Question

Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 $\mu F$ is :

• A. $\frac{32}{23}\mu F$
• B. $\frac{31}{23}\mu F$
• C. $\frac{33}{23}\mu F$
• D. $\frac{34}{23}\mu F$

Answer 1

The correct option is A $\frac{32}{23}\mu F$

Between points $E$ and $D$ :

$12\mu F$ and $6\mu F$ are connected in series, So replace them with a single capacitor $C^{\prime }$

$\therefore$ $C^{\prime }=\frac{12\times 6}{12+6}=4\mu F$

Now this $4\mu F$ is connected in parallel with another $4\mu F$ capacitor.

Equivalent capacitance between $E$ and $D$ $C_{ED}=4\mu F+4\mu F=8\mu F$

Now this $8\mu F$ is connected in series with another $1\mu F$ capacitor.

Equivalent capacitance between $R$ and $D$ $C_{RD}=\frac{1\times 8}{1+8}=\frac{8}{9}\mu F$

Between points $P$ and $Q$ :

Equivalent capacitance between points $P$ and $Q$, $C_{PQ}=2\mu F+2\mu F=4\mu F$

This $4\mu F$ is connected in series with another $8\mu F$ capacitor.

Equivalent capacitance between $R$ and $Q$ $C_{RQ}=\frac{4\times 8}{4+8}=\frac{8}{3}\mu F$

Equivalent capacitance between $R$ and $B$ $C_{RB}=\frac{8}{3}\mu F+\frac{8}{9}\mu F=\frac{32}{9}\mu F$

Equivalent capacitance between $A$ and $B$ $C_{AB}=\frac{C\times \frac{32}{9}}{C+\frac{32}{9}}$

$\therefore$ $1\mu F=\frac{\frac{32}{9}C}{C+\frac{32}{9}}$ $⟹C=\frac{32}{23}\mu F$