Obtain the equivalent capacitance of the following network in Fig. For 300 V supply, determine the charge and voltage across each capacitor. :

Explanation

`C_(1) = 100 pF, C_(2) = C_(3) = 200 pF, C_(4) = 100 pF`
Here `C_2` and `C_3` are in series combination and let their net capacitance be `C_5`.
`therefore 1/C_(5) = 1/C_(2) + 1/C_(3) = 1/200 + 1/200`
`1/C_(5) =(1+1)/200 = 2/200 = 1/100`
`rArr C_(5) = 100 pF`
Now `C_5` and `C_1` are in parallel combination and let their net capacitance be `C_(6)`.
So, `C_(6) = C_(1) + C_(5)`
`rArr C_(6) = 100 + 100`
`rArr C_(6) = 200 pF`
Now `C_4` and `C_6` are in series combination and let their net capacitance be C.
So, `1/C = 1/C_(4) + 1/C_(6)`
`1/C =1/100 + 1/200 = (2+1)/200 = 3/200`
`rArr C = 200/3 = 66.67 pF`
Charge supplied by the source =q=CV
`q =200/3 xx 10^(-12) xx 300`
`=2 xx 10^(-8) C`
Charges across `C_4` and `C_6` is same as they are in series combination.
So chft-ge across `C_(4) = 2 xx 10^(-8) C`