#### Question

Each capacitor shown in figure has a capacitance of 5⋅0 µF. The emf of the battery is 50 V. How much charge will flow through AB if the switch S is closed?

#### Solution

Initially, when the switch S is open, the equivalent capacitance is given by

`C_(eq) = (2CxxC)/(3C)`

`⇒ C_("eq") = 2/3 C = 2/3 xx 5.0 "uF"`

The Charge supplied by the battery is given by

`Q = C_(eq) xx V`

`⇒ Q = 2/3 xx (5.0 "uF") xx (50 "V")`

`⇒Q = 500/3 "uC"`

When the switch S is closed, no charge goes to the capacitor connected in parallel with the switch.

Thus, the equivalent capacitance is given by `C_(eq) = 2C = 2 xx 5.0 xx 10 = uF`

The charge supplied by the battery is given by

`Q = 10 "uF" xx 50 = 500 "uC"`

The initial charge stored in the shorted capacitor starts discharging."?

Hence, the charge that flows from A to B is given by

`Q_"net" = 500 "uC" - 500/3 "uC"`

`⇒ Q_"net" = 3.3 xx 10^-4 C`