The effective capacitances of two capacitors are 3 μ F and 16 μ F , when they are connected in series and parallel respectively. The capacitance of two capacitors are:

Correct answer: C. 12 μ F, 4 μ F given: C_S=3 M f aligned & C_P=16 M f & as C_S=[(C_1 C_2) /(C_1+C_2)]=3 & C_P=C_1+C_2=16(2) & from (1), [(C_1 C_2) /(16)]=3 & C_1 C_2=48 & from (2), C_2=16-C_1 & C_2=16-(48 / C_2) & C_2+(48 / C_2)=16 & C_2^2+48-16 C_2=0 & C_2=12 μ F, C_2=4 μ F aligned ) i.e. capacitance are (4 μ F \& 12 μ F)

NEET 2022

Exam: NEET
Year: 2022
Subject: Physics
Chapter: Capacitance

NEET 2022 · Physics · Capacitance

The effective capacitances of two capacitors are 3 μ F and 16 μ F ,...

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:

Options

A $10 \mu \mathrm{F}, 6 \mu \mathrm{F}$
B $8 \mu \mathrm{F}, 8 \mu \mathrm{F}$
C $12 \mu \mathrm{F}, 4 \mu \mathrm{F}$
D $1.2 \mu \mathrm{F}, 1.8 \mu \mathrm{F}$

Correct answer

C. $12 \mu \mathrm{F}, 4 \mu \mathrm{F}$

Step-by-step solution

given:

C_S=3 M f

\begin{aligned} & C_P=16 M f \\ & \text{as } C_S=\left[\left(C_1 C_2\right) /\left(C_1+C_2\right)\right]=3 \\ & C_P=C_1+C_2=16(2) \\ & \therefore \text{from (1), }\left[\left(C_1 C_2\right) /(16)\right]=3 \\ & C_1 C_2=48 \\ & \text{from (2), } C_2=16-C_1 \\ & C_2=16-\left(48 / C_2\right) \\ & \therefore C_2+\left(48 / C_2\right)=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 \mathrm{F} \& 12 \mu \mathrm{F})$

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