Question

Each capacitor shown in figure is $2\mu F.$ Then the equivalent capacitance between A and B is :

• A. $2\mu F$
• B. $4\mu F$
• C. $6\mu F$
• D. $8\mu F$

Answer 1

Answer: (A)

$2\mu F$

Equivalent capacitance of upper arms in series

$C_1=\frac{2\times 2}{2+2}=1\mu F$

In lower arms $C_2=1\mu F$

Now $C_1$ and $C_2$ are in parallel

$\therefore$ Equivalent capacitance $C=C_1+C_2=2\mu F$