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

  • 0
What are you looking for?