Kirchhoff problem 2 solution

At t=0, the cap[[Image:acitor acts like a short circuit and the inductor acts like an open circuit, and we can simply the cicuit into:

$V_{R1}=6.818V$

$V_{R2}=0V$

$V_{R3}=V_{R4}=8.182V$

$I_{R1}=13.64mA$

$I_{R2}=0A$

$I_{R3}=5.455mA$

$I_{R4}=8.182mA$

At $t=\infty$, the capacitor acts like an open circuit and the inductor acts like a short circuit. We can then simplify the circuit:

$V_{R1}=9V$

$V_{R2}=36V$

$V_{R3}=0V$

$V_{R4}=6V$

$I_{R1}=18mA$

$I_{R2}=12mA$

$I_{R3}=0A$

$I_{R4}=6mA$