Question

In the circuit of Figure 7.1, let $\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{125}\mathbf{\Omega }$,$\mathbf{L}\mathbf{=}\mathbf{10}\mathbf{mH}$and the source voltage is $\mathbf{e}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{151}\mathbf{sin}\mathbf{(}\mathbf{377}\mathbf{t}\mathbf{+}\mathbf{a}\mathbf{)}$Determine the current response after closing the switch for the following cases: (a) no dc offset or (b) maximum dc offset. Sketch the current waveform up to $\mathrm{t}=0.10\mathrm{s}$corresponding to parts (a) and (b).

Short answer

(a) Thecurrent response after closing the switch for no dc offset is $40\sin \left(377\mathrm{t}\right)\mathrm{A}$.

(b) The current response after closing the switch for maximum dc offset is$-40\cos \left(377t\right)+40\cos \left({90}^{°}\right)e\frac{1}{0.08}A.$

The plot of current response after closing the switch for no dc offset

The plot of the current response after closing the switch for maximum dc offset.

Step-by-step solution

Write the given data from question.

$$\begin{gathered} \mathrm{The}\mathrm{source}\mathrm{voltage},\left(\mathrm{t}\right)=151\sin (377\mathrm{t}+\propto )\mathrm{v} \\ \mathrm{The}\mathrm{rms}\mathrm{input}\mathrm{voltage},\mathrm{v}=\frac{151}{\sqrt{2}}\mathrm{volts} \\ \mathrm{The}\mathrm{inductance},\mathrm{L}=10\mathrm{mH} \\ \mathrm{The}\mathrm{resistance},\mathrm{R}=0.125\mathrm{\Omega } \\ \mathrm{Natural}\mathrm{frequency},\mathrm{\omega }=377\frac{\mathrm{rad}}{\mathrm{s}} \end{gathered}$$

Determine the formulas to calculate the current response.

The equation to calculate the impedance of the R-L circuit is given as follows.

$\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L}\right)}^2}$

The equation to calculate the symmetrical fault current is given as follows.

${\mathrm{l}}_{\mathrm{ac}}=\frac{\mathrm{v}}{\mathrm{z}}$

The equation to calculate the time constant is given as follows.

$\mathrm{T}=\frac{\mathrm{L}}{\mathrm{R}}$

The equation to calculate the angle $\theta$is given as follows.

$\mathrm{\theta }={\tan }^{-}\left(\frac{\mathrm{\omega L}}{\mathrm{R}}\right)$

The equation to calculate the total current in the circuit is given as follows.

$$\begin{gathered} \mathrm{i}\left(\mathrm{t}\right)={\mathrm{l}}_{\mathrm{ac}}\left(\mathrm{t}\right)+{\mathrm{l}}_{\mathrm{ac}}\left(\mathrm{t}\right) \\ \mathrm{i}\left(\mathrm{t}\right)=\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}\sin \left(\mathrm{\omega t}+\mathrm{a}-\mathrm{\theta }\right)-\frac{\sqrt{2}\mathrm{V}}{\mathrm{Z}}\sin (\mathrm{a}-\mathrm{\theta }){\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{T}}} \end{gathered}$$

Calculate the current response after closing the switch for no dc offset.

(a)

Consider the electrical circuit.

Calculate the impedance of the circuit.

$$\begin{gathered} \mathrm{Substitute}0.125\mathrm{\Omega }\mathrm{for}\mathrm{R},\frac{\mathrm{rad}}{\mathrm{s}}\mathrm{for}\mathrm{\omega }\mathrm{and}2\mathrm{mH}\mathrm{for}\mathrm{L}\mathrm{into}\mathrm{equation}\left(1\right) \\ \mathrm{Z}=\sqrt{0.{125}^2+{(377\times 0.01)}^2} \\ \mathrm{Z}=\sqrt{0.0156+0.14.213} \\ \mathrm{Z}=\sqrt{14.228} \\ \mathrm{Z}=3.77\mathrm{\Omega } \\ \mathrm{Calculate}\mathrm{the}\mathrm{rms}\mathrm{symmetrical}\mathrm{fault}\mathrm{current}. \end{gathered}$$

Substitute $3.77$$\Omega$for $Z$and $\frac{151}{\sqrt{2}}$$volts$for $V$into equation (2).

$$\begin{gathered} {\mathrm{l}}_{\mathrm{ac}}=\frac{{\displaystyle \frac{151}{\sqrt{2}}}}{3.77} \\ {\mathrm{l}}_{\mathrm{ac}}=\frac{151}{\sqrt{2}\times 3.77} \\ \mathrm{Calculate}\mathrm{the}\mathrm{time}\mathrm{constant}\mathrm{of}\mathrm{the}\mathrm{circuit}. \end{gathered}$$

Substitute $10\mathrm{m}\mathrm{H}$for $L$and $0.125\Omega$for $R$into equation (3).

$$\begin{gathered} T=\frac{10\times {10}^{-3}}{0.125} \\ T=0.08sec \\ Calculatetheangle\theta . \\ Substitute0.125\Omega forR,377\frac{rad}{s}for\omega and10mHforLintoequation\left(6\right) \end{gathered}$$

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{377\times 10\times {10}^{-3}}{0.125}\right) \\ \theta ={\tan }^{-1}(3.016) \\ \theta =88.{10}^{°} \end{gathered}$$

Calculate the total current flowing through the circuit

Substitute $\frac{151}{\sqrt{2}}$for V , $3.77\Omega$for $Z$, $88.{10}^{°}$for $\theta$, $0.08sec$for $T$and $377\frac{rad}{s}$for $\omega$into equation (5).

$$\begin{gathered} i\left(t\right)=\frac{\sqrt{2}\times {\displaystyle \frac{151}{\sqrt{2}}}}{3.77}\sin \left(377t+\propto -88.{10}^{°}\right)-\frac{\sqrt{2}\times {\displaystyle \frac{151}{\sqrt{2}}}}{3.77}\sin \left(\propto -88.{10}^{°}\right)e^{-\frac{t}{0.08}} \\ i\left(t\right)=\frac{151}{3.77}\sin \left(377t+\propto -88.{10}^{°}\right)-\frac{151}{3.77}\sin \left(\propto -88.{10}^{°}\right)e^{-\frac{t}{0.08}} \end{gathered}$$

Consider the no dc offset, therefore the switch should be closed at $\propto =88.{10}^{°}$.

$$\begin{gathered} \mathrm{i}\left(\mathrm{t}\right)=\frac{151}{3.77}\sin \left(377\mathrm{t}+88.{10}^{°}-88.{10}^{°}\right)-\frac{151}{3.77}\sin \left(88.{10}^{°}-88.{10}^{°}\right){\mathrm{e}}^{-\frac{\mathrm{t}}{0.08}} \\ \mathrm{i}\left(\mathrm{t}\right)=\frac{151}{3.77}\sin \left(377\mathrm{t}\right)-\frac{151}{3.77}\sin \left(0\right){\mathrm{e}}^{-\frac{\mathrm{t}}{0.08}} \\ \mathrm{i}\left(\mathrm{t}\right)=\frac{151}{3.77}\sin \left(377\mathrm{t}\right)-0 \\ \mathrm{i}\left(\mathrm{t}\right)=40\sin \left(377\mathrm{t}\right)\mathrm{A} \end{gathered}$$

Hence thecurrent response after closing the switch for no dc offset is $40\sin \left(377t\right)A$.

Calculate the current response after closing the switch for maximum dc offset.

For the maximum dc offset,

$$\begin{gathered} \propto =88.{10}^{°}-{90}^{°} \\ \propto =1.9^{°} \end{gathered}$$

Consider the equation,

localid="1655207717886" $i\left(t\right)=\frac{151}{3.77}\sin \left(377t+\propto -88.{10}^{°}\right)-\frac{151}{3.77}\sin \left(\propto -88.{10}^{°}\right)e^{-\frac{t}{0.08}}$

Substitute localid="1655207724697" $1.9^{°}$for localid="1655207728166" $\propto$into above equation.

localid="1655207732489" $$\begin{gathered} i\left(t\right)=\frac{151}{3.77}\sin \left(377t-1.9^{°}-88.{10}^{°}\right)-\frac{151}{3.77}\sin \left(1.9^{°}-88.{10}^{°}\right)e^{-\frac{t}{0.08}} \\ i\left(t\right)=40\sin \left(377t-{90}^{°}\right)-40\sin \left(377t-{90}^{°}\right)e^{-\frac{t}{0.08}} \\ i\left(t\right)=-40\cos \left(377t\right)+40\sin \left({90}^{°}\right)e^{-\frac{t}{0.08}} \end{gathered}$$

Hence the current response after closing the switch for maximum dc offset islocalid="1655207713300" $i\left(t\right)=-40\cos \left(377t\right)+40\sin \left({90}^{°}\right)e^{-\frac{t}{0.08}}A$.

The plot of current response after closing the switch for no dc offset

The plot of the current response after closing the switch for maximum dc offset.