Kirchhoff problem 2 solution: Difference between revisions

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Latest revision as of 16:50, 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:

$V_{R 1} = 6.818 V$

$V_{R 2} = 0 V$

$V_{R 3} = V_{R 4} = 8.182 V$

$I_{R 1} = 13.64 m A$

$I_{R 2} = 0 A$

$I_{R 3} = 5.455 m A$

$I_{R 4} = 8.182 m A$

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

$V_{R 1} = 9 V$

$V_{R 2} = 36 V$

$V_{R 3} = 0 V$

$V_{R 4} = 6 V$

$I_{R 1} = 18 m A$

$I_{R 2} = 12 m A$

$I_{R 3} = 0 A$

$I_{R 4} = 6 m A$

@@ Line 1: / Line 1: @@
 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>