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

The effective capacitances of two capacitors are 3 $\mu \mathrm{F}$ and $16 \mu \mathrm{F}$, when they are connected in series and parallel respectively. The capacitance of two capacitors are:
  1. $10 \mu \mathrm{F}, 6 \mu \mathrm{F}$
  2. $8 \mu \mathrm{F}, 8 \mu \mathrm{F}$
  3. $12 \mu \mathrm{F}, 4 \mu \mathrm{F}$
  4. $1.2 \mu \mathrm{F}, 1.8 \mu \mathrm{F}$

Solution

Given: $C_S=3 \mu f$, $\begin{aligned} & C_P=16 \mu f \\ & \text{as } C_S=\frac{C_1 C_2}{C_1+C_2}=3 \\ & C_P=C_1+C_2=16 \\ & \therefore \text{from (1), } frac{C_1 C_2}{16}=3 \\ & C_1 C_2=48 \\ & \text{from (2), } C_2=16-C_1 \\ & C_2=16-\frac{48}{C_2} \\ & \therefore C_2+\frac{48}{C_2}=16 \\ & \therefore C_2^2+48-16 C_2=0 \\ & \therefore C_2=12 \mu F, C_2=4 \mu F \end{aligned}$ i.e. capacitance are $4 \mu F$ and $12 \mu F$.

Asked in: NEET 2022 (Phase 2)

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