Obtain the equivalent capacitance of the network in Fig. For a 300 V supply. Determin the change and voltage across each capacitor.

Explanation

Here `C_(2)` and `C_(3)` are in series
`:. (1)/(C_(1)) = (1)/(200) + (1)/(200) = (2)/(200) = (1)/(100)`
`C_(s) = 100 pF`
Now `C_(s)` and `C_(1)` are in parallel
`:. C_(P) = C_(s) + C_(1) = 100 + 100 = 200 pF`
Again `C_(P)` and `C_(4)` are in series
`:. (1)/(C_(s)) + (1)/(C_(P)) + (1)/(C_(4)) = (1)/(200) + (1)/(100) = (3)/(200)`
`:. C = (200)/(2) pF = 66.7 xx 10^(-12) F`
As `C_(P)` and `C_(4)` are in series
`:. V_(P) + V_(4) = 300`
charge on `C_(4)` is `q_(4)`
`= CV = (200)/(3) xx 10^(-12) xx 300 = 2 xx 10^(-8) C`
Potential difference across
`C_(4)` is `V_(4) = (q_(4))/(C_(4)) = (2 xx 10^(-8))/(100 xx 10^(-12)) = 200 V`
from (i) `V_(P) = 300 - V_(4) = 300 - 200 = 100V`
Potential difference across
`C_(1)` is `V_(1) = V_(P) = 100 V`
Charge on `C_(1) = q_(1) = C_(1) V_(1)`
`= 100 x 10^(-12) xx 100 = 10^(-8) C`
Potential diff across
`C_(2)` and `C_(3)` in series = 100 W
charge on `C_(2)` ,
`q_(2) = C_(2) V_(2) = 200 xx 10^(-12) xx 50 - 10^(-8) C`
Charge on `C_(3)`.
`q_(3) = C_(3)V_(3) = 200 xx 10^(-12) xx 50 = 10^(-8) C`