Question

What is the equivalent capacitance of the
capacitors between A and B​

Answer 1

Answer:

8C/5

Explanation:

Right corners is in series so there equivalent will be

$\frac{c \times c}{c + c} = \frac{c}{2}$

Now c/2 and c are in parallel so there equivalent

3c/2 again it's in series with c .

Equivalent

$\frac{c \times \frac{3c}{2} }{c + \frac{3c}{2} } = \frac{3c}{5}$

And it's parallel with c so

c + 3c/5 = 8c/5

Answer 2

Answer:8c/5

Explanation:

from right side of the given circuit i.e from point B if we move upward 1st two capacitors are in series so total capacitance is (c×c)÷(c+c) = c² / 2c = c/2. we know that when two capacitors say c1 and c2 are in series then equivalent capacitance or total capacitance is cs =  (c1×c2)/(c1+c2).since here both the capacitors are same value say "c", simply cs = c/2.

now c/2 is parallel with another capacitor c .now total capacitance = c + c/2 = 3c/2 (when two capacitors c1 , c2 are in parallel total capacitace = c1+c2). again 3c/2 is in series with 4th capacitor "c". so total capcitance now = (3c/2 × c)/ (3c/2 + c) = 3c/5. now 3c/5 is in parallel with 5th capacitor I.e most lefft hand capacitor . so total capacitance of the circuit CAB = CAPACITANCE FROM A TO B = 3c/5 + c = 8c/5.