Question

t_n_k

Joined Mar 6, 2009
5,455
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.
 

mik3

Joined Feb 4, 2008
4,843
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.
 
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