In the following circuit, the resultant capacitance between A and B is 1,mu F. Then value of C is

Q: In the following circuit, the resultant capacitance between $A$ and $B$ is $1\,\mu F$. Then value of $C$ is

d (d) $12 \,\mu F$ and $6\,\mu F$ are in series and again are in parallel with $4\,\mu F$. Therefore, resultant of these three will be $ = \frac{{12 \times 6}}{{12 + 6}} + 4 = 4 + 4 = 8\,\mu F$ This equivalent system is in series with $1 \,\mu F.$ Its equivalent capacitance $ = \frac{{8 \times 1}}{{8 + 1}} = \frac{8}{9}\,\mu F$ ....$(i)$ Equivalent of $8\,\mu F, 2\,\mu F$ and $2\,\mu F$ $ = \frac{{4 \times 8}}{{4 + 8}} = \frac{{32}}{{12}} = \frac{8}{3}\,\mu F$ .....$(ii)$ $(i)$ and $(ii)$ are in parallel and are in series with $C$ $\frac{8}{9} + \frac{8}{3} = \frac{{32}}{9}$ and ${C_{eq}} = 1 = \frac{{\frac{{32}}{9} \times C}}{{\frac{{32}}{9} + C}}$ $==>$ $C = \frac{{32}}{{23}}\,\mu F$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

In the following circuit, the resultant capacitance between $A$ and $B$ is $1\,\mu F$. Then value of $C$ is

A$\frac{{32}}{{11}}\,\mu F$
B$\frac{{11}}{{32}}\,\mu F$
C$\frac{{23}}{{32}}\,\mu F$
D$\frac{{32}}{{23}}\,\mu F$

Diffcult

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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