Question

The equivalent capacitance between the points A and B in the circuit shown in figure :

• A. $1F$
• B. $9F$
• C. $1.5F$
• D. $3.38F$

Answer 1

Answer: (D)

$3.38F$

Adding $8\mu F$ and $4\mu F$ in series :
$C^{\prime }=\frac{8\mu F+12\mu F}{3}=\frac{8\mu F}{3}$
Adding $C^{\prime }$ and $4\mu F=\frac{8\mu +12\mu F}{3}=\frac{20\mu F}{3}$
Adding $C^{\prime \prime },12\mu F.$ and $16\mu F$ in series :
$\frac{1}{C_{eq}}=\frac{1}{C"}+\frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{C_{eq}}$ = $\frac{3\times 12\times 16+20\times 16\times 1+12\times 20\times 1}{20\times 12\times 16}$
$\frac{1}{C_{eq}}=\frac{576+320+240}{3840\mu F}=\frac{1136}{3840\mu F}$
$C_{eq}=\frac{3840\mu F}{1136}=3.38\mu F$