Question

The equivalent capacitance $C_{AB}$ of the circuit shown in the figure is :

• A. $\frac{5}{4}C$
• B. $\frac{4}{5}C$
• C. $2C$
• D. $C$

Answer 1

Answer: (A)

$\frac{5}{4}C$

The given circuit can be reconstructed as shown (As the potential difference between A and B has to same through all possible paths)

Using the laws of combination of capacitors,

$C_{net}=C_1+C_2+C_{3}where$

$C_2=\frac{C}{2},\frac{1}{C_1}=\frac{1}{C}+\frac{1}{C}+\frac{1}{(C+\frac{C}{2})}=\frac{2}{C}+\frac{2}{3C}=\frac{8}{3C}\Rightarrow C_1=\frac{3C}{8}$

Similarly, $C_3=\frac{3C}{8}$

$C_{net}=\frac{6C}{8}+\frac{C}{2}=\frac{5C}{4}$