Laplace transforms: Simple Electrical Network: Difference between revisions

Revision as of 21:00, 30 November 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 = 60 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

$0.5 \frac{d i_{1}}{d t} + 60 i_{2} = 50$

$60 (1 0^{- 4}) \frac{d i_{2}}{d t} + i_{2} - i_{1} = 0$

Applying the Laplace transform to each equation gives

$\frac{1}{2} (s ℒ {i_{1}} - i_{1} (0)) + 60 ℒ {i_{2}} = 50$

$0.006 (s ℒ i_{2} - i_{2} (0)) + ℒ {i_{2}} - ℒ {i_{1}} = 0$

$\Rightarrow \frac{1}{2} s ℒ {i_{1}} + 60 ℒ {i_{2}} = \frac{50}{s}$

$- 200 ℒ {i_{1}} + (s + 200) ℒ {i_{2}} = 0$

Solving for $ℒ {i_{1}}$ gives

@@ Line 10: / Line 10: @@
 Substituting numbers into the equations, we have
-<math>0.5\frac{di_1}{dt}+60i_2=0</math>
+<math>0.5\frac{di_1}{dt}+60i_2=50</math>
 <math>60(10^{-4})\frac{di_2}{dt}+i_2-i_1=0</math>
@@ Line 16: / Line 16: @@
 Applying the Laplace transform to each equation gives
-<math>\frac{1}{2}(s^2\mathcal{L}{I_1}(s)-i_1(0))+60\mathcal{L}\left\{i_2\right\}</math>
+<math>\frac{1}{2}(s\mathcal{L}\left\{i_1\right\}-i_1(0))+60\mathcal{L}\left\{i_2\right\}=50</math>
-<math></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>\Rightarrow\frac{1}{2}s\mathcal{L}\left\{i_1\right\}+60\mathcal{L}\left\{i_2\right\}=\frac{50}{s}</math>
+<math>-200\mathcal{L}\left\{i_1\right\}+(s+200)\mathcal{L}\left\{i_2\right\}=0</math>
+Solving for <math>\mathcal{L}\left\{i_1\right\}</math> gives

Laplace transforms: Simple Electrical Network: Difference between revisions

Revision as of 21:00, 30 November 2009

Problem Statement

Solution

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