Question

answer the given question ​

Answer 1

Answer:

The equivalent capacitance between A and B is

$\green{C_{eq} = 1 \mu \: F}$

Explanation:

→ Here capacitors c, d, e are in series.

So,

$\frac{1}{C_1} = \frac{1}{c} + \frac{1}{d} + \frac{1}{e} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} \\ \therefore \: C_1= 1 \mu \: F \:$

→ $C_1$ and f are parallel

$C_2 = C_1 + f = 1 + 2 = 3 \mu \: F \:$

→b, $C_2$ , h are series

$\frac{1}{C_3} = \frac{1}{b} + \frac{1}{C_2} + \frac{1}{h} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} \\ \therefore \: C_3 = 1 \mu \: F \:$

→ $C_3$ , g are in parallel

$C_4 = C_3 + g = 1 + 2 = 3 \mu \: F \:$

→ a,$C_4$, i are in series

$\frac{1}{C_{eq}} = \frac{1}{a} + \frac{1}{C_4} + \frac{1}{i} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} \\ \therefore \: C_{eq} = 1 \mu \: F$