Solution


The capacitor and inductor in one branch pose some difficulty. We handle it by using an auxilary node, V2. Now we write KCL at node V1 (currents leaving):

 

The second equation is KCL at node V2 (currents leaving):


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© Calvin College, 2004
This page was written and is maintained by Steve VanderLeest.
It was last modified on 17-mar-04 .