Each capacitor shown in figure has a capacitance of 5.0 mu F , The emf of the battery is 50 V , How much charge will flow though AB if the switch S is closed ?

Explanation

Initially when the switch S is not connected
` C_eq = 2C xx C / 3C `
`= (2 / 3) C = 2 / 3 xx 5.0 mu F `
` charge `= C_eq V `
`2 / 3 xx 5.0 mu F xx 50`
`= 500 / 3 mu C`
When the switch S is closed
` C_eq = 2C = 2 xx 5.0 = 10 mu F`
charge `= 10 mu F xx 50 = 500mu C`
Hence the charge flow
` AB = 500 mu C - 500 / 3 mu C`
= 3.3 xx 10^(-4) C`.