Difference between revisions of "Kirchhoff problem 2 solution"

Latest revision as of 15:50, 20 June 2006

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:

$V_{R1}=6.818V$

$V_{R2}=0V$

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

$I_{R1}=13.64mA$

$I_{R2}=0A$

$I_{R3}=5.455mA$

$I_{R4}=8.182mA$

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

$V_{R1}=9V$

$V_{R2}=36V$

$V_{R3}=0V$

$V_{R4}=6V$

$I_{R1}=18mA$

$I_{R2}=12mA$

$I_{R3}=0A$

$I_{R4}=6mA$

@@ Line 1: / Line 1: @@
-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:
+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:
-[[Image:kirchhoff_problem2_solution_1.jpg]]
+[[Image:kirchhoff_problem2_solution_1.gif]]
 <math>V_{R1}=6.818V</math>
@@ Line 20: / Line 20: @@
 At <math>t=\infty</math>, the capacitor acts like an open circuit and the inductor acts like a short circuit.  We can then simplify the circuit:
-[[Image:Example.kirchhoff_problem2_solution_2.jpg]]
+[[Image:Example.kirchhoff_problem2_solution_2.gif]]
 <math>V_{R1}=9V</math>