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Just had a clarification question. I understand why the i0 is 1 mA (current divider rule), but why is v0 = 0V? My thinking is that the voltage across R2 is equal to the voltage across the cap.

$i_{(0_{∘})}=R_{2}+R_{1}I_{s}R_{1} =I_{mA}V_{(0_{∘})}=0V $ b) $V_{L}(t)−V_{(t)}+V_{R_{3}}=−dtdi −V(t)+i(t)R_{3}=i(t)=−cdtdv dtdi =−cdt_{2}dv_{2} $

The answer to your question is as follows:

Why (VC(0-) = 0V: The answer for this is as the inductor is exactly connected in parallel to the series combination of R3 and the capacitor.ie

At t=(0-) instant the circuit is under steady state conditions. since under steady state conditions