When the switch is thrown to position 2 @ time t=0 in the attached circuit, is the current through R1 4.0A?

 

No the current is 12/11 Amps. At the instant the switch closes consider the capacitor to be a short circuit and the inductor an open circuit.

 

The current will be (12-2)/11=0.91 Amps

If you assume that with the switch at position 1 the circuit was in a steady state then the voltage across the capacitor will be 2V. At the instant the switch moves to position 2 the capacitor will act a short circuit and the inductor as an open circuit. However, the voltage across the capacitor is 2V and it will subtract from the 12V battery voltage due to its polarity.

 

I agree with mik3. I disagree with t n k.

 

Yes t_n_k is wrong!