Physics capacitor question

Question

Two capacitors give an equivalent capacitance of 9.20 pF when connected in parallel and an equivalent capacitance of 1.73 pF when connected in series. What is the capacitance of each capacitor?

smaller capacitor=____________pF
bigger capacitor=____________pF

Answer 1

Parallel capacitors combine using:

C₁ + C₂ = C_total

Series capacitors combine using: 1/C_total = 1/C₁ + 1/C₂ which simplifies to:

C_total = (C₁•C₂)/(C₁ + C₂)

Plugging in the numbers we get:

C₁ + C₂ = 9.20 meaning C₁ = 9.20 - C₂

and (C₁•C₂)/(C₁ + C₂) = 1.73

Plugging in "9.20 - C₂" in place of C1 into the second equation gives us:

((9.20 - C₂)•C₂)/((9.20 - C₂) + C₂) = 1.73

(9.20C₂ - C₂²)/9.20 = 1.73

(9.20C₂ - C₂²) = 15.916

0 = 15.916 - 9.20C₂ + C₂²

C₂² - 9.20C₂ + 15.916 = 0

Using the quadratic formula:

we can solve for C2 (which is our "x") using a = 1, b = -9.2, and c = 15.916

x = (9.2 ± √20.976)/2

x = 6.89 and x = 2.31

So C₂ is either 6.89 pF or 2.31 pF

When C₂ is 6.89 pF, using C₁ = 9.20 - C₂ we calculate C₁ = 2.31 and when C₂ is 2.31, we calculate C₁ = 6.89

So the small capacitor is 2.31 pF and the bigger capacitor is 6.89 pF

Answer 2

The capacitance for parallel capacitors is given by C_p = C1 + C2.

For series capacitors, the equation is C_s = C1 x C2 / (C1 + C2).

Since you are given C_p and C_s, you are left with two equations with two unknowns. You can then solve for C1 and C2.