Laplace transforms: Simple Electrical Network: Difference between revisions

Revision as of 16:01, 1 December 2009

Problem Statement

Using the formulas

 $E (t) = L \frac{d i_{R}}{d t} + R i_{C}$ 
 $R C \frac{d i_{R}}{d t} + i_{R} - i_{C} = 0$

Solve the system when V0 = 50 V, L = 0.5 h, R = 80 Ω, C = 10^-4 f, and the currents are initially zero.

Solution

Solve the system when V0 = 50 V, L = 0.5 h, R = 60 Ω, C = 10^{-4} f, and the currents are initially zero. Substituting numbers into the equations, we have

$0.5 \frac{d i_{1}}{d t} + 80 i_{2} = 50$

$80 (1 0^{- 4}) \frac{d i_{2}}{d t} + i_{2} - i_{1} = 0$

Applying the Laplace transform to each equation gives

$\frac{1}{2} (s ℒ {i_{1}} - i_{1} (0)) + 80 ℒ {i_{2}} = 50$

$0.008 (s ℒ i_{2} - i_{2} (0)) + ℒ {i_{2}} - ℒ {i_{1}} = 0$

$\Rightarrow \frac{1}{2} s I_{1} (s) + 80 I_{2} (s) = \frac{50}{s}$

$- 125 I_{1} (s) + [s + 125] I_{2} (s) = 0$

Solving for $I_{1} (s)$ gives

$I_{s} (s) = \frac{100 s + 12500}{s (s^{2} + 125 s + 20000)}$

@@ Line 10: / Line 10: @@
 Substituting numbers into the equations, we have
-<math>0.5\frac{di_1}{dt}+60i_2=50</math>
+<math>0.5\frac{di_1}{dt}+80i_2=50</math>
-<math>60(10^{-4})\frac{di_2}{dt}+i_2-i_1=0</math>
+<math>80(10^{-4})\frac{di_2}{dt}+i_2-i_1=0</math>
 Applying the Laplace transform to each equation gives
-<math>\frac{1}{2}(s\mathcal{L}\left\{i_1\right\}-i_1(0))+60\mathcal{L}\left\{i_2\right\}=50</math>
+<math>\frac{1}{2}(s\mathcal{L}\left\{i_1\right\}-i_1(0))+80\mathcal{L}\left\{i_2\right\}=50</math>
-<math>0.006(s\mathcal{L}{i_2}-i_2(0))+\mathcal{L}\left\{i_2\right\}-\mathcal{L}\left\{i_1\right\}=0</math>
+<math>0.008(s\mathcal{L}{i_2}-i_2(0))+\mathcal{L}\left\{i_2\right\}-\mathcal{L}\left\{i_1\right\}=0</math>
-<math>\Rightarrow\frac{1}{2}sI_1(s)+60I_2(s)=\frac{50}{s}</math>
+<math>\Rightarrow\frac{1}{2}sI_1(s)+80I_2(s)=\frac{50}{s}</math>
-<math>-200I_1(s)+[s+200]I_2(s)=0</math>
+<math>-125I_1(s)+[s+125]I_2(s)=0</math>
-Solving for <math>\mathcal{L}\left\{i_1\right\}</math> gives
+Solving for <math>I_1(s)</math> gives
+<math>I_s(s)= \frac{100s+12500}{s(s^2+125s+20000)}</math>

Laplace transforms: Simple Electrical Network: Difference between revisions

Revision as of 16:01, 1 December 2009

Problem Statement

Solution

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