Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

10 \mu F, 6 \mu F

8 \mu F, 8 \mu F

12 \mu F, 4 \mu F

1.2 \mu F, 1.8 \mu F

Step-by-Step Solution

Formulas:

Parallel combination: $C_p = C_1 + C_2$
Series combination: $C_s = \frac{C_1 C_2}{C_1 + C_2}$

Given Data:

$C_p = 16 \text{ \mu F}$ (Sum of capacitances)
$C_s = 3 \text{ \mu F}$

Set up Equations:

From parallel: $C_1 + C_2 = 16$
From series: $\frac{C_1 C_2}{16} = 3 \implies C_1 C_2 = 48$

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 $x = 4$ .
Verification:

Sum: $12 + 4 = 16$ (Matches $C_p$ )
Series: $\frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3$ (Matches $C_s$ )

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