A network of four 10 µF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.)

Explanation

Draw a effective network as follow

Let effective capacitance of series combination of `C_(1) , C_(2) , C_(3), ` is `C^(1)`
`:. (1)/(C^(1))=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))`
`= (1)/(10)+(1)/(10)+(1)/(10)`
` (3)/(10)`
`:. C^(1)=(10)/(3) mu F `
Now network obtain as below

Effective capacitance of `C^(1) ` and `C_(4)` is
`C_(AD)= C^(1) +C=(10)/(3) +10 =(40)/(3) muF`
`:. C_(AB)= 13.3 mu F`
Now `C_(1), C_(2)` and `C_(3)` are in series hence charge on each of the capacitor are same. Suppose this same charge is Q. Now, the potential difference across AB is `(Q)/(C_(1))` across CD is `(Q)/(C_(3))` we have
`(Q)/(C_(1))+(Q)/(C_(2))+(Q)/(C_(3))` =500
`(Q)/(C_(4))` =500 where Q is the charge on `C_(4)`
`:. Q[(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))]= 500`
`:. Q[ (1)/(C_(1))] = 500`
`:. Q = 500xx C^(1)`
`= 500xx(10)/(3)`
`= (5000)/(3) xx10^(-6) C `
`= 1.7xx10^(-3)C ` is the charge on `C_(1) , C_(2) , C_(3)`
If Q is the charge on `C_(4)` then `Q^(1) = VC_(4)`
`Q^(1)= 500xxC_(4)`
`= 500 xx10xx10^(-6)`
`= 5xx10^(-3) C`