Kirchhoff problem 2 solution

From Mech

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

$V R 1 = 6.818 V$

$V R 2 = 0 V$

$V R 3 = V R 4 = 8.182 V$

$I R 1 = 13.64 m A$

$I R 2 = 0 A$

$I R 3 = 5.455 m A$

$I R 4 = 8.182 m A$

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

$V R 1 = 9 V$

$V R 2 = 36 V$

$V R 3 = 0 V$

$V R 4 = 6 V$

$I R 1 = 18 m A$

$I R 2 = 12 m A$

$I R 3 = 0 A$

$I R 4 = 6 m A$