,000,000 * -19,999,990)] / (2 * 2,000,000)

x [20,000,001 ± √(400,000,042,000,000 15,999,992,000,000)] / 4,000,000

x [20,000,001 ± √559,999,962,000,000] / 4,000,000

x [20,000,001 ± 236,230,996] / 4,000,000

x 256,230,997 / 4,000,000 ≈ 63.057 μF (not a feasible solution)

x -192,230,995 / 4,000,000 ≈ -48.058 μF (not feasible)

Therefore, the valid solution is:

x 10 - y 10 - 7.236 ≈ 2.764 μF

y 7.236 μF

Alternative Solution

If you consider the measurements in scientific notation, the larger capacitor should be very close to 10 μF while the smaller one should be very close to 2 μF. This is based on the fact that the smaller capacitor mostly determines the series connection, while the larger one dominates the parallel connection:

C1 10,000 / (2,500 * sqrt(624,999,5)) ≈ 2.00000004 μF

C2 (10,000 - 1) / (2,500 * sqrt(624,999,5)) ≈ 999,9997.9 μF

Note that these values are theoretical and capacitors with such exact values are rare.

However, the solution provided by solving the equations accurately gives us the values of the capacitors, verifying that the series capacitance is less than the individual capacitances, and the parallel capacitance is their sum.