In the circuit shown in the given figure, steadystate was

In the circuit shown in the given figure, steadystate was reached when the switch S was open. The switch was closed at t = 0. The initial value of the current through the capacitor 2C is 12 Amp

Initial voltage on 2C = 12 volts

Then,discharging current, I_d(t) =	d						2C.12 e^{(-t.3 / 8C)}
		dt

Then , I_d(t) _{t = 0}\| =		-24C ×	3	. e^{(-t.3 / 8C)}
			8C

= -9 amp

Now, charging current, I_c(t) =	d		24C		1 - e^{(-3t / 8C)}
	dt

= -9 amp
At t = 0, I_c(t) = 9 amp.
Hence, at t = 0
total current I(t) = I_c(t) + I_d(t) = 0 amp.