Question

each capacitor shown in the figure has a capacitance of 5 microfarad the EMF of the battery is 50 volt how much charge will flow through AB if the switch S is closed​

Answer 1

Given :

The capactiance of each capacitor = 5μF

EMF of the battery = 50V

To Find :

The charge flow through AB when switch is closed

Solution :

• The initial capacitance of the circuit when switch is closed is

         $C_{eq} =\frac{2C\times C}{3C} = \frac{2C}{3} =\frac{10}{3} \mu F$

• The charge flow in the circuit is

         $Q=CV$

         $Q=\frac{10}{3} \times 50$

         $Q=\frac{500}{3} C$

• The equivalent capacitance of the system when switch is closed

        $C_{eq} =2C=2\times 5=10\mu F$

• The charge flow in the circuit is

         $Q=CV$

         $Q={10} \times 50$

         $Q=500C$

• The charge flow through the switch AB is

         $Q_{net} =500-\frac{500}{3}$

         $Q_{net} =3.3\times10^{-4}C$

The charge flow through the switch is 3.3×10⁻⁴ C.