Question

two capacitor of capacitance 4mf and 2mf are joined series with a battery of EMF 50 V the connection are broken and the like terminals of the capacitor are then joined find the final charge on each capacitor​

Answer 1

Answer:

First thing charge flowing through both capacitor must be same.

q is constant..

Charging and discharging of the capacitor depends on the time constant of the capacitor.

Voltage across the 2uf capacitor will be more than voltage across 4uf capacitor.

Total voltage = 100 .

voltage across C1 ( 4uf ) = 2×100/6 = 33.33V .

voltage across C2 ( 2uf ) = 400/6 = 67.67V.

charging time of 4uf capacitor = (4×10^-6× resistance of the wire )

charging time of 2uf capacitor = (2×10^-6×resistance of the wire )

suppose resistance of the wire is negligible. So we can determine the time constant.

{(q/4)+(q/2)}=100×10^-6

6q/8= 100×10^-6

q = ( 100×10^-6) × 8/6.

This is the total charge

now if the connection is broken and the path is shorted then current flows through the short circuit path discharging of current direction will be opposite to the charging current.

At Final or steady state condition charge will be there . discharging of the capacitor takes place until the charges on the capacitor places are same . Imbalance of the charges between two plates causes flowing of displacement current . Conduction current flows when voltage difference takes place ..