In the given circuit, Fig, if the points b is connected to earth and a potential of `+1200` volt is given to the point b is connected to earth and a potential of `+1200` volt is given to the point a, calculate the charges acquired by each of the capacitors and the potentials of the point C ?

Explanation

In Fig, the capacitors `C' = C_(2) + C_(3) = 4+2 = 6 muF`
The equivalent capacitance C' and the capacitance `C_(1)` are in series. Therefore, if C be the equivalent capacitance of the whole circuit, then
`(1)/(C ) = (1)/(C') + (1)/(C_(1)) = (1)/(6) + (1)/(3) = (3)/(6) = (1)/(2)`
`:. C = 2 muF`
The charge on the whole combination of the capacitors is
`q = CV = (2xx10^(-6) farad) xx (1200 vol t) = 2.4xx10^(-3)` coulumb,
The same will be the charge on `C_(1)` alone and also on the combination of `C_(2) and C_(3)`.
Thus, `q_(1) = 2.4xx10^(-3)` coulomb.
Now, `V_(ac) = (q_(1))/(C_(1)) = (2.4xx10^(-3))/(3xx10^(-6)) = 800 volt :' V_(ac) = V_(a) - V_(C)`
`:. V_(C) = V_(a) - V_(ac) = 1200 - 800 = 400 vol t`
Now, `V_(Cb) = V_(C) - V_(b) = 400 - 0 = 400 vol t`
`:. q_(2) = C_(2) V_(Cb) = (4xx10^(-6)) xx400 = 1.6xx10^(-3)` coulomb
and `q_(3) = C_(3) V_(Cb) = (2xx10^(-6)) xx400 = 0.8xx10^(-3)` coulomb