We will use the general formula,
Ic(t) = Isc + (I(0) – Isc)*exp(
- t*Rth/L) (1)
When the switch is closed we have the KVL for the left loop:
2*ia = 6*ia + 12 so ia = - 3A
Isc is given by the right loop, with L = 6H we get
2*ia = 3*Isc + 6*dIsc/dt
So we have 2*ia = 3*Isc so Isc = - 2A
I(0) is given just before the switch closes, so we have the circuit
with only the two resistors (which are equivalent to a 9Ω resistor) and
the voltage source. Thus I(0) = U/R = 12/9 = 4/3
And moreover Rth is 3Ω so that Rth/L = 3/6 =
0.5
So we finally have from eq. (1) Ic(t) =
- 2 + (4/3 + 2)*exp( - 0.5*t) A for t > 0
So we get Ic(t) = - 2 + 10/3*exp( - 0.5*t) A for t > 0