fig. 1 find the equivalent capacitance

fig 2 find potrential diff. between a and b.....

(HCV ques from capacitors....25 and 26....)

sorry for shabby diagrams...

by loop eqn,(if u name the loops properly),

(qa, qc ,qb are the 3 charges assumed)

u'll get 1 eqn.-

qa/1+qc/4-qb/3=0  ..............(i)

=> 12qa+3qc-4qb=0

or, 12qa-4q2=0     ..........(i)

another eqn,

(qa-qc)/3-(qb-qc)/1-qc/4=0

or,4qa-4qc-12qb-12qb-12qc-3qc=0

or, 4qa-12qb=19qc    ...............(ii)

therefore from (i) & (ii)

qa=-7/8qc

qb=-15/8qc

since, p.d. =qa/1 + (qa-qc)/3 = (4qa-qc)/3

hence eq. capacitance

=> total charge / p.d

=(qa+ qb) / {4qa-qc)/3}

= 11/6 micro farad!!

thanx aratrika....

but can we do this question without using the loop law???

i mean by simple parallel and series combination funda???

and in the first eq i m getting -qb/3+qa/1-qc/4=0....

can u please give the direction of charge flow????(diagram if possible)

im getting equi.capacitance = 2 F

is it right???? plz nudge me

its possible by parallel & series combination!! bt  that will make the sum unnessecerily much more complicated!!

& u'r getting the negative charge flow from that of mine coz u've taken the opposite convention from that of mine!! the  equation that u've got , if u proceed with that one then u'll get the answer as -11/6!! no need to worry abt that!!