Laplace transforms:Series RLC circuit: Difference between revisions

Revision as of 21:11, 19 October 2009

Laplace Transform Example: Series RLC Circuit

Problem

Given a series RLC circuit with $R = 10 O h m s$ , $L = 0.1 H$ , and $C = 1 0^{- 5} F$ , having power source $v (t) = 10 c o s (20 t)$ , find an expression for $i (t)$ if $i (0) = 0 A$ and $v_{c} (0) = 0 V$ .

Solution

We begin with the general formula for voltage drops around the circuit:

$v (t) = R i + L \frac{d i}{d t} + \frac{1}{C} \int i d t$

Substituting numbers, we get

$10 c o s (20 t) = 10 i + 0.1 \frac{d i}{d t} + 1 0^{5} \int i d t$

$\Rightarrow c o s (20 t) = i + 0.01 \frac{d i}{d t} + 10000 \int i d t$

Now, we take the Laplace Transform and get

$\frac{s}{s^{2} + 2 0^{2}} = I + 0.01 [s I - i (0)] + 10000 \frac{I}{s}$

Using the fact that $i (0) = 0 A$ , we get

$\frac{s}{s^{2} + 400} = I + 0.01 s I + 10000 \frac{I}{s}$

$\Rightarrow \frac{s^{2}}{s^{2} + 400} = s I + 0.01 s^{2} I + 10000 I$

$\Rightarrow \frac{s^{2}}{s^{2} + 400} = (0.01 s^{2} + s + 10000) I$

$\Rightarrow I (s) = \frac{s^{2}}{(s^{2} + 400) (0.01 s^{2} + s + 10000)}$

Using partial fraction decomposition, we find that

$I (s) = \frac{1.0004 - 4.003 * 1 0^{- 8} s}{0.01 s^{2} + s + 10000} + \frac{4.003 * 1 0^{- 6} s - 0.04002}{s^{2} + 400}$

$\Rightarrow I (s) = \frac{100.04 - 4.003 * 1 0^{- 6} s}{s^{2} + 100 s + 1000000} + \frac{4.003 * 1 0^{- 6} s - 0.04002}{s^{2} + 400}$

$\Rightarrow I (s) = \frac{100.04 - 4.003 * 1 0^{- 6} s}{(s + 50)^{2} + 997500} + \frac{4.003 * 1 0^{- 6} s - 0.04002}{s^{2} + 400}$

$\Rightarrow I (s) = \frac{100.038}{(s + 50)^{2} + (50 \sqrt{399})^{2}} - \frac{4.003 * 1 0^{- 6} s + . 002}{(s + 50)^{2} + (50 \sqrt{399})^{2}} + \frac{4.003 * 1 0^{- 6} s}{s^{2} + 2 0^{2}} - \frac{0.04002}{s^{2} + 2 0^{2}}$

$\Rightarrow I (s) = \frac{10.038}{50 \sqrt{399}} \frac{50 \sqrt{399}}{(s + 50)^{2} + (50 \sqrt{399})^{2}} - 4.003 * 1 0^{- 6} \frac{s + 50}{(s + 50)^{2} + (50 \sqrt{399})^{2}} + 4.003 * 1 0^{- 6} \frac{s}{s^{2} + 2 0^{2}} - \frac{0.04002}{20} \frac{20}{s^{2} + 2 0^{2}}$

Finally, we take the inverse Laplace transform to obtain

$i (t) = 0.01 e^{- 50 t} s i n (998.8 t) - (4.003 * 1 0^{- 6}) e^{- 50 t} c o s (998.8 t) + (4.003 * 1 0^{- 6}) c o s (20 t) - 0.002 s i n (20 t)$

which is our answer.

@@ Line 1: / Line 1: @@
 ==Laplace Transform Example: Series RLC Circuit==
 ===Problem===
-Given a series RLC circuit with <math>R=10 Ω</math>, <math>L=0.08 H</math>, and <math>C=10^{-5} F</math>, having power source <math>v(t)=50cos(20t)</math>, find an expression for <math>i(t)</math> if <math>i(0)=0 A</math> and <math>v_c(0)=0 V</math>.
+Given a series RLC circuit with <math>R=10 Ohms</math>, <math>L=0.1 H</math>, and <math>C=10^{-5} F</math>, having power source <math>v(t)=10cos(20t)</math>, find an expression for <math>i(t)</math> if <math>i(0)=0 A</math> and <math>v_c(0)=0 V</math>.
 ===Solution===
@@ Line 10: / Line 10: @@
 Substituting numbers, we get
-<math>50cos(20t)=10i+0.08\dfrac{di}{dt}+10^{5}\int{i dt}</math>
+<math>10cos(20t)=10i+0.1\dfrac{di}{dt}+10^{5}\int{i dt}</math>
-<math>\Rightarrow cos(20t)=0.2i+0.0016\dfrac{di}{dt}+20000\int{idt}</math>
+<math>\Rightarrow cos(20t)=i+0.01\dfrac{di}{dt}+10000\int{idt}</math>
 Now, we take the Laplace Transform and get
-<math>\dfrac{s}{s^2+20^2}=0.2I+0.08[sI-i(0)]+20000\dfrac{I}{s}</math>
+<math>\dfrac{s}{s^2+20^2}=I+0.01[sI-i(0)]+10000\dfrac{I}{s}</math>
 Using the fact that <math>i(0)=0A</math>, we get
-<math>\dfrac{s}{s^2+400}=0.2I+0.08sI+20000\dfrac{I}{s}</math>
+<math>\dfrac{s}{s^2+400}=I+0.01sI+10000\dfrac{I}{s}</math>
-<math>\Rightarrow \dfrac{s^2}{s^2+400}=0.2sI+0.08s^2I+20000I</math>
+<math>\Rightarrow \dfrac{s^2}{s^2+400}=sI+0.01s^2I+10000I</math>
-<math>\Rightarrow \dfrac{s^2}{s^2+400}=(0.08s^2+0.2s+20000)I</math>
+<math>\Rightarrow \dfrac{s^2}{s^2+400}=(0.01s^2+s+10000)I</math>
-<math>\Rightarrow I(s)=\dfrac{s^2}{(s^2+400)(0.08s^2+0.2s+20000)}</math>
+<math>\Rightarrow I(s)=\dfrac{s^2}{(s^2+400)(0.01s^2+s+10000)}</math>
 Using partial fraction decomposition, we find that
-<math>I(s)=\dfrac{1.0016-1.605*10^{-8}s}{0.08s^2+0.2s+20000}+\dfrac{2.046*10^{-7}s-.02003}{s^2+400}</math>
+<math>I(s)=\dfrac{1.0004-4.003*10^{-8}s}{0.01s^2+s+10000}+\dfrac{4.003*10^{-6}s-0.04002}{s^2+400}</math>
-<math>\Rightarrow I(s)=\dfrac{6.24*10^7-s}{4.98*10^6s^2+1.25*10^7s+1.25*10^{12}}+\dfrac{
+<math>\Rightarrow I(s)=\dfrac{100.04-4.003*10^{-6}s}{s^2+100s+1000000}+\dfrac{4.003*10^{-6}s-0.04002}{s^2+400}</math>
+<math>\Rightarrow I(s)=\dfrac{100.04-4.003*10^{-6}s}{(s+50)^2+997500}+\dfrac{4.003*10^{-6}s-0.04002}{s^2+400}</math>
+<math>\Rightarrow I(s)=\dfrac{100.038}{(s+50)^2+(50\sqrt{399})^2}-\dfrac{4.003*10^{-6}s+.002}{(s+50)^2+(50\sqrt{399})^2}+\dfrac{4.003*10^{-6}s}{s^2+20^2}-\dfrac{0.04002}{s^2+20^2}</math>
+<math> \Rightarrow I(s)=\dfrac{10.038}{50\sqrt{399}}\dfrac{50\sqrt{399}}{(s+50)^2+(50\sqrt{399})^2}-4.003*10^{-6}\dfrac{s+50}{(s+50)^2+(50\sqrt{399})^2}+4.003*10^{-6}\dfrac{s}{s^2+20^2}-\dfrac{0.04002}{20}\dfrac{20}{s^2+20^2}</math>
+Finally, we take the inverse Laplace transform to obtain
+<math> i(t)=0.01e^{-50t}sin(998.8t)-(4.003*10^{-6})e^{-50t}cos(998.8t)+(4.003*10^{-6})cos(20t)-0.002sin(20t) \,</math>
+which is our answer.

Laplace transforms:Series RLC circuit: Difference between revisions

Revision as of 21:11, 19 October 2009

Laplace Transform Example: Series RLC Circuit

Problem

Solution

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