Consider the circuit shown in the figure .Capacitors A and B ,each have capacitance C=2 F .The plates of capacitor A are shorted using a wire of resistance R =1 ohm while the left plate of capacitor B is given an initial charge Q=+4C.The switch is closed at time t=0.What will be the initial current drawn from the battery immediately after the switch is closed?
Let q be the charge flow from the cell at time t.
Solving mesh 1
i1 - q/C = 0
=> i1 = ½q ---1.
Solving mesh 2
i + i1 + (q+4)/C = E ---2.
=> i + q + 2 = 4×4/2 = 8
=> i + q = 6
At time t = 0 no charge will build up so q = 0
=> i = 6 A
Mathematically lets see this result
=> dq/dt + q = 6