Obtain the equivalent capacitance of the network in Fig. 2.33. For a 300 V supply, determine the charge and voltage across each capacitor.

Explanation

Drawing equivalent network as follow,

Given circuit in example look like as shown in above figure.
Here `C_(1)= C_(4)` = 100 pF and `C_(2)=C_(3)= 200 pF`
Equivalent capacitance of series connection of `C_(2)` and `C_(3)`
`(1)/(C_(23)) =(1)/(C_(2))+(1)/(C_(3))implies C_(23)= (C_(2)C_(3))/(C_(2)+C_(3))`
`(1)/(C_(23))=(200xx200)/(200+200)`
`= (200xx200)/(400)= 100 ` pF
Now `C_(1)` and `C_(23)` are in parallel. Its equivalent capacitance C
`:. C = C_(1)+C_(23)= 100 +100 = 200 `p F
Now `C_(4)` and C are in series its equivalent capacitance C
`(1)/(C)=(1)/(C_(4))+(1)/(C) implies C = (C.C_(4))/(C+C_(4))`
`:. C = (200xx100)/(200+100)=(200xx100)/(300)=(200)/(3)` pF
`:. C = (200)/(3) pF`
Total charge is Q then C = `(Q)/(V)`
`:. Q = CV`
`= ((200)/(3) xx10^(-12))(300)`
`:. Q = 2xx10^(-8) ` C
Since `C_(4)` and C are in series the charge Q on both capacitor are same .
`:. ` Charge on `C_(4)=2xx10^(-8) C =0.02 muC ` and charge on C =` 2xx10^(-8) ` C .
But C is for parallel connection of `C_(1)` and `C_(23)` and both capacitors are same hence both capacitor having same charge .
`:.` Charge on `C_(1)`
`q_(1)=1xx10^(-8) C = 0.01 muC`
and charge of `C_(23) = 1xx10^(-8)` C
But `C_(23)` is the equivalent capacitance of series connection of `C_(2)` and `C_(3)` . Hence `C_(2)` and `C_(3)` having same charge and it is equivalent to the charge of `C_(23)`
`:. ` Charge on `C_(2) = 1xx10^(-8) C = 0.01 muC `
and charge on `C_(3) = 1 xx10^(-8) C = 0.01mu C `
Voltages across each capacitor :
`V _(1)=(q_(1))/(C_(1))= (10^(-8))/(10^(-10))= 100` V
`V_(2) = (q_(2))/(C_(2))= (10^(-8))/(2xx10^(10))=50 V `
`V_(3) = (q_(3))/(C_(3))= (10^(-8))/(2xx10^(-10))= 5- V `
`V_(4) = (q_(4))/(C_(4))= (2xx10^(-8))/(10^(-10))= 200V `