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(Solved): Just had a clarification question. I understand why the i0 is 1 mA (current divider rule), but why ...


\[ \begin{array}{l} i_{\left(0^{\circ}\right)}=\frac{I_{s} R_{1}}{R_{2}+R_{1}}=I_{m A} \\ V_{\left(0^{\circ}\right)}=0 \mathr

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.

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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
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