Question

The equivalent capacitance between points $A$ and $B$ for the given figure is

• A. $1F$
  
• B. $2F$
  
• C. $3F$
  
• D. $4F$
  

Answer 1

The correct option is A $1F$

$C_1,C_2,C_3$ are in series The equivalent capacitance of the series combination is $\frac{1}{C_x}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$ $\therefore C_x=\frac{3}{3}=1pF$

Now $C_{x}isparallelto{C}_4$ The equivalent capacitance of the parallel combination is $C_y=C_x+C_4$

$\therefore C_y=1+2=3pF$

Further $C_y,C_5,C_6$ are in series The equivalent capacitance of the series combination is $\frac{1}{C_z}=\frac{1}{C_y}+\frac{1}{C_5}+\frac{1}{C_6}$

$\therefore C_z=\frac{3}{3}=1pF$

Next $C_{z}isparallelto{C}_7$ The equivalent capacitance of the parallel combination is $C_w=C_z+C_7$

$\therefore C_w=1+2=3pF$

$C_w,C_8,C_9$ are in series and the resultant capacitance is $\frac{1}{C_{eq}}=\frac{1}{C_w}+\frac{1}{C_8}+\frac{1}{C_9}$

$\therefore C_{eq}=\frac{3}{3}=1pF$

Hence the resultant capacitance for the given series/parallel combinations of capacitor is $1pF$