91Ó°ÊÓ

To find the steady-state current one needs to find a particular solution to the equation;

\({\bf{L}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{I}}}}{{{\bf{dt}}}}{\bf{ + R}}\frac{{{\bf{dI}}}}{{{\bf{dt}}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{I = }}\frac{{{\bf{dE}}}}{{{\bf{dt}}}}\)

In our case \({\bf{E(t) = 10cos20t\;V,R = 120\Omega ,L = 4H}}\) and \({\bf{C = (2200}}{{\bf{)}}^{{\bf{ - 1}}}}{\bf{\;F}}.\)

So, looking for a particular solution;

\(\begin{aligned}{c}{\bf{4I'' + 120I' + 2200I = }}\frac{{\bf{d}}}{{{\bf{dt}}}}{\bf{(10cos20t)}}\\{\bf{4I'' + 120I' + 2200I = - 200sin20t}}\\{\bf{I'' + 30I' + 550I = - 50sin2t}}\end{aligned}\)

One will find a particular solution using the method of undetermined coefficients. Assume that\({{\bf{I}}_{\bf{p}}}{\bf{(t) = Acos20t + Bsin20t}}\). Then,

\(\begin{aligned}{c}{\bf{I}}_{\bf{p}}^{}{\bf{'(t) = - 20Asin20t + 20Bcos20t,}}\;\\{\bf{I}}_{\bf{p}}^{}{\bf{''(t) = - 400Acos20t - 400Bsin20t}}\end{aligned}\)

Substituting this into the previous differential equation one will get

\(\begin{aligned}{c}{\bf{I}}_{\bf{p}}^{}{\bf{''(t) + 30I}}_{\bf{p}}^{}{\bf{'(t) + 550}}{{\bf{I}}_{\bf{p}}}{\bf{(t) = - 400Acos20t - 400Bsin20t + 30( - 20Asin20t + 20Bcos20t) + 550(Acos20t + Bsin20t)}}\\{\bf{ = - 400Acos2t + 60Bcos20t + 550Acos20t - 400Bsin20t - 60Asin20t + 550Bsin20t}}\\{\bf{ = 150(A + 4B)cos20t + 150(B - 4A)sin20t}}\\{\bf{ = - 50sin20t}}\end{aligned}\)

Comparing the multiplying coefficients, one will get;

\({\bf{150}}\left( {{\bf{A + 4 B}}} \right){\bf{ = 0}}\)and\({\bf{150}}\left( {{\bf{B - 4 A}}} \right){\bf{ = - 50}}\)

\({\bf{A + 4 B = 0}}\)and \({\bf{3}}\left( {{\bf{B - 4 A}}} \right){\bf{ = - 1}}\)

\({\bf{A = - 4B}} \Rightarrow {\bf{3(B + 16B) = - 1}}\)

\({\bf{B = - }}\frac{{\bf{1}}}{{{\bf{51}}}}\)and \({\bf{A = }}\frac{{\bf{4}}}{{{\bf{51}}}}\)

So now one has that the steady-state current is\({{\bf{I}}_{\bf{p}}}{\bf{(t) = }}\frac{{\bf{4}}}{{{\bf{51}}}}{\bf{cos20t - }}\frac{{\bf{1}}}{{{\bf{51}}}}{\bf{sin20t}}\).

When solving the differential equation which has a form of

\({\bf{L}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{I}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ + R}}\frac{{{\bf{dI}}}}{{{\bf{dt}}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{I = }}{{\bf{E}}_{\bf{0}}}{\bf{cos}}\gamma {\bf{t}}\)one defines\({\gamma _{\bf{r}}}\) with\({\gamma _{\bf{r}}}{\bf{ = }}\sqrt {\frac{{\bf{1}}}{{{\bf{LC}}}}{\bf{ - }}\frac{{{{\bf{R}}^{\bf{2}}}}}{{{\bf{2}}{{\bf{L}}^{\bf{2}}}}}} \)and then the resonance frequency is\({{\bf{f}}_{\bf{0}}}{\bf{ = }}\frac{{{\gamma _{\bf{r}}}}}{{{\bf{2\pi }}}}\).

Substituting the given values for \({\bf{L,C}}\) and \({\bf{R}}\)one has that \({\gamma _{\bf{r}}}\) is:

\({\gamma _{\bf{r}}}{\bf{ = }}\sqrt {\frac{{{\bf{2200}}}}{{\bf{4}}}{\bf{ - }}\frac{{{\bf{14400}}}}{{{\bf{32}}}}} {\bf{ = }}\sqrt {{\bf{550 - 450}}} {\bf{ = }}\sqrt {{\bf{100}}} {\bf{ = 10}}\)and then the resonance frequency is\({{\bf{f}}_{\bf{0}}}{\bf{ = }}\frac{{{\gamma _{\bf{r}}}}}{{{\bf{2\pi }}}}{\bf{ = }}\frac{{{\bf{10}}}}{{{\bf{2\pi }}}}{\bf{ = }}\frac{{\bf{5}}}{{\bf{\pi }}}\)