Q 3.5-1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

An RL circuit with a 5  resistor and a 0.05-H inductor carries a current of 1 A at t = 0, at which time a voltage source E(t) = 5 cos 120t V is added. Determine the subsequent inductor current and voltage.

Step-by-Step Solution

Answer

• The subsequent inductor current is $\mathbf{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{cos}\mathbf{\left(}\mathbf{120}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{2} \mathbf{sin}\mathbf{\left(}\mathbf{120}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{44}{\mathbf{e}}^{\mathbf{-}\mathbf{100}\mathbf{t}}}{\mathbf{2}\mathbf{.}\mathbf{44}}$.

 

• The subsequent inductor voltage  ${\mathbf{E}}_{\mathbf{L}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{6}\mathbf{sin}\mathbf{\left(}\mathbf{120}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{cos}\mathbf{\left(}\mathbf{120}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{100}\mathbf{t}}}{\mathbf{2}\mathbf{.}\mathbf{44}}$

Step 1: Determine the subsequent inductor current

Here given   $\mathrm{R}=5\Omega ,L=0.05\text{ H},\text{ I}\left(\text{t}\right)=1,\text{ E}\left(\text{t}\right)=5 \cos \left(120\mathrm{t}\right)\mathrm{V}$

Apply 

$$\begin{gathered} \mathrm{I}\left(\mathrm{t}\right)={\mathrm{e}}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\left[\int {\mathrm{e}}^{\frac{\mathrm{Rt}}{\mathrm{L}}}\frac{\mathrm{E}\left(\mathrm{t}\right)}{\mathrm{L}}\mathrm{dt}+\mathrm{K}\right] \\ \mathrm{I}\left(\mathrm{t}\right)={\mathrm{e}}^{\frac{-5\mathrm{t}}{0.05}}\left[\int {\mathrm{e}}^{\frac{5\mathrm{t}}{0.05}}\frac{5 \cos \left(120\mathrm{t}\right)}{0.05}\mathrm{dt}+\mathrm{K}\right] \\ \mathrm{I}\left(\mathrm{t}\right)={\mathrm{e}}^{-100\mathrm{t}}\left[\int {\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)\mathrm{dt}+\mathrm{K}\right] \end{gathered}$$

Step 2: Solving the integral part ∫ e 100 t cos ( 120 t ) dt .

$$\begin{gathered} \int {\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)\mathrm{dt}=\frac{{\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)}{100}+\int 120 \sin \left(120\mathrm{t}\right)\frac{{\mathrm{e}}^{100\mathrm{t}}}{100} \\ =\frac{{\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)}{100}+\frac{6}{5}\int \sin \left(120\mathrm{t}\right){\mathrm{e}}^{100\mathrm{t}}\mathrm{dt} \\ =\frac{{\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)}{100}+\frac{6}{5}\left[\frac{{\mathrm{e}}^{100\mathrm{t}}\sin \left(120\mathrm{t}\right)}{100}-\int 120 \cos \left(120\mathrm{t}\right)\frac{{\mathrm{e}}^{100\mathrm{t}}}{100}\mathrm{dt}\right] \\ \frac{{\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)}{100}+\frac{6}{5}\left[\frac{{\mathrm{e}}^{100\mathrm{t}}\sin \left(120\mathrm{t}\right)}{100}-1.44\int \cos \left(120\mathrm{t}\right) {\mathrm{e}}^{100\mathrm{t}}\mathrm{dt}\right] \end{gathered}$$

Put  $A=\int {\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)\mathrm{dt}$

$$\begin{gathered} \mathrm{A}=\frac{{\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)}{100}+\frac{6}{5}\frac{{\mathrm{e}}^{100\mathrm{t}}\sin \left(120\mathrm{t}\right)}{100}-1.44\mathrm{A} \\ \mathrm{A}+1.44\mathrm{A}=\frac{{\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)}{100}+\frac{6}{5}\frac{{\mathrm{e}}^{100\mathrm{t}}\sin \left(120\mathrm{t}\right)}{100} \\ 2.44\mathrm{A}=\frac{{\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)}{100}+\frac{6}{5}\frac{{\mathrm{e}}^{100\mathrm{t}}\sin \left(120\mathrm{t}\right)}{100} \\ \mathrm{A}=\frac{{\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)+1.2{\mathrm{e}}^{100\mathrm{t}}\sin \left(120\mathrm{t}\right)}{244} \end{gathered}$$

$$\begin{gathered} \mathrm{I}\left(\mathrm{t}\right)={\mathrm{e}}^{-100\mathrm{t}}\left[100(\frac{{\mathrm{e}}^{100\mathrm{t}}\cos \left(120\mathrm{t}\right)+1.2{\mathrm{e}}^{100\mathrm{t}}\sin \left(120\mathrm{t}\right)}{244})+\mathrm{K}\right] \\ \mathrm{I}\left(\mathrm{t}\right)=\frac{\cos \left(120\mathrm{t}\right)+1.2 \sin \left(120\mathrm{t}\right)}{2.44}+{\mathrm{Ke}}^{-100\mathrm{t}} \end{gathered}$$

Put I(0) = 1 ,then K = 1.44/2.44

$$\begin{gathered} \mathrm{I}\left(\mathrm{t}\right)=\frac{\cos \left(120\mathrm{t}\right)+1.2 \sin \left(120\mathrm{t}\right)}{2.44}+\frac{1.44}{2.44}{\mathrm{e}}^{-100\mathrm{t}} \\ \mathrm{I}\left(\mathrm{t}\right)=\frac{\cos \left(120\mathrm{t}\right)+1.2 \sin \left(120\mathrm{t}\right)+1.44{\mathrm{e}}^{-100\mathrm{t}}}{2.44} \end{gathered}$$

Hence, the subsequent inductor current is $\mathbf{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{cos}\mathbf{\left(}\mathbf{120}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{2} \mathbf{sin}\mathbf{\left(}\mathbf{120}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{44}{\mathbf{e}}^{\mathbf{-}\mathbf{100}\mathbf{t}}}{\mathbf{2}\mathbf{.}\mathbf{44}}$.

Step 3: Evaluate the subsequent indicator voltage

$$\begin{gathered} {\mathrm{E}}_{\mathrm{L}}\left(\mathrm{t}\right)=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}} \\ =0.05\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\cos \left(120\mathrm{t}\right)+1.2 \sin \left(120\mathrm{t}\right)+1.44{\mathrm{e}}^{-100\mathrm{t}}}{2.44}\right] \\ =0.05\left[\frac{-120 \sin \left(120\mathrm{t}\right)+7.2 \cos \left(120\mathrm{t}\right)-7.2{\mathrm{e}}^{-100\mathrm{t}}}{2.44}\right] \\ =\frac{-6 \sin \left(120\mathrm{t}\right)+7.2 \cos \left(120\mathrm{t}\right)-7.2{\mathrm{e}}^{-100\mathrm{t}}}{2.44} \end{gathered}$$

Hence, the subsequent inductor voltage ${\mathbf{E}}_{\mathbf{L}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{6}\mathbf{sin}\mathbf{\left(}\mathbf{120}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{cos}\mathbf{\left(}\mathbf{120}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{100}\mathbf{t}}}{\mathbf{2}\mathbf{.}\mathbf{44}}$