Question

When two capacitors are connected in parallel the resulting combination has capacitance 10 $\mu$F. The same capacitors when connected in series results in a capacitance 0.5 $\mu$F. The respective values of individual capacitors are

• A. 1.9 $\mu$F and 0.2 $\mu$F
• B. $(8+2\sqrt{5})$ $\mu$F and $(2-2\sqrt{5})$ $\mu$F
• C. $(5+2\sqrt{5})$ $\mu$F and $(5-2\sqrt{5})$ $\mu$F
• D. 12 $\mu$F and 1.7 $\mu$F
• E. 5 $\mu$F and 2 $\mu$F

Hint:

When sum and product of two values are known ($S=10, P=5$), they are roots of $x^2 - Sx + P = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ gives the answers immediately.

Answer 1

The Correct Option is C

Concept:
For two capacitors $C_1$ and $C_2$: [itemsep=8pt]
• Parallel Combination: $C_p = C_1 + C_2$
• Series Combination: $C_s = \frac{C_1 C_2}{C_1 + C_2}$

Step 1: Set up the equations from the given values.
Given $C_p = 10$ $\mu$F and $C_s = 0.5$ $\mu$F. \[ C_1 + C_2 = 10 \quad \cdots (1) \] \[ \frac{C_1 C_2}{C_1 + C_2} = 0.5 \implies \frac{C_1 C_2}{10} = 0.5 \implies C_1 C_2 = 5 \quad \cdots (2) \]

Step 2: Solve for $C_1$ and $C_2$.
We use the quadratic relation $(C_1 - C_2)^2 = (C_1 + C_2)^2 - 4C_1 C_2$: \[ (C_1 - C_2)^2 = 10^2 - 4(5) = 100 - 20 = 80 \] \[ C_1 - C_2 = \sqrt{80} = 4\sqrt{5} \quad \cdots (3) \] Adding (1) and (3): $2C_1 = 10 + 4\sqrt{5} \implies C_1 = 5 + 2\sqrt{5}$
Subtracting (3) from (1): $2C_2 = 10 - 4\sqrt{5} \implies C_2 = 5 - 2\sqrt{5}$