The effective capacitances of two capacitors are $3 ext{ \mu F}$ and $16 ext{ \mu F}$, when they are conne

Question

The effective capacitances of two capacitors are $3 	ext{ \mu F}$ and $16 	ext{ \mu F}$, when they are connected in series and parallel respectively. The capacitance of two capacitors are:

Accepted Answer

1. **Formulas:** 
   - Parallel combination: $C_p = C_1 + C_2$
   - Series combination: $C_s = \frac{C_1 C_2}{C_1 + C_2}$
2. **Given Data:**
   - $C_p = 16 	ext{ \mu F}$ (Sum of capacitances)
   - $C_s = 3 	ext{ \mu F}$
3. **Set up Equations:**
   - From parallel: $C_1 + C_2 = 16$
   - From series: $\frac{C_1 C_2}{16} = 3 \implies C_1 C_2 = 48$
4. **Solve:**
   We need two numbers that add up to 16 and multiply to 48. Using the quadratic equation $x^2 - (sum)x + (product) = 0$:
   $x^2 - 16x + 48 = 0$
   $(x - 12)(x - 4) = 0$
   Therefore, $x = 12$ or $x = 4$.
5. **Verification:** 
   - Sum: $12 + 4 = 16$ (Matches $C_p$)
   - Series: $\frac{12 	imes 4}{12 + 4} = \frac{48}{16} = 3$ (Matches $C_s$)

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