capacitors

write q = cv for all of them and eliminate..

ty nasiko but im hoping some detailed explaination please.

The main thing to do in any sum of this type is to do the charge distribution. Use symmetry arguments wherever possible.

Now, we have 3 unknowns and hence we need three equations.

The basic idea behind Kirchhoff's Loop Law is that when you take a loop and start from a point and come back to it, the potential difference will be zero, which is also quite obvious.

Now, lets apply it here. Take the smallest top left loop, we move clockwise and write the potential drop,

Q₁2C - Q₃-Q₁
2C - Q₂C = 0 ...(i)

Now take the larger upper loop, and do the same thing and take the largest loop consisting of 2C and 3C capacitors in the end and get an equation.. You'll have three equations and three unknowns. All you'll need now is some good maths skills and you'll be done. [1]

Thats what.. You use this expression and find it out.

ξ - Q₁2C - Q₃3C = 0 [By KLL]

ξ = EMF of the battery = Q₁+Q₂+Q₃C_eq
think why!

And well you can't use that expression Q1+Q2+Q3 = 0 !

Because Q1+Q2+Q3 is the charge being stored by the equivalent(imaginary) capacitor. It has some finite value.

im getting answer as 5c/8 ..can u pls check if ur getting the same

I somehow seem to be getting 3C. Answer doesnt matter, I could have made a mistake in calculations. But if you've done it the right way, you'll get the answer.. You dont have the answeR?