Question

Physics jygyiguigyjcfdggsyrdmutgiuhiu

Answer 1

Good afternoon

Answer is

C=8/3

Explanation:

Applying wheatstone bridge principal

[tex] \huge \rightrightarrows \frac{3}{1} = \frac{6}{2} [/tex]

$\huge \rightrightarrows \frac{c1}{c3} = \frac{c2}{c4}$

Hence no current will flow through middle wire

that's why centred capacitors can be ignored

now C1 ,C2 are in series and C3 ,C4 are in series there equivalent=

$c \: of \: upper \: wire =\huge 2 \mu f$

$c \: of \: lower \: wire = \huge \frac{2}{3} \mu f$

Total capacitance as now both wires are in parallel

$\huge \: \frac{2}{3} + 2 = \frac{8}{3} \mu \: f$